' This book is DUB on the last date stamped below OCT 31 1924 ;. ia ^ AJOV low * 47 XJ JAN 3 til 1927 1 AUG 2 9 1938 L*^-^' MAY 2 5 192T WAR ft 1940 ; Oo* .; &. We extend our field of numbers by removing these limitations. Thus if you wish to have a number x which multiplied by 8 gives 5 you do not find it among the positive integers ; you may then decide to create such a number, calling it , the two symbols indicating the definition and genesis of the new number. Such extensions of the number field are briefly indicated in the next section. 4. Rational numbers ; zero, fractions, and negative integers. These fundamental equalities and definitions from I to X are now extended by removing all limitations (except one, as noted below) upon the numbers, a, b, c, and x. Note that only the operations of addition, subtraction, multiplication, and division are included at this point. Extension of IV. x + 6 = a, when b = a defines zero, written 0. By definition then, + a = a. x + b = 0, defines a negative number which is written b. The negative here is a sign of quality ; by definition 6 is the result of subtract- ing 6 from 0, and b -f- b = 0. x -f b = a, when b > a, defines the negative number a b, which is the negative of b a. Subtracting a negative number can now be shown to be equivalent to adding a positive number, and similarly the other rules of elementary algebra relating to the addition and sul> traction of positive and negative quantities. That the product of two numbers with like signs is positive and the product of two numbers with unlike signs is negative follows from the above development. A negative number a is placed in our line of reference symmetrically to the corresponding positive number a, with respect to the origin ; of two negatives, the one toward the right is called the greater. Extension of X. b - = a, for all values of b except 0. b This extension of - to mean a number which multiplied by NUMBERS OF ALGEBRA 7 b gives a introduces new numbers of the type - , rational frac- tions in which a and b are positive or negative integers. Integers are included in this definition if & is a factor of a or if b is equal to one. Division by is explicitly excluded. A rational number is any number which can be expressed as the quotient of two integers, the denominator not to be zero. All of our rules for operating with fractions follow from the definition of - and from the preceding development. Thus a a a -b Further, by definition, ft f* ->-, when both are positive if ad > be ; b d fl (* s - = , when both are positive if ad = be ; and 6 d - < - , when both are positive if ad < be. b d Positive fractions can thus be arranged in a determined order upon our line of reference ; the value of the fraction determines the position and a graphical method of locating - on the b scalar line is indicated in the next section ; negative fractions are placed symmetrically to the corresponding positive frac- tions, with respect to the origin. Of any two rational numbers a and b, a is greater than b (a > 6) if a 6 is positive ; for when a b is positive, a posi- tive length must be added to b to give a, and consequently a must lie to the right of b. If on the line of reference two points aj x and x 2 are taken (fixed points), x 2 x l gives the dis- tance from the first point to the second ; this expression is positive if x 2 > a? l7 and negative if a? 2 3 x b a T a , and 6' -b 3. Is 3 > 2 ? Which is greater, or 3 ? Explain. 4. What is the difference between 4 and 3? 4 and 3? 4 and 11 ? 5. What fundamental law is assumed in the common process of multiplication, e. g. as in 325 by 239 and also x7 by x2? Is there a corresponding assumption in division ? 6. Which is greater, 4- or f ? Is 1 greater or less than f| ? Explain. 7. What is the product of by 7 ; by - 8 ; by 3 ; by ^? If a product is zero, what limitation is imposed upon the factors ? 5. Representation of a rational number, On cross-section 6 paper any rational fraction can be represented, using ruler and compass. Using 5 divisions to represent unity, each division NUMBERS OF ALGEBRA 9 represents 4- of a unit. To represent f-, one measures off 13 units, OA, on the line of reference, and 7 units, OB, on a second line through the origin (for convenience, use cross-sec- -r-B-T, 3 -'"4 _ 7 .- 12 : Graphical division \ to $ of 13 represented on horizontal line of reference. tion paper). Connect the ends, AB, and through the point U, one unit from on the second line, draw a line parallel to AB. The intersection point on the reference line represents the fraction - a T 3 -. Similarly any fraction ^ can be represented. The series of parallels to AB through the first 7 unit points on the vertical axis will cut off (plane geometry theorem) 7 equal parts of 13 on the horizontal axis. Graphical division to -y- expressed decimally. 10 UNIFIED MATHEMATICS On cross-section paper a somewhat better method of indicat- ing any quotient - is to move out on the line of reference 6 b units and up 1 unit ; connecting this point with the origin gives a straight line which can be used to read the desired quotients. Thus, since OB = 7 units, EC 1, and OA = 13, AP it follows that = BC OB' = , whence AP equals . 8.5 To obtain -^-, you find the point 8.5 units from 0, and the 7 vertical distance to the oblique line represents -^-, or 1.2. 7 6. Irrational numbers. s? = 2 is a simple and familiar illus- tration of a relation which is not satisfied by any rational number, -, with a and b integers ; geometrically, the diagonal of a square with side unity is not represented by any rational number. If you wish the length of this diagonal for any practical purpose, you use i, or -u ( 3 > O r \fo, or if, or liii or 1^42. The car . penter uses 1 foot 5 inches, or 1^ feet, in the diagonal for every foot of side, with an error of ^ of one per cent. The series of rational , which can be indefinitely ex- tended always increasing, and the series, always decreasing, 2> TT> TF! > TW' TOOOO> w ^h a constantly decreasing difference of limit between corresponding terms, together define the irrational number called the square root of 2. No rational number satisfies the relation ; no number - is at the end of b either series, but either series determines a definite point on .2 .4 .6 .8 1 1.2 1.4 Graphical representation of V2 numbers 1, 14, | NUMBERS OF ALGEBRA 11 our line, and algebraically defines our number, which we will call the square root of 2. Proof of the irrationality of V2. Assume that V2 =-> a rational fraction in lowest terms, with p and q integers. Both p and q cannot be even numbers, either p is odd or q is odd. If p is odd, squaring and clearing of fractions, 2 3* = p z ; but p is odd and you have an even number equal to an odd number. Hence p cannot be an odd number. Kow assume that p is even and that q is odd, and further let p = 2 ra. Then 2 q- = 4 m*, q 9 = 2 m*, and again we have an odd num- p ber equal to an even number. Our assumption that V 2 = - leads to an absurdity, that an odd number equals an even number. Describe about the origin with a radius 10 a circle, and using a protractor measure an angle of 20 degrees. The length of the perpendicular and the part cut off by the perpen- dicular from the end of this line are definite and precise points which can be computed to any degree of accuracy de- sired. No rational number rep- resents these lengths, which are trigonometric irrationalities. A series of constantly increasing rational numbers can be found, such that there is no greatest of the series, to represent lines which are always shorter than the given line ; and another series of terms constantly decreasing, but approaching to the terms of the first series, can be found. No largest number can be found in the first series and no smallest in the second ; both sequences together define, we may say, an irrational number. A trigonometric irrational 12 UNIFIED MATHEMATICS EXERCISES 1. Write 6 terms of the decreasing series defining V2; V3; V5. 2. How is the series for defining the length of the circum- ference of a circle obtained ? What assumption is made ? 3. Find the series defining the square root of 3673 by suc- cessive approximations. 4. Inscribe a square and an equilateral triangle in the circle of radius 10 ; find the sides. 7. Constants and variables. Every fixed point on the line of reference is at a fixed distance from the point of reference, ; the distance is constant. We can think of a point as mov- ing on the line OX in either direction. The distance from then varies and we speak of the distance as a variable. Thus, also, the price of wheat during a term of years or in different parts of the world is a variable ; the weight of an animal at different ages is a variable. We think of the variable quantity as taking a series of values under diverse conditions. We can represent the variable distance of a point from on our line by the single letter cc, which may then in the various possible positions on the line be thought of as positive, or negative, or zero, as rational or irrational. This letter x represents then a variable quantity and is essentially a number, subject to all the operations on algebraic quantities as noted above. In gen- eral, we designate the variable point by the single letter P, the distance from is OP, of which the length and direction from is indicated by the number or variable x. A point on the line X f X is represented by a single letter x, called the abscissa of the point. The fixed points on the line are fre- quently represented by the letters a, b, c, d, or by x 1} x z , 3, , each of which may represent any point upon the line. 8. Historical note. Modern algebra with the systematic employment of literal coefficients, letters to represent general constants, was introduced by the great French mathematician NUMBERS OF ALGEBRA 13 and statesman, Francois Viete (1540-1603) ; Viete used the consonants to represent known quantities and vowels to re- present unknowns, using capitals for both. The use of the symbols of operation in equations dates also from about the same time ; our equality sign was introduced by Robert Recorde, an English physician and mathematician, whose " Whetstone of Witte," 1557, is the first treatise in the Eng- lish language on algebra. The -f and symbols are due to a German, Widrnann, and date from 1489. 9. Geometrical equivalents of the four fundamental operations, a. Addition. The operation of addition of x l and x 2 is repre- sented graphically by placing the length OP 2 upon the line 1:17 OP 3 = + PjP 3 = OPi + OP 2 = xi + xa. from the point PI in the direction of OP 2 . Physically, addition is the result in general of two different causes. Thus a weight of 3 pounds -f a weight of 5 pounds ; a vertical velocity due to the action of gravity (on. a falling body) + a vertical velocity due to some other force ; a transportation (translation) from one point to another + another translation in the same direction ; two successive rotations of a wheel about its axis ; these are familiar examples of addition. 6. Subtraction. OPi + PiP 2 = OP 2 ; P^P 2 = OP 2 - OP^ Whatever the relative positions on the line OX of P r and P 2 , with respect to the position of 0, OP t + P^P 2 = OP 2 , all of these representing directed line segments. In words, the equality PiP 2 = OP 2 OPi states that the distance from any point PI on a directed line to a second point on the line is given by the abscissa of the second point minus the abscissa of 14 UNIFIED MATHEMATICS the first, with respect to any third point O, on the line, as origin. If we represent the distance from P l to P 2 by the letter d, we have Xi + d = x, 2 , or d = x-^ x t . Subtraction is represented by the distance from the first point to the second point, which OPi + PjP 2 = OP 2 . Fundamental property of any three points on a directed line equals OP 2 OPi- Since in physical problems the distance represents the rise or fall in numerical value of physical values, graphical subtraction is more frequently noted than addition. We say the temperature has risen 10 degrees or fallen 10 degrees, having in mind the original and the final reading ; the iron bar has expanded an inch, has a precise and valuable physical meaning when the initial and the final length of the bar are known. The operation involved in measuring the expansion is strictly subtraction. c. Multiplication and division. Graphical multiplication and division upon cross-section paper, involving theorems con- cerning similar triangles, lends itself to numerical work where great accuracy is not necessary. Some convenient length is taken to represent 1 unit on the horizontal scale and a con- venient length as unit on the vertical scale ; for three-place accuracy a sheet of cross-section paper considerably larger than the page of this book would be necessary. If multiples and quotients of 27 (2.7 or 270 or 2700 ...) are desired, take the unit on the horizontal scale so as to use 8 to 10 inches to represent the numbers up to 10 ; on a vertical line above the 10 on the horizontal scale represent 27. On our NUMBERS OF ALGEBRA 15 diagram below we have used 1 horizontal unit to repre- sent 10 on the vertical scale; greater accuracy would be attained by using a longer length as unit on the vertical scale. The tenths of 27 are represented by the vertical lengths from the points 1, 2, 3, 4, ..., 9 on the horizontal scale to the line joining the intersection of the two scales to the point 27 units above the point 10 on the horizontal scale. To multiply 2.7 by 3.6 you find the vertical length above the point 3.6 on the Graphical multiplication Tenths and hundredths of 27 represented by the vertical lines. horizontal scale and this length is 9.7 ; for 27 x 36 you inter- pret this as 970 and similarly for .27 x .36 you interpret this as .097. Evidently MP : 3.6 = AG : 10, or MP : 3.6 = 27 : 10, whence MP = 3.6 x 2.7. In division the length MP corre- sponds to the dividend and 27 to the divisor ; the quotient is then given as the horizontal distance corresponding to the given dividend (MP). Thus 20 -f- 27 is obtained graphically by finding the horizontal distance at which- the line 20 units above our horizontal scale cuts the given oblique line; this appears to be at 7.4 units and must be interpreted as .74; the position of the decimal point must be fixed by the com- puter. The explanation would be a trifle easier if the line of length 27 were placed directly above the unit distance, but a slight error in the position of the upper point would produce large errors at the right end of the line ; a line to be used as indicated must be kept on the paper and it must be drawn between points which are not close together. 16 UNIFIED MATHEMATICS PROBLEMS 1. Represent the square root of 2 geometrically, taking 10 quarter-inches as 1. Represent the square root of 3 on coordi- nate paper, using the same scale. Represent the square root of 5, 6, 7, and 8. 2. On coordinate paper take a horizontal length of 10 half- inches and at the right extremity draw a vertical line 7 half- inches long. Draw the line connecting the extremity of the vertical line to the left extremity of the horizontal line drawn. Use the divisions of the cross-section paper to read tenths of 7. Find .43 x ~. What change must you make to read 4.3 x 7 ? or 43 x 7 ? 3. Make the vertical line 7.4 in length, and read tenths of 7.4. Find approximately .83 X 7.4. Interpret as 8.3 x 7. .4 and as 83 x 74. 4. On the oblique line of problems 2 and 3 find where the vertical distance above the horizontal line is 5. How far out is this on the horizontal line ? This, read as tenths per half- 5 5 inch, represents the quotient of - and - respectively. Find the quotient of 5.3 divided by 7. Find 5.3 divided by 7.4. 5. Draw a semicircle on a diameter of 10 half-inches. Xote that the perpendicular at any point on this diameter is, by plane geometry, a mean proportional between the segments of the diameter. Read the mean proportional between 1 and 9, as the vertical line drawn at the point on the diameter 4 units out from, the center. Read the mean proportional between 2 and 8, similarly ; between 3 and 7 ; between 4 and 6. 6. Regard the preceding circle as having a radius 10 quar- ter-inches. Find approximately, from it, V19, V36, V51, V64, V75, V84, V91, V96, V99, and V100. These are the vertical lengths at the points dividing the diameter in the ratio 19 to 1 ; 18 to 2 ; 17 to 3 ; - 10 to 10. NUMBERS OF ALGEBRA 17 7. Take a circle of diameter 12 half-inches ; from it ap- proximate Vll, V20, V27, V32, V35, and_ V36. Note that V20=2V5, V27 = 3V3, and V3^ = 4V2; from these approximate V2, V3, and V5. See diagram on page 72. 8'. On the preceding circle check the geometrical fact that either side of a right triangle is a mean proportional between the whole hypotenuse and the adjacent segment of the hypote- nuse cut off by the perpendicular from the vertex of the right angle. CHAPTER II APPLICATION OF ALGEBRA TO ARITHMETIC 1. Arithmetic in science. In science definite progress is usu- ally intimately associated with the arithmetical investigation of the data of the science. Even in medicine, biology, sociology, and chemistry, arithmetic plays a large role ; in physics, astron- omy, and engineering, computation is absolutely essential. The student of mathematics must take account of the numerical side of mathematical work, constantly applying the theoretical work to definite, numerical problems. The accuracy of computed results in scientific work depends upon the accuracy of the observation measurements upon which these computations are based. This can be taken ae almost axiomatic. Whenever numbers are given representing meas- urements, the computations involving these numbers should not be carried beyond the point which these measurements justify. 2. Arithmetical application of algebraic formulas. Algebraic formulas and methods can frequently be applied to arithmeti- cal problems with a great saving of labor; practice with nu- merical examples is absolutely essential for success. The student is expected to use constantly the rules given in elementary arithmetic for multiplying and dividing by 5, 25, 12.5, 334, and by these numbers multiplied by integral powers of 10. Thus, to multiply by 5, multiply by 5 as ^-, annexing a zero and dividing by 2 ; similarly, 25, either as multiplier or divisor, is considered as J-^, and so with the other aliquot parts of 100 which have been mentioned. It is better, in general, to use these rules with the few numbers mentioned, and their 18 APPLICATION OF ALGEBRA TO ARITHMETIC 19 decimal multiples, rather than to extend this work to other aliquot parts whose occurrence is less frequent. 9350 ILLUSTRATION. 25 x 374 is written 4)37400 ; 2.5 x 375 is written 937.5 4)3750.0 ; 173 H- 250 is written .173 x 4 = .692 ; 75 x 375 is written : 18750 (50 times 375, dividing 375 by 2 and extending 2 places.) 9375 (25 times 375, or \ of 50 times.) 28125 (75 times 375.) The four formulas of elementary algebra which enjoy the widest use are undoubtedly : (x + a) 2 = x* + 2 ax + a 2 . (x a) 2 = a; 2 2 ax + a 2 . (x + a)(x a) = a 2 a 2 , (a; + a)(x + 6) = x 2 + (a + b)x + ab. (x + a)(x + 6) gives a simple rule for the product of two "-teens," e.g. 19 x 17. Thus, (10 + a)(10 + &) = 10 2 + (a+6)10 + aft, or =10(10+ a+b) + ab. Put into words, this formula states that the product of two numbers between 10 and 20 is equal to the whole of one plus the units of the other ; this sum is to be multiplied by 10 ; to this product is to be added the product of the units. RULE. To find the product of two " -teens," add the whole of one to the units of the other and annex a zero; to this number add the product of the units. 19 17_ 260 63 323 If x is taken as 20, 30, 40, 50, ..., the corresponding rule for the product of two two-place numbers having the same tens' digit is to add to the one of two numbers the units of the other ; the sum is to be multiplied by the tens' digit, and a 20 UNIFIED MATHEMATICS zero annexed to the product ; to this number add the product of the units. Thus, (37 x 35) = 30 x 42 + 35 = 1260 + 35 = 1295. Such products are most easily found, evidently, if the two units' digits sum to 10. 87 X 83 = 8 x 9 x 100 + 21 = 7221. 64 x 66 = 6 x 7 x 100 + 24 = 4224. In mental work with numbers work from left to right, and not from right to left, dealing first with the numbers of greater significance. (x 4- a) 2 and (x a) 2 are particularly useful in the computa- tion of squares of numbers of three places beginning with 1 or9. (10.7) 2 = 100 + 2 x 10 x .7 + .49 = 114.49. (11.3) 2 = 121 + 6.6 + .09 = 127.69, or = 100 + 26 + 1.69 = 127.69. (1.57) 2 = 2.25 + .21 + .0049, where .21 is obtained as 1.5 x .14 by the rule for the product of two "-teens." (.97)2 _ (i.oo - .03)2 _ i _ .06 + .0009 = .9409. (8.70)2 = (io _ 1.3)2 _ 100 - 26 -f 1.69 = 75.69. Frequently it is more convenient to use these formulas re- arranged as follows : (x + a) 2 = x(x + a + a) + a 2 , (x a) 2 = x(x a a) + a 2 . Thus, (84)2 _ 100(100 - 16 - 16) + 16" = 100(84 - 16) + 16 2 = 6800 -|- 256 = 7056. (25.7)2 = 20(25.7 + 5.7) + (5.7)2 = 628 + 32.49 = 660.49. (25.7)2= 25(25.7 + .7) + .49 = 25(26.4) + .49 = 660.0 (since 25 = *)+ .49. APPLICATION OF ALGEBRA TO ARITHMETIC 21 The square of any number between 25 and 75 is obtained from (x + a) 2 , as follows : (50 a)2 = 2500 100 a + a? = 100 x (25 a) + at. Thus, (37)2 _ 2500 - 1300 + 169 = 1369. RULE. To find the square of any number between 25 and 75 ; find the difference between the given number and 50; add, if the given number is greater than 50, or subtract, if the given number is less than 50, this difference from 25 and annex to this two zeros. Add to this number the square of the difference. Thus, (65) 2 =(25 + 15) x 100 + 152 = 4225. For numbers between 75 and 150 the squares may be ob- tained as 100(100 - a - a) + a 2 or (100) (100 + a + a) + a 2 , noting that 100 a or 100 -f a is your, given number whose square is sought. Thus, 1122 = 12,400 + 144. 13.7 2 = 174.00 + 3.72 = 174.00 + 13.20 + .49 = 187.69. Frequently, of course, only three or four significant figures are desired, and the methods mentioned give the significant figures first. (x + d)(x a) may also be used for squares, thus : x 2 = (z + a) (x a) + a 2 . (87)2 _ ( 8 7 + 13 ) ( 87 _ 13) + 13 2. (2.33)2 =(2.33+ .17)(2.33-.17)+.172 = 5.4 + .0289. (41.7)2 =(41.7 + 8.3) (41.7 - 8.8) + (8.8) = 50 x 33.4 + 8.32 = 1670 + 68.89 = 1738.89. (41.72)2 _ 1738.89 + (.04) (41. 7) + .0004 = 1738.89 + 1.6684 = 1741 to units, or 1740.6 to tenths. 22 UNIFIED MATHEMATICS PROBLEMS 1. Multiply mentally 19 x 18, 17 x 15, 18 x 14. 2. Use the rule given above to give the table of 18's from 18 x 11 to 18 x 19. 3. Multiply mentally 12 x 13, 36 x 34, 45 X 45, 82 x 88, 91 x 99. 4. Multiply mentally 27 x 25, 34 x 32, 54 x 58, 92 x 98. 5. What is the product of 44 x 36 or (40 + 4) X (40 4), 58 x 62, 44 x 37 or (40 + 4) x (40 - 3) ? 6. What are the first three figures of (114) 2 , (107) 2 , (131)*, and (118) 2 ? Note (114)2 is 12,800 + 14 2 , and the first three figures 129 ; in (116)'- to 13,200 you must add (16) 2 , which in- creases the first 13,200 to 13,400. 7. From the preceding answers in 6 write the first three figures of (1.14) 2 , (.107) 2 , (1.31)2, (H80) 2 . 8. Write the squares of 9.7, 88, 940, 8.7, and 9.2. 9. Approximately how much greater is (9.71) 2 than (9.7) 2 ? (88.2) 2 than (88) 2 ? (941) 2 than (940) 2 ? (8.75)' than (8.7) 2 ? (9.26) 2 than (9.2)2? Note that (88.2) 2 differs from (88) 2 first by .4 x 88, or by a little more than 35 units ; the .04 is usually negligible. 10. Square 43, 47, 52, 63, and 62 by using the difference between these numbers and 50 according to the rule. 11. Using the preceding answers, square 4.3, .47, .052, 630, and 6.2. NOTE. Use common sense rules to determine the position of the decimal point. 12. Using the formulas for (50 a) 2 , (a a) 2 , (100 a) 2 , write the following 25 squares. Time yourself on writing simply the answers ; the exercise should be completed in 6 minutes. APPLICATION OF ALGEBRA TO ARITHMETIC 23 62* = 67 2 = 82 2 = 49 2 = 13. Using the results of the preceding exercise, compute to four significant figures the following squares, timing yourself. 57 2 = 17 2 = 24 2 = 33 2 = 63 2 = . 42 2 = 39 2 = 52 2 = 87 2 = 63 2 = 59 2 = 7I 2 = 43 2 = 10.8 2 = 21 2 = 66 2 = 98 2 = 16 2 = 92 2 = 55 2 = 57.1 2 = 17.3 2 = 42\5 2 = 24.5* = 33.1* = 62.4' = 673 2 = 63.2 2 = 39.7 2 = 52.9* = 87.4* = 63.? = 59.2 2 = 71.8 2 = 82.9 Z = 43.5- = 10.82 2 = 21.4 2 = 66.7 Z = 1.92 2 = 98.6 2 = 16.6* = 92.6 2 = 55.3' 2 = 49.8 2 = 14. Employing the formula for (# -(- a) (x + 6) write the following products ; the exercise should be completed in 6 minutes. 16 x 19 22 x 24 32 x 38 51 x 52 66 x 64 15 x 14 23 x 26 . 43 x 42 33 x 31 88 x 82 13 x 18 24 x 29 46 x 44 27 x 24 97 x 93 17 x 12 28 x 28 54 x 59 24 x 22 57 x 53 16x18 36x33 82x87 17x13 79x71 3. Extraction of roots. In extraction of square root, the method of successive approximation should frequently be employed. Thus, V179.63 > 13 and < 14. Vl69 + 10.63 = 13 + a, wherein a must be a number such that 2 a x 13 equals approximately 10.6. A glance shows that .4 x 13 equals 5.2, which doubled gives 10.4. Hence, (13.4)2 = 169 + 10.4 + .16. (13.4)2 = 179.56, or 179.63 - .07. (13.4 + a)2 = 179.56 + 2 a x 13.4 + a. a now is less than .01 ; hence, a 2 is less than .0001 ; a is to be a number of hundred ths or thousandths, evidently, so that 2 a x 13.4 is approxi- mately .07 ; a is roughly .003, slightly too large. (13.403)2 = 179.56 + .0804 + .000009 = 179.6413. 24 The rule is commonly given to take x - '- as first approxi- 2 x mation of Vo? a, wherein a is small as compared with x. The process illustrated follows this rule, but suggests thinking multiplication instead of division. Thus in the square root of 300, as V17 2 + 11, approximately -^ is to be added to 17 ; however it is easier to think 34 X a = 11, whence a = .3, or not quite .33 ; trying .32 (since the a 2 term is to be added) gives 17.32 of which the square is 289 + 10.88 (or .32 x 34) + (.32) 2 = 299.9829. Similarly, V3000 = V3025-25 = V55 2 -25 =55 -a, wherein a must be a number such that 2 x 55 x a will give approxi- mately 25 ; a is evidently .2 to one decimal place or .23 to two ; 54.77 is correct to four significant figures as given. 4. Approximate roofs. Another method of approximating square root is to divide the given number by the first approxi- mation, then to use the arithmetic mean of the two numbers as a second approximation. Thus, 179.63-^-13=13.82; taking -\ o I -i o 09 - as the approximate root gives 13.41 as a second a approximation. 179.63 -j- 13.41 gives 13.3952 and the average 13.4026 is within .0001 of the correct value. Similarly the cube root may be obtained. Thus in 179.63, 5 is the first approximation. 5 2 = 25 ; 179.63 -H 25 = 7.2 nearly. Taking the average of 5, 5, and 7.2 gives 5.7 as second approximation ; (5.7) 2 = 32.49 ; 179.63 -=- 32.49 = 5.529 ; the average of 5.7, 5.7, and 5.529 gives 5.643, which is correct within .001. PROBLEMS 1. What is the approximate square root of 1.26 ? 128 ? 2. Is the square root of 1.35 nearer to 1.16 or to 1.17 ? NOTE. (1.16)2 = 1.32 + .0256 and (1.17) 2 = 1.34 + .0289. APPLICATION OF ALGEBRA TO ARITHMETIC 25 3. By successive approximations find Vl.26 to four decimal places and compare with ordinary process of extraction of root. HINT. Use 1.12 as first approximation. 4. Find the cube root of 1.26 by this process, using 1.08 as first approximation. 5. Find the approximation to one decimal place of the square roots of 65, 63, 8.30, 8.76, and 27.32. HINT. Regard 8.30 as (3 x) 2 , whence x must be roughly .12. 6. Write the square roots of the following numbers, correct to 2 decimal places. Time yourself. 9.9 35 65 140 200 16.8 34 68 150 300 17.2 37.2 78 125 10.4 25.8 39.4 85 108 20.8 28. 48 90 112 30.6 5. Synthetic division and remainder theorem. The ordinary process of division, and particularly the abbreviated process by detached coefficients, can frequently be applied to com- putation. 3x 2 + 4x + 15 x 2)3 x3 2 x 2 + 7 x 5 3x3 -6x 2 + 4x2 + 4a 2 8x + 15x + 15X-30 + 25 Notice that the 3 of 3 x 2 has been written three times ; the 4 of the 4 x 2 has been written three times ; the 15 of the 16 x has been written three times. Each of these, 3, 4, and 15, represents not simply the coefficient of the highest remaining term, but also the coefficient of the corresponding term in the quotient. Further, note that 2 has been multiplied by 3 x 2 , by 4x, and by 15, and the product with sign changed has been added to the corresponding term of the dividend. x-2 + 2)3-2+ 7 - 5 + 6+ 8+30 3 + 4+16(+26 26 UNIFIED MATHEMATICS This represents the division of 3x 3 2x 2 + 7 x 5 by x 2 ; the 2 has been replaced by +2, subtraction has been replaced by change of sign and addition ; the remainders have not been written but are merged in the quotient, which is written below. Divide -2)5+ 0+ 0-(- 8 + 7 _ 10 + 20 - 40 +64 5 -10 + 20 -32 (+71 Interpreted : ox 4 + 8x + 7 divided by x + 2 gives as quotient 6x 3 10 x 2 + 20 x 32 and 71 as remainder. The above operations may also be interpreted : 1. 3x3-2x2 + 7x-5= (z-2)(3x 2 + 4x+15) +25. 2. 5x + 8x + 7 = (x + 2)(5x3-10x 2 + 20x-32) + 71. Substituting in (1) x = 2, the right-hand member reduces to the con- stant term, and the left-hand member is the original expression with 2 substituted forx. Similarly in (2), substituting 2 for x shows that the remainder obtained by dividing by x + 2 is equal to the original expres- sion with 2 substituted for x. If a rational, integral, algebraic expression in x is divided by x a (a may be negative), the division being continued until the remainder does not contain x, the remainder is equal to the original expression with a put for x. Let us represent by f(x) an expression of this kind, i.e. an expression consisting only of the algebraic sum of positive integral powers of x with constants as coefficients. Then, you can divide /(#) by x a, obtaining a quotient, Q(x~), also a function of x, and a remainder which does not contain x ; by definition of division, f(x) = (x a) Q(x) + R ; substituting a for a, /(a) = (a a) Q(a) + It. /(a) = + R, or R =/(a). Q. E. D. The common " check by nines " may readily be proved by the remainder theorem : When 12,738, which may be written 1 x 10< + 2 x 10 3 + 7 x 102 + 3 x 10 + 8, APPLICATION OF ALGEBRA TO ARITHMETIC 27 in powers of 10, is divided by 10 1, the remainder is equal to the original expression with 1 put for 10. Hence the re- mainder when 12,738 is divided by 9 is 1 + 2 + 7 + 3 + 8, or 21, or 3 (since 21 divided by 9 gives 3 as remainder, or 2 + 1). Thus, a x 10" + b x 10 n -i + c x 10 n ~ 2 + g x 10 + h divided by 10 1 gives a -f b + c + + g + h as remainder. Division by 11, 10 + 1, gives as remainder the sum of the odd coefficients less the sum of the even coefficients, counting from units' place ; a sum to 11 can, of course, be dropped as it occurs, or 11 can be added to make the remainder a positive number. Illustrative examples. Dividing by the method of detached coefficients : 1. Divide 5x 3 + 3a5 2 + 2a; + 6bya; 2 and by x + 2. 2)5 + 3 + 2 + 6 4-10 + 26 +56 5 5 + 13 + 28(+62 = (x-2)(5x 2 + 13 x+ 28) + 62. x + 2 -2)5 + 3+ 2 + 6 10 + 14 - 32 5x3 + 3x 2 + 2x + 6 6- 7 + 16(-26~ = (x + 2)(5x 2 -7x + 16) -26. 2. Divide 5326 by 98 and 102 ; i.e. by 100 - 2 and 100 + 2. 5326 = 98 x 54 + 34. Note that the 500 is subtracted mentally, and since 5x2 too much has been taken away, 5x2, or 10, is added to the remainder 32 ; again, 4x2 too much has been taken away, hence, 8 is added to the remainder. o4 100 + 2 52 5326 = 102 x 52 + 22. Here, when 500 is taken away, there still remains 5 x 2 to be subtracted, since the divisor is 102. 4 ~22 28 UNIFIED MATHEMATICS 3. Divide 127,384 by 96, by 103, by 124. 96 1326 127,384 = 96 x 1326 + 88. 4)127384 4 313 12 258 8 664 24 88 rem. i A little practice will enable one to perform the additions mentally, writing: 06 1326 4)127384 31 25 66 88 rein. 103 1236 103 1236 -3)127384 or, abbreviated, -3)127384 -3 24 243 37 -6 69 378 76 rem. -9 694 - 18 76 rem. 120 + 4 1027 4)127384 In this division the remainders when dividing 4 by 120 must be noted mentally ; thus, 330 less 240 338 gives 90, whence 98 as remainder, from which 8 is 8 to be subtracted. Similarly, 900 less 840 gives 60, 904 to which the 4 is added mentally, and 28 sub- 28 tracted leaves 36 as remainder. 36 rem. A number of two places ending in 9 or 8 can be used as divi- sor in short division form. Thus, APPLICATION OF ALGEBRA TO ARITHMETIC 29 Think 60 in 127, 2, with remainder 7 ; add 2159 with 3 rem. 2 to 7, giving 9 ; 93 by 60 gives 1, with re- 59)127384 mainder 33 ; 33 + 1 = 34 ; 348 divided by 60 gives 5, with remainder 48 ; 48 + 5 = 53 ; 534 by 60 is 9 less 6; add 9 to 6 gives + 3. 88 1447, Think of 90 2 as divisor. 2)127384 127 -*- 90 = 1, remainder 37 ; 37 + 2 = 39. J*_ 393 393 -T- 90 = 4, remainder 33 ; 33 + 8 (since 4 x 2 too 8 much has been subtracted) =41. 418 418 -* 90 = 4, remainder 58 ; 58 + 8 = 66. _8_ 664 664 -f- 90 = 7, remainder 34 ; 34 + 14 = final re- 14 mainder, 48. 48 renio PROBLEMS 1. Divide 2x 3 +3x 2 9 a Sbycc 1, x2, x 3, x + l, x + 2, and by x 3, employing synthetic division, and inter- pret results. + 2)2 + 3- 9-- 5 _2)2 + 3-9-5 + 4 + 14 +10 -4 + 2 +14 2 + 7+ 5(+ 5 2-l-7(+ 9 ' 2af + Sec* - 9z - 5 = (x - 2)(2a? + 7a;+ 5) + 5. When x = 2, 2x* + 3x* - $x - 5 = + 5. 2^ + 3^ 9 a; 5 =(x + 2)(2z 2 x - 7) + 9. When z=-2,2z 3 + 3a; 2 -9a;-5 = + 5. 2. Divide x 2 8x z + 2 x 5 by x 1, x 2, and x 3. What is the value of this function of x for x = 1, x = 2, a; = 3 ? 3. Divide # 4 3x 2 18 by x 2 and by x 4. 4. Divide y? - 1,289,000 by x - 100 and x - 110. What is the value of x 3 - 1,289,000 for x = 100 and x = 110 ? 5. Divide af' + Sz 2 1 x 21oy x Iby synthetic division. 6. Divide 2 or 5 - 5x* + 7* - 10 by x - 2. 7. Divide 38,942 by 96 (as 100 - 4). 30 UNIFIED MATHEMATICS 8. Divide 38,942 by 59 (as 60 - 1) by short division. 9. Divide 38J942 by 104 (as 100 + 4). 10. Divide 3 i* - 2 t 3 - 18 P + 6 t - 11 by t - 1. Find the quotient and remainder. 11. Find the value of z 3 179.63 when x = 5.6. 12. Find the square root of 321.62 by repeated division and approximation, as indicated. 13. What is the approximate value of the square root of (a) 145; (6) 147; (c) 150; (d) 26; (e) 2615? 14. Approximate the cube root of 126 ; 8.1 ; 27.4 ; 64.2 ; 127 ; 218 ; 350 ; 520 ; 735 ; and 1004. 15. Perform the following divisions by synthetic division ; give quotient and remainder ; time yourself ; 20 minutes is an ample allowance. a. a? - 3 x 2 + 7 x - 5 by x - 2. b. 2 a 3 + 4 z 2 7 a; + 8 by a - 3. c. 3 x 3 - 2 a; 2 - 8 x + 10 by x + 2. d. 4 ^-2^ + 8 a; -5 by a .5. e. & 7 a; 2 + 8 x 5 by x 2. f. tf - 18,700 by x - 25. g. 2 a? - 100 x z - 18,700 by x - 300. h. y? - .2 x - .05 by x - .8. t. a 5 - 3 ;r 2 - 100 by x-3. j. X s - 2,000,000 by x - 140. 16. Divide by the method corresponding to synthetic divi- sion ; time yourself ; carry the division to four significant risrures a. 1375 by 100 - 2 (or 98). b. 1375 by 60 -2. c. 4356 by 100 + 2. d. 8.736 by 5.9. e. 6248 by 7.9. APPLICATION OF ALGEBRA TO ARITHMETIC 31 6. Percentage of error. "When any measurement of physi- cal objects is given, the measurement has a certain limit of accuracy, determined in part by the instruments and methods of measurement and in part by the very nature of the thing measured. In measuring the distance to the sun from the earth, at some fixed time, the measurement may be given as 93,000,000 miles 1,000,000 miles, or 93,000,000 miles, within a million miles ; the thickness of a watch spring may be meas- ured as .014 inch with a possible error of one thousandth of an inch, or .014 .001 inch. However, from the point of view of the physicist and mathematician, the distance to the sun is more accurately given than the thickness of the watch spring, for the percentage of error ratio of possible error to measured value in the case of the sun's distance is slightly more than 1 % of the distance, while in the other case it is more than 7 % of the thickness of the spring. Every number which represents a measurement involves this type of error. Obviously, in any computations with such numbers, results are significant only within limits determined by the percentage of error. "7. Significant figures. The significant figures in any num- ber representing a measurement are those which are given by the measurement, and do not include those initial or terminal zeros which are determined by the unit in which the measure- ment is made. The terminal zeros in 93,000,000 are not sig- nificant figures, as the unit of measurement here is evidently a million miles ; as the measurement can be made to one further place, the distance may be written, in " standard form," 9.3 x 10 7 miles or 9.30 x 10 7 miles in which only significant digits appear in the first factor combined with powers of 10. In the thickness .014 inch, the initial zero is not a significant figure, as it is apparent that the measurements are made in thousandths of an inch ; in " standard form," this is 1.4 x 10~ 3 inches. 8. Measurement computations. Products. If the length of a rectangle is measured with an error of less than 1 % of its 32 UNIFIED MATHEMATICS true value, and if the breadth is given absolutely, the true area will be given with the same percentage of error as the length. But if the breadth is also only approximately meas- ured, the possible error in the area obtained as the product will be greater than if only one factor involves a possible error. The graph represents the right- hand end of a rectangle whose length and breadth are measured as 21.5 cm. and 12.2 cm. respec- tively, where it is understood that the measurement only pre- tends to give these dimensions to within one millimeter, one tenth of one centimeter. The meaning of these figures then is that the length lies between 21.4 cm. and 21.6 cm., and the breadth between 12.1 cm. and 12.3 cm. The first is an error of less than ^ of 1 % and the sec- ond of less than 1 %. The un- certainty in area due to the pos- sible error in length is indicated by the areas at the right end with dimensions .1 cm. by 12.1 cm. or .1 cm. by 12.3 cm. ; the uncertainty in area due to the breadth measurement is of di- mensions .1 cm. by 21.4 cm. or .1 cm. by 21.6 cm. The area uncertainty is then at most .1 cm. by (21.6 + 12.2) cm. or 3.39 sq. cm. of area. This error may evidently affect the third figure in our computation of the area and hence in the product the figures beyond the third place are not significant, and give no real information concerning the actual area in question. Note that the area as the product of 12.2 by 21.5 is 262.30 Itf 17 18 19 20 21 Measurement of an area A rectangle measured as 21.5 cm. by 12.2 cm. APPLICATION OF ALGEBRA TO ARITHMETIC 33 sq. cm., but the inaccuracy of measurement of the length means that there is an uncertainty of area at the right-hand end amounting to 1.22 sq. cm. (cm 2 .), and similarly at the top an uncertainty of 2.16 cm 2 . ; the total uncertainty of area amounts to more than 3 cm 2 ., and should be given as 3.38 cm 2 ., or 3.4 cm 2 . Commonly, of course, the measure- ments 12.2 cm. and 21.5 cm. mean that the area has been measured to one half of the last unit given ; thus this area actually falls between rectangles of dimensions 12.25 by 21.55 and 12.15 by 21.45 ; even in this case the area uncertainty is greater, by precisely similar reasoning, than 1.5 cm 2 ., and is approximately 1.7 cm 2 . To give 262.30 as the area of this measured rectangle is giving nonsense in the last two places ; it should be given as 262 or 262.3 1.7 cm 2 . Let k and k' represent measured quantities given with pos- sible errors of i % and e J respectively, e and i being assumed as smaller than unity (in common practice) ; the absolute / i \ values of these measured quantities lie between k 1 1 H -- j and "t- and between *' 1+ The true product lies, then, between kk'( 1-f-^-^H -- ) and kk'( 1 t-? -j--^_ V in other words, the true product \ 100 looooy may vary by l from the computed product ; is 100 10000 disregarded, if i and e are less than 1, since the fraction is less than 1 o/o of 1 % of kk 1 . In the graph the product *f x kk' lUuOO is represented by one of the small corner squares with dimen- sions .1 cm. by .1 cm. Illustrative example. The product of 987 by 163 wherein each num- ber is correct to within a unit need be computed only to the fourth significant figure as the percentage error may be as great as ^ of 1 % + ^ of 1 %, since in 987 parts is approximately ^^ or ^ of 1 %, and J in 34 UNIFIED MATHEMATICS 163 is greater than y^ or T 3 ff of 1 %. The error in the product may be as great as (^ + T %) of 1 % or 2 7 5 of 1 % ; but 2 7 ff of 1 % of any number certainly affects the fourth place and probably affects the third place in the number. Hence there is no point whatever in carrying this compu- tation beyond four places, 987 163 98700 59220 2961 160881 163 987 1467 130 11 160800 am. 987 163 987 begin with 100 x 987. 692 take 6 x 98, carrying however the 4 from 6x7. 29 take 3x9, carrying the 2 of 3 x 8. 160800 987 + J 987 - i 163+ \ 163 - \ 160881 + |(987 + 163) + $ 160, 881 - $(087 + 163) + $ = 161456$ = 160306| The product of 987 x 163 is 160,881 ; 987| x 163 gives 161,456$ ; 986| x 162 gives 160,306$ ; the actual area, if these represent dimensions of a rectangle measured to three significant figures lies between 160,306$ and 161,456$. In practice we take the product 987 x 163 to four signifi- cant figures, which gives the area slightly more accurately than our measurements justify. 9. Abbreviated multiplication of decimals. The abbreviated process of multiplication applies particularly well to decimal fractions, but the method can be extended to integers quite as well. To find .9873 x .1346 correct to four decimal places. .9873 begin with the highest digit of the multiplier ; first x fourth decimal place gives fifth decimal place. continue with 8x7 (second x third place), carrying the 2 from 8x3. take 4x8 (third x second place) carrying 3 from 4 x 7, or 28, which is more than 2 units in the fifth place. 6 x 9, or 54, + 5 carried from the 6x8. It assists in the process to cross out the last upper digit as it is used ; thus here 3 would be crossed out first, then 7, then 8, and finally 9. 9873 7898 395 _59 18225 APPLICATION OF ALGEBRA TO ARITHMETIC 35 If a check is desired, multiply again, reversing the order of the factors ; thus : .1846 .9873 16614 begin with 9x6. 1477 take 8x4, adding 5 from the 8 x 6 product. 129 take 7x8, adding 3 from the 7x4 product. 6 take 3x1, adding 2 from the 3x8 product. .18225 Read this as .1823 to four places. Obviously, if these were integers, you could proceed in the same way, writing the final product with four zeros, as 18,230,000. A similar abbreviation can be effected in division by drop- ping each time the last figure of the divisor used, and using the^ remaining part of the original divisor as new divisor. Thus, to divide .18225 by .9873 or by .1846 you proceed as follows : ^1846 . .9873 .18225 .1846). 18225 9873 16614 8352 1611 7898 1477 454 134 395 129 59 5 59_ 5 Here 9873 is used as the first divisor ; then 987 is used, but to the partial product, 8 x 987, is added the tens' digit of 8 x 3, the digit just crossed out ; then 98 is taken as divisor and to the product is added the tens' digit of 4 x 7 (28 is taken as giving a tens' digit of 3) ; then 9 is used and 5 carried over from 6x8. 10. Percentage effect of errors in divisor. If a divisor is known to be too large or too small by a definite percentage of itself, the quotient will be respectively smaller or larger than the correct quotient, for small per cents, by approximately the same per cent. 36 IXIFIED MATHEMATICS By division, - = 1 i + i* i 3 -f- i 4 1 4- i - For values of i less than .05, i 2 is less than .0025, or ^ of 1 % ; i, ?' 4 , and t are less than .000125, .00000625, and .0000003125, respectively. Hence an error of from 1 % to 5 % of excess in the divisor means an error of deficiency vary- ing also from 1 % to 5 %, within \ of 1 %, or from .99 % to 4.75%, or from .9901% to 4.7625% in the quotient. For values of i between \ of 1 % and 1 %, an error of deficiency in the divisor means the same error of excess in the quotient, within -j-J-j of this error. The meaning in physical measure- ments of these results is that when the divisor is correct only to the third significant figure, with a possible error of to 1 unit in the third place, the quotient will be correct to about the same degree of accuracy. For three-place numbers the divisor may vary from 100 to 999. The possible error of 1 unit, ^, means that 100 must be replaced by (100 .5) or 100(1 .005) and 999 by 999 .5, or approximately 999(1 + .0005) ; the quotient will vary from 1 .005 to 1 .0005 times the obtained quotient. Hence the quotient obtained is valuable at most to the fourth place, and frequently not beyond the third place. Illustration. Given that 76,430 is divided by 180; what variation ic /iQft i KC\ AOA a. * n tne ( l uoti ' eu t would a change of 1 in the divisor produce ? Suppose that instead of 180, 170 should have been used. What is the error in the "quotient ? 180 1 180(1 .006), the error in the divisor is more than .5 % and less than .6 Jo of the divisor ; the error in the quotient is no more than 2.5 and no less than 2.1, since 1 % of 424.6 is 4.246 and .6 /o and .5 % are respectively 2.5 and 2.1 ; the quotient may be taken as 427.1, whereas 426.9 is obtained by actual division. Even an error of \ a unit in the divisor 180 affects the third place in the quotient. In obtaining .5 % and .6 % of 424.6, there is no point in carrying the work beyond two places ; the values show that the error is between 2.1 APPLICATION OF ALGEBRA TO ARITHMETIC 37 and 2.5, and further places add nothing to the accuracy. The fraction 3 J 5 , or in the cass of the \ unit error of s ^ 5 , might just as well be used as per cents. This gives in the latter case ^-^ of 424.6, or + 1.2 as cor- rection, giving 425.8 as quotient ; the actual quotient is 425.71)4. Do not carry divisions and multiplications beyond the degree of accu- racy warranted by the data. Illustrative examples. You can multiply 3.14159 by 140.8 and obtain the result numerically correct to six decimal places. But if the 140.8 represents the diameter of a circle, measured correctly to the tenth of an inch (or of a foot, or of a meter) the product of 140.8 by 3.14159 gives a valuable result only to the first decimal place ; the circumference cannot be computed correctly to any further percentage of accuracy than that with which the diameter is measured. The area can be com- puted here with any meaning only to four significant figures ; in fact an error of .05 inch in the diameter makes a possible error of 10 square inches in the area. It is convenient to write the products from left to right, dropping work beyond the second decimal place. 140.8 140.8 31 3.14159 the .00059 is of no use as it 422 . 4 422 . 4 does not figure i n the prod uct. 20.1 14.1 442.5 circumference 5.6 1 442.2 circumference For most practical purposes 3^ is sufficiently accurate, as in finding this area, wr 2 : 70.4 70.4 4928 28 4956 only 4 places to be retained. 4956 3.14159 14868 496 15576 area. 198 5 15567 area. Area as found to correspond to data, 15570. 38 UNIFIED MATHEMATICS PROBLEMS 1. The distance of the earth from the sun varies between 91.4 x 10 6 miles and 94.4 x 10 6 miles. The length of the earth's orbit lies between circles having these lengths as radii. Between what values does this orbit lie ? What is the ap- proximate orbital speed of the earth in miles per hour ? 2. The mean distance of the earth from the sun is 92.9 x 10 6 miles. Compute the circumference and the mean speed and compare by percentages with the preceding. 3. Compute the speed of a point on the earth due to the rotation, taking that at latitude 45 the radius of the circle of latitude is 3050 miles. Compare the rotational speed with the revolutional speed. 4. What effect on the computed velocity would it have to take 365.25 instead of 365 ? How would you correct your division for 365.25 as divisor after having obtained the quo- tient, dividing by 365? what change would using 365.26 in- stead of 365.25 effect in the computed velocity? 5. The distance of the moon from the earth varies between 221,000 and 260,000 miles, mean 238,000 ; discuss the length of the path of the moon and the velocity of the moon which has a periodic time of 27.32 days. 6. A man whose salary is S 3000 pays $ 480 for rent. What per cent is this of his salary ? Suppose that he earns $ 275 in addition to his salary, what per cent is the rent paid of his income ? Compute only to tenths of one per cent. 7. If a man with an income $ 3275 pays $ 1100 per annum for food, $ 630 for clothes, $ 240 for life insurance, S 200 for "higher life," and saves the balance, compute his budget by per cents. 8. Given that a pendulum of length I cm. makes one beat, one oscillation, in t seconds, connected by the relation, T '980' find the length I to two decimal places when t = 1. APPLICATION OF ALGEBRA TO ARITHMETIC 39 9. What effect on / does a change from 980 to 981 pro- duce ? What decimal place in I would be affected ? 10. What error would the use of 3j- instead of 3.14159 introduce ? 11. The number n of vibrations of a pendulum of length 99.39 cm. is 86400, when g = 980.96 ; g is the acceleration due to gravity, and the formula for the number of vibra- tions is given by the formula, 86400 la n = \/,> 7T * or for the seconds' pendulum, when I = 99.39, g = 980.96, it is n = 86,400. Suppose that at the top of a mountain (g di- minishes) this pendulum of length 99.39 loses 86 beats per day, what is the approximate percentage of loss in n ? The percentage of loss in g is approximately double this since V2 * Vl i = 1 - - -. What is the approximate loss in g? Take 86 as T V of 1 % of 86,400. 12. Given g 980, I = 50, compute n in the formula of problem 11. What maximum effect on n would a change from I = 50 to / = 50.5 cm. produce ? 13. Compute the weight of a table top, hardwood, dimen- sions correct to .05 foot, top 48.1 x 36.4 x 2.1 inches. Weight of wood 48 pounds per cubic foot. 14. If a table top similar to the above weighs 97 pounds, compute the weight per cubic foot of the wood. 15. If water weighs 62.4 pounds per cubic foot, compute the specific gravity of each of the preceding woods. wt. of cubic foot of wood 8 = wt. of cubic foot of water 16. The path of the earth is approximately a circle of radius 92.9 x 10 6 miles, of which the center is approximately 11 mil- lion miles from the sun. Compute this circumference and compare with the results in problems 1 and 2. CHAPTER III EXPONENTS AND LOGARITHMS 1. Exponent laws. For convenience the product of a by itself, a x a, is represented by a 2 , a x a x a by a 3 , ..., and a a a a to m factors by a m . In this notation m is called the ex- ponent and a the base. The following laws evidently hold : I. a m ' a" = a m+n . II. = a m ~ n , when m > n. a" III. (a m ) n = a m ' n . IV. (a 6) m = a m b m . In the definition as given, m represents the number of factors and is assumed to be a positive integer. However, it is found possible to define a m for all real values (fractional, negative, zero, irrational) of m so as to have the resulting numbers combine according to the four laws given above. Thus, a a m = a *"* = a m , if Law I is to continue to hold ; hence, a must be defined to equal 1, since multiplying a number, a m , it gives that number. To be justified in using a zero exponent with this meaning the other exponent laws must be shown to Hold when either m or n is zero, but in II only n could be zero at this point. For a negative integer, n, if Law I is to hold, a must be defined as such a number that a" a~ n = a~ n+n = a = 1 ; hence we define a~ n as the reciprocal of a", a~" = All the laws a" I to IV can be shown to hold under this extension of the meaning of a". 40 EXPONENTS AND LOGARITHMS 41 p Similarly, a q , if Law III is to hold, must represent a number ^ which raised to the gth power equals a p ; a q is thus defined as the qth root of the pth power of a. Taking this definition of p a 9 , Laws I to IV can be shown to hold with this extension in possible values of m and n ; p and q are assumed to be integers. p p For fractional exponents a fifth law is introduced, a* = a "*. For irrational values of n, a n is defined by a limiting process. Thus, d^* is defined as the limit of the series a 1 - 4 , a 1 - 41 , a 1 - 414 , ..., wherein the successive rational exponents define the square root of 2. The operations of elementary algebra with radicals are made subject to the exponent laws. Thus, 2V3 = 2-3* = (2 2 )* . 3* =(2 2 . 3)* = 12* = Vl2, by successive application of V, III, and IV. The operation of raising to a power indicated by a m , with m integral, is called involution. The inverse operation of finding x when x m is given equal to a is called evolution. 2. Logarithms. A logarithm is an exponent. The relation x = a m may be written m = log a x. m is the exponent which applied to a gives x; m is the logarithm of x to the base a. 3. Fundamental laws of logarithms, a. Logarithm of a product. If x = a m and y = a", x y = a m a" = a m+n . log u (x - y) = m + n = log u x + log a y. 42 UNIFIED MATHEMATICS In the language of logarithms and translated into ordinary language this theorem is as follows : I. log a (*!/) = log a x + Iog y ; in words, the logarithm of a product is the sum of the logarithms of the factors. b. Logarithm of a quotient. - = a *- log a x - = m n = log a x log a y. y a n y II. log a - = log a x log a y ; in words, the logarithm of a quotient is the logarithm of the dividend minus the logarithm of the divisor. c. Logarithm of a power. If x = a m , x n ( m ) n = a mn ; log a x" = m n = n log a x. III. log a x n = n log a x ; the logarithm of a power of a number is the index of the power times the logarithm of the number. Since our exponent laws hold for all values of m and ?i, these theorems hold for all values of m and n. x and y are assumed to be positive numbers and for compu- tation purposes 10 is commonly taken as the base. We assume that as the logarithm increases the number increases. This can be readily proved from the fact that 10 m 10" = 10 m+n ; no matter how small the n is, as a .positive quantity, 10" is greater than 1. For n any positive fraction, , 10" represents the gth root of the pth power of 10, wherein p and q are integers. Now 10* will be an integer 10, or greater than 10, and the S the amount at the end of n intervals. Given interest compounded at rate t per annum for n years. At the end of 1 year you have P -f iP = P(l + 1). At the end of 2 years you have p(i + o + /P(i + = P(i + O 2 - At the end of 3 years you have P(i + i) 2 + *p(i -f i) 2 = P(i + O 3 - At the end of n years you have 3 = P(l + i) n . EXPONENTS AND LOGARITHMS 55 Or you may say that since the interest for 1 year in- creases the principal P to (1 + *)P, then in 1 further year this new principal P(l + i) will be increased in the same ratio, giving P(l +i) x (1 + *) or P(l-f i) 2 , and for each further year the factor (1 + i) is introduced. Hence the amount at the end of n years is P(l + i) n . If interest is compounded at the end of every three months, i ? or every six months, you substitute for i, - or - , and for n, 4 n or 2 n, since the number of intervals of three months in n years is 4 n and of six months is 2 n. f j\mn The formula S = P( 1 + ) is used for an interest rate V mj given as j per annum, but compounded m times per annum, at rate for each interval. m Problems in compound interest lend themselves to solution by logarithms. Given, S = P(I + t). log S = log P + log (1 + 0". log 8 = log P + n log (1 + i). log P = log S - n log (1 + i). log (1 + i) Note that it is better not to use these as formulas, memoriz- ing them, but rather to go back to the fundamental relation, S = P(l -+- i) n . Note also that the formula holds for other than integral values of n ; thus at 6 % per annum the interest on the amount P for six months or eight months is defined as P(l + .06)i - P or P(l + .06)1- P, respectively. Hence for n-f4 years the amount would be 56 UNIFIED MATHEMATICS PROBLEMS 1. Find the amount of $ 1000 at interest 4 % annually, compounded for 20 years. Find the amount when compounded semi-annually at a nominal rate of 4 % per annum, i.e. 2 % semi-annually. 2. In how many years will money double itself at 4 %, 5 %, 6 lo interest, compounded annually '.' 3. Given that at the end of 20 years $ 1000 amounts to $ 1480, what is its approximate rate of interest ? 4. Given that at 5 % interest, compounded annually, $ 1000 amounts to $ 1480, what is the approximate number of years ? 5. Find the compound amount of $ 1400 at 5 % interest, compounded semi-annually, for 10 years, 11 years, 12 years, up to 20 years. 6. If $ 100 is left to accumulate at 3 % interest, com- pounded annually, what will it amount to in 100 years ? Solve by logarithms. What amount put at 3 % interest will amount in 100 years to $ 1,000,000 ? What is the present equivalent of $ 1000 to be paid at the end of 100 years, money worth 3 J , compounded annually ? 7. Solve problem 6 for 4%, 5%, and 6% interest, com- pounded annually. For 6 J per year, compounded semi- annually, for 50 years. 8. Benjamin Franklin, who died in 1790, left 1000 pounds to '"the town of Boston" and the same to the city of Phila- delphia. His will directed that this amount should be loaned at interest to young artisans, and thus accumulated for 100 years until the principal should have increased to 130,000 pounds. He directed further that at that time the major por- tion of this amount should be expended for some public im- provement and the residue left to accumulate, similarly, for another hundred years. What rate of interest did Franklin assume that his money would earn ? In Boston the amount, EXPONENTS AND LOGARITHMS 57 $ 5000 approximately, accumulated to about $ 400,000. Find the average rate of interest earned annually. Assuming that $ 5000 was kept aside in 1891, as directed, what will this amount to in 1991, compounded at 4 % annually ? 9. Find the amount at the end of 200 years of $5000, interest at 4 %, 5 $>, and 6 %, compounded annually. 10. If a business doubles its capital, out of earnings, in 12 years, what rate of interest on capital invested does this represent per year ? If in 20 years the capital is doubled, find the rate of interest earned. 11. The United States has increased in population from 7.2 million in 1810 to 101.1 million in 1910 ; find the approximate rate of increase per year, and for each ten-year period. Com- pare with the figures on page 65. 12. The city of New York increased in population from 120,000 to 4,769,000 in the interval from 1810 to 1910. Com- pute the average annual rate of increase, using the formula, 120,000(1 + i} m = 4,769,000. Compute the average ten-year increase and compare with the actual statistics on page 66. CHAPTER IV GRAPHICAL REPRESENTATION OF FUNCTIONS 1. Functional relationships. Expressions of the form 3 x -\- 5, ax + b are called linear or first degree functions of the variable x ; in elementary algebra such expressions have been combined according to the fundamental operations and subject to the laws given in a preceding chapter. Further, some atten- tion is given in elementary algebra to expressions of the form ax* + bx -f c, the general quadratic function of x, and expressions involving higher powers of x. The expression ax n + bx n ~ l + is called an algebraic function of x of the nth degree when n is a positive integer and the coefficients a, b, are constants. This represents of course a number for any value of x. F(x), G(x), (%), A(#), are methods of representing functional relationships ; F(x), (read F of x or F function of x), means that this expression assumes various values as x varies, these values being determined by some law. In the equation, y = 3 x + 5, y is explicitly given as a function of x ; y is here a linear function of x. In the equation, y = x z + 4 x 5, y is an explicit function of x ; as a; varies, so does y. In x 2 -^ y z = 25, as x ta.kes on different values so does y, but one must solve for the corresponding values of y. Here y is called an implicit function of x. When two variable quantities are so related that the varia- tion of one of these depends upon the variation of the -other, either is said to be a function of the other. Thus the pro- duction of wheat in the United States from 1900 to 1915 is a variable quantity depending upon the year of production. The height of a given tree is a function of its age ; to each number expressing in any convenient unit of time the age of 58 GRAPHICAL REPRESENTATION OF FUNCTIONS 59 the tree corresponds a given number expressing the height of the tree. Similarly the weight of a tree is a function of the age of the tree. This type of relationship cannot be ex- pressed algebraically. It may be exhibited by the two series of numbers, or it may be expressed graphically. 2. Graphical representation of statistics. Since two variable quantities are to be represented, two sets of numbers must be indicated ; this could be done by placing the two sets upon two lines straight or curved, drawn parallel to each other. This is the form used upon grocers' scales wherein the vari- ables of weight and corresponding price are placed upon con- centric circular arcs ; corresponding numbers are cut by the pointer. Series of corresponding numbers graphically represented It is commonly more convenient to place the two scales for representing the two variable quantities upon two lines per- pendicular to each other. Upon the following figure the tem- perature and barometric pressure are indicated by the diagram for the week, March 4-11, 1918, at Ann Arbor, Michigan. 60 UNIFIED MATHEMATICS' Temperature and barometric chart by moving pointer The sharp break in barometer curve corresponds to a violent rainstorm. The horizontal displacement of any point on either graph, located by the vertical rulings, indicates the time of the observation ; the correspond- ing temperature or pressure is indicated by the vertical displacement. ILLUSTRATIVE EXERCISES 1. Production and price of wheat in the U. S. from 1895 to 1916 are given in statistical form and graphically. TEAR PRODUCTION EXPOBTS PRICE YEAR PRODUCTION EXPORTS PRICK Millions of Bushels Cents Millions of Bushels Cents 1895 467 126 50.9 1907 634 163 87.4 1896 428 145 72.6 1908 665 114 92.8 1897 530 217 80.8 1909 737 87 98.6 1898 675 222 58.2 1910 635 69 88.3 1899 547 186 58.4 1911 621 80 87.4 1900 522 216 61.9 1912 730 143 76.0 1901 748 235 62.4 1913 763 146 79.9 1902 670 203 63.0 1914 891 332 98.6 1903 638 121 69.5 1915 1,026 243 91.9 1904 552 44 92.4 1916 640 160.3 1905 693 98 74.8 1917 1906 735 148 66.7 Statistics from the Yearbook of the U.S. Department of Agriculture GRAPHICAL REPRESENTATION OF FUNCTIONS 61 Graphical representation of (broken line) wheat production (continuous line) and price in the United States, 1895-1917 Note that the graphical form of these statistics brings out several points of interest. In the first place the maximum price paid for wheat in the interval is immediately found, and so also the minimum price of 51 ^ (50.9^) in 1895. Further, the diagram shows very pointedly that a large production under normal circumstances is accompanied by a fall in price, and an immediate diminution of production. In 1917, under war conditions, both production and price increased greatly. 2. The weight of water per cubic foot, or 60 pints, is 62.4 pounds. For cylindrical vessels filled to a height of 12 inches the weight for an area 144 square inches in the base would be 62.4 pounds ; for 72 square inches in the base the weight of 62 UNIFIED MATHEMATICS water would be 31.2 pounds ; for square inches the weight is pounds. On coordinate paper represent square inches of base on the horizontal line, taking 1 major division to repre- sent 10 square inches, and represent weight on the vertical line. Weiyltt 10 20 30 40 50 60 70 80 90 100 110 120 130 140 Square Inches in base Weight of water in cylindrical vessels with varying base when filled to a height of 12 inches or 10 inches Note that the weight of 12 inches of water in a vessel with a base con- taining 50 square inches is 21.5 pounds, and conversely if a cylindrical vessel contains 21.5 pounds in 12 inches of height, the base contains 50 square inches, and similarly, of course, for other values. PROBLEMS 1. Plot the temperature, as vertical lengths, and the time, by hours, as horizontal lengths, for 24 hours. 2. Plot the contents in pints of cylindrical vessels 12 inches in height, with varying bases ; take that with base 144 square inches, the capacity is 60 pints ; with 72 square inches GRAPHICAL REPRESENTATION OF FUNCTIONS 63 in the base, 30 pints ; with base, pints. The straight line joining these points can be used to give the base in square inches of any cylindrical vessel whose capacity for a height of 12 inches is known. What would be the base of a vessel that contains 10 quarts when filled to the height of 12 inches? 3. Plot cubic inches against pints, taking 1728 cubic inches as 60 pints. INCREASE IN VOLUME WITH TEMPERATURE INCREASE As liquids are heated the volume changes, generally increasing ; thus water increases in volume when heated except between and + 4 C. Given 1000 cu. cm. of water at 4 C. and 1000 cu. cm. of mercury at C., the volume at other temperatures is given by the following table : TEMPERATURE VOLUME OF WATER VOLUME OF MEKCTRY 1000.13 1000.00 1 1000.06 1000.2 4 1000.00 1000.9 8 1000.13 1001.4 10 1000.27 1001.8 15 1000.87 1002.7 20 1001.77 1003.6 25 1002.94 1004.5 30 1004.35 1005.4 35 1005.98 1006.3 40 1007.82 1007.2 4. Plot the increase above 1000 cu. cm., or decrease, in cu. cm. in volume of the water, using 1 half-inch for 1 on hori- zontal axis and 1 half-inch for 1 cu. cm. on the vertical axis. Note that by adding 1000 to the given readings, actual volumes can be read. 5. Plot the same curve for the increase in volume of the mercury. It is evident that the increases in volume of the mercury are approximately proportional to the increases in temperature. 64 UNIFIED MATHEMATICS STATISTICS ON WEKJHT AND HKKMIT From an investigation of 'the statistics giving characteristics of a group of over 200,000 men and 130,000 women, the following facts are obtained on average height. The facts are given for groups of 1000. HEIGHT FRK^I EN< v mi No. IN GBOirp Men Women 4' 9" 1 4' 10" 4 4' 11" 10 5' 0" 2 40 5' 1" 2 55 5' 2" 5 107 5' 3" 12 135 5/ 4 " 30 184 5' 5" 55 167 5' 6" 90 134 5' 7" 127 83 6' 8" 169 48 5' 9" 145 18 5' 10" 147 8 5' 11" 104 3 6' 0" 66 1 6' 1" 22 6' 2" 11 6' 3" 3 6' 4" 1 WEIGHT (TO XKARKST I.VIKi.KK IN 5 OR 0) OK MEN ; AGKS 3/>-#J ; HEIGHT 5' lu" FREyfENCY OR Nl'MIIKR IN Qmuvr 125 4 130 14 135 33 140 60 145 78 150 114 155 95 160 106 165 90 170 87 175 72 180 59 185 48 190 37 195 25 200 32 205 12 210 14 215 4 220 8 225 3 230 1 In any such group the number of individuals having any given char- acteristic is called the frequency corresponding to the given characteristic. 6. Plot the frequency curve of heights of men and women, taking ^ inch as corresponding to 1 inch of height on the horizontal axis and taking ^ inch vertical for 30 individuals. This curve represents very nearly what is termed a normal symmetrical distribution. GRAPHICAL REPRESENTATION OF FUNCTIONS 65 7. Plot the frequency curve for weights of men between 35 and 39. AGRICULTURAL STATISTICS In the Yearbook of the Department of Agriculture statistics of produc- tion and prices of standard crops and farm products are given, covering a period frequently of 50 years. Use this Yearbook to obtain the data for the following curves : 8. Plot the curve showing the production of corn in the United States from 1866 to the present time. Use 200,000,000 bushels as a vertical unit, taking ^ inch as the unit ; take one year as y 1 ^ of an inch. 9. Plot prices on the diagram of 8, using a right-hand scale. 10. Plot similarly statistics for the amount and price of sugar produced in the United States. 11. Plot the average price in the United States of eggs by months for the current year ; plot butter prices similarly. POPULATION STATISTICS The population statistics of the United States by 10-year intervals as given by the 1914 Statistical Atlas of the U. S. Bureau of Census are as follows : DATE U.S. (Millions) NEW YOKK (Thousands) TEXAS (Thousands) 1790 3.9 340 1800 5.3 589 1810 7.2 959 1820 9.6 1,373 1830 12.9 1,919 1840 17.1 2,429 1850 23.2 3,097 213 1860 31.4 3,881 604 1870 38.6 4,383 819 1880 50.2 5,083 1,592 1890 63.0 6,003 2,236 1900 77.3 7,263 3,049 1910 101.1 9,114 3,897 66 UNIFIED MATHEMATICS 12. Plot the curve of population of the United States. 13. Plot the population curve for Michigan, and estimate the population for the 5-year periods. DATE MICHIGAN (Thousands) DATE NEW YOBK CITY (Thousands) DATE MlCHH. \ v (Thousands) XF.W YOKK CITY (Thousands) 1837 175 1790 49 1864 804 1800 79 1870 1,184 1 V 478 1810 120 1874 1,334 1820 152 1880 1,637 1,912 1830 242 1884 1,854 1890 2,094 2,507 1840 212 391 1894 2,242 1900 2,421 3,437 1845 303 1904 2,530 1910 2,810 4,769 1850 398 696 1854 507 1860 749 1,175 14. Plot the population curve of New York City. What has been the average rate of increase for 10-year intervals and for yearly intervals, approximately, since 1810? Note that this requires the solution of the equations 120(1 + /) 10 = 4769, and 120(1 -f *) 100 = 4769 ; solve by taking the logarithm of both sides. 15. Discuss the increase of the population of the United States from 1810 to 1910 as in problem 15 the population of New York City is discussed. 3. Graphical representation of algebraic functions. To repre- sent a point on a given line only one number is necessary with a point of reference and some unit of length. To every num- ber corresponds one point and only one and .conversely to every point corresponds one number and only one number. GRAPHICAL REPRESENTATION OF FUNCTIONS 67 The distance cut off from 0, the origin, by any point on this line may be called the abscissa of the point ; a moving point upon this line may be designated by the variable x, which is then thought of as assuming different values, corresponding to the different positions of the point upon the line. i i I i I I I _7 _e -5 -4 -3 -2 -1 I I 8 9 If another scalar line, OF, be taken intersecting OX at 90, the two lines may be conveniently used to represent the position of any point in the plane of the two lines. The two lines of reference are called commonly the osaxis and the y- axis respectively. The position of a point on the earth's surface is given by a pair of numbers representing in degrees longitude and lati- tude ; the + and of our numbers are replaced by E. and W. in longitude, and by N. and S. in latitude. If we agree to give longitude first, then + and could, in both terms, replace the letters, and position on the earth's surface of any point can be given by a pair of numbers. The system of represent- ing points in a plane is not essentially different. Given any point in the plane as P, a perpen- dicular is dropped to the horizontal line. The dis- tance cut off on this horizontal line is called the abscissa or cc-coordi- nate of P; the distance cut off on the vertical line OF by a perpen- dicular from P to F is called the ordinate or ^-coordinate of the point P and it is evidently equal to the _L PM dropped Location of points in a plane 68 UNIFIED MATHEMATICS to the axis OX. The two numbers together, abscissa given first, serve to locate the point ; thus a point PI 1 units to the right of Y and 5 units above OX is located on our diagram. To this point corresponds the pair of numbers (7, 5) (read " seven, five ") and to the pair of numbers (7, 5) corresponds point PI. The point P 4 symmetrical to PI with respect to OX, is (7, 5) the negative ordinate indicating that the point is below the #-axis. P 2 ( 7,5) and P 3 (7, 5) are located upon the diagram. A moving (or a variable) point P in the plane is designated by (x, y"), which is read "x, y" (not "x AND ?/"), and the coordinates, abscissa and ordinate, of P are a different pair of numbers for each position of the point, i.e. x and y are variable. Every point represents a pair of numbers, and consequently a series of points will represent a series of pairs of numbers. In the statistical diagrams the pairs of numbers are numbers functionally related. In an algebraic function, y 3 x + 5, we have involved a relationship corresponding to a mass of statistical information, and the pairs of numbers can be repre- sented upon a diagram just as before. Corresponding num- bers, a pair of numbers, are obtained by giving a value to x and computing the value of y. The points are seen to lie upon a straight line, which we shall see includes all points and only those points whose coordinates, abscissa and ordinate, when substituted for x and y, respectively, satisfy our given equation. This line is called the graph of the function, 3 #+5, or the locus of the equation, y = 3 x -f- 5 ; the operation of locating the points and connecting them is termed plotting the graph. To represent on cross-section paper any equation in two variables x and y, t and s, u and v, or by whatever letters designated, two intersecting scales as axes of reference OX and OF, OTand OS, or OC/"and OF are taken, and pairs of values which satisfy the functional relationship are plotted as above. GRAPHICAL REPRESENTATION OF FUNCTIONS 69 4. Historical note. The invention, or more properly the discovery, of analytical geometry was made in the early part of the seventeenth century. The first work directly on the subject was published by Rene Descartes in 1637, La G6om6- trie, a work small in compass but great in its effect upon the development of mathematics and science. Almost simultane- ously another Frenchman, Pierre Fermat, also discovered the methods independently of Descartes. The idea of coordinates, called Cartesian after Cartesius (Latin form of Descartes) was not new ; in fact, as we have noted, this idea is found in the latitude and longitude of Hip- parchus (200 B.C.). The idea of coordinates for drawing simi- lar figures was known even to the early Egyptians, and this idea was used for surveying purposes by Heron of Alexandria (c. 100 B.C.). The idea of fundamental properties of any curve as related to its axis or axes or to tangent lines and diameters was also not new. The new point was to combine these ideas, referring several curves and straight lines to axes geomet- rically independent of the curves, using letters to represent constant and variable distances associated with the curves and lines involved ; the graphical representation of negative quantities is a vital part of the analytical geometry. These developments were made both by Descartes and by Fermat. Modern mathematics begins with this analytical geometry and with the calculus which was developed within a century after Descartes by Newton and probably independently by Leibniz. 5. Industrial applications. At the present time the graphical representation of statistics is playing an increasingly important role in many industrial enterprises. Curves derived from ob- servations, empirical curves, are expressed in graphical form for convenience of reference and, frequently, for interpolation between observed values. The normal distribution curve is employed not only by statisticians but also in the production departments in many factories in the classification of their products. 70 UNIFIED MATHEMATICS ILLUSTRATIVE EXAMPLES To plot a function of x, give x values, find the corresponding values of y, or conversely, and plot the points. Connect by a smooth curve passing through all the points in succession moving continuously from left to right. 1. Plot the graph of the function of+ 4 x 5, i.e. plot the locus of the equation, y=x* -\-kx- 5. Give to x the values from to 3 and from to - 6 ; beyond these values in either direction the values of y evidently become very large. The curve is evi- dently symmetrical "with re- spect to a line parallel to the axis and 2 units to the left. The points where this curve crosses the z-axis rep- resent solutions of the equa- tion x 2 + 4 x 5 = 0. POINTS POINTS Graph of y x 2 + 2. Plots = 20 +50. -5 250 200 150 100 50 1 70 2| 90 3110 4130 5150 6170 7190 8210 9230 10 150 1 70 1.1 72 1.274 1.876 1.478 1.580 1.6 82 1.784 1.886 1.988 290 Upper graph, s = 20 1 + 50 from t = to t = 10, upper and left-hand scales Lower graph, s = 20 1 + 50 from t = 1 to t 2, lower and right-hand scales GRAPHICAL REPRESENTATION OF FUNCTIONS 71 The lower and right-hand scales would be used if you were interested in the behavior of the function in the interval from t = 1 to t = 2. By the tenfold enlargement you can read values to the third significant figure. This may represent the motion of a body which starting at a point 50 feet from the given point of reference moves away from that point in a straight line at the rate of 20 feet per second. The units might be miles and hours, so that the speed would be given as 20 miles per hour ; this may represent then the motion of a train. 3. Plot y = a? 2& 18a;+24. The values of y are so large that the figure occupies too much space vertically. To obviate this difficulty one square on the axis of y is taken to represent ten units of y and one square on the x-axis is taken to - lKi>44) r |::iiE|;i|!|;i|j|y:j|E|;ii| POINTS ON THE __ ::::::::::: CURVE S"/ -- ^80 :.:.. 4 g 1 '/I/ I I V 3 Y - \\- . ':.:_./;:::|:_::: ^ -1 ~f ? Y \ \ l : /:> t " ~/|:: -- : sl % .-T--'- i / 10- 3 ^/ c~ : ~~:l 2 ' 7; , / >n - :_.5t_,z: ::::: ::.:: .:::: ::.:: 3 f (^ V' 'i /I /. . c u_ ::;;: fc: --- -30 1 :::: 5 1J_LJ A Illl Graph of y = r - 2 x 2 - 18 x + 24 represent one unit of x. This serves to compress or telescope the curve, but the essential peculiarities are preserved. In particular the points at which the curve crosses the x-axis, the values of x which make x 3 2x 2 18 x + 24 = 0, remain unchanged. These values, the roots of x 3 2X 2 18 x + 24 = 0, are seen to be 4, 1.3, approximately, and 4.8 approximately. In general an algebraic equation of this type is not likely to have a rational root, such as the 4 above. 72 UNIFIED MATHEMATICS 4. Plot a- 2 + y 2 - 36 = 0. Graph of x- + y- 36 = In drawing the graph of this function of x (implicit), it is important to note that there are two values of y corresponding to each value of x, and that these two values are symmetrically distributed with respect to the x-axis. Similarly this curve is symmetrical with respect to the ar-axis, since any value of y gives two corresponding values of x, numerically equal but opposite in algebraic sign. The points when located are con- nected by a smooth curve which is here a circle. To this diagram reference has been made in problem 7, page 17. As a circle of radius 6 the ordinates at x = 1, 2, 3, 4, and 5, respectively, give graphically the square roots of 35, 32, 27, 20, and 11. The more complete discussion of equations of this type is given in Chapter XIV. GRAPHICAL REPRESENTATION OF FUNCTIONS 73 5. Plot y = x z for a? = 0, .1, .2, , 1.0, 1.1, 1.2, 1.3. Note that precisely the same curve is obtained if units 0, 1, 2, 3, 15 are taken instead of tenths on the x-axis and tens in the place of tenths on the vertical axis, as indicated on the lower and right-hand scales. POINl THEC I.E HAND, UPI SCA X -.2 1 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 ir__ 8 ON THE CURVE, RIGHT- _ II , ' i / : HAND, AND ii;"i :::_t-in- LOWER : 12::: SCALES IIIIIIIIIII : ~~~~,~ nit _ / ti v :--- : x y ...... :x.-_i-oo- .04 - 2 4 : / 111' 01 " " i / - -H)0- :: - r .\ \ \- m S *-y ---80- .01 .1 1 i / .04 ;i AQ 2 4 I::::::::::::::::::: .16 -Q 4 16 : iri: ^50- .25 : 5 25 " (1, 8), (2, ^), (13, 0), (-'/, 1), (-1, - 2 3 8 -), ~. By substi- tuting or 1 or 2 or 1, 2, , for x and solving for y, or conversely, points are obtained whose coor- dinates satisfy the given equation ; similarly every point on the second line is such that its coordinates satisfy the equation, Graphs -^ y w w the point of intersection satisfies both equations, and its coordinates can be obtained by solving the two equations as simultaneous. The argument is THE LINEAR AND QUADRATIC FUNCTIONS 81 entirely similar for the points of intersection of any two loci, representing algebraic equations ; the points of intersection satisfy both equations, and give a graphical method of ap- proximating the solutions of the equations regarded as simul- taneous. To review this demonstration, answer the questions below, and read the discussion. What is true concerning the coordinates of every point on the first line ? on the second line ? What is true concerning the point of intersection so far as the two given equations are concerned? The drawing shows that ( 11, 16) satisfies both the equations, and substitution shows that this is precisely correct. In general the graphical solution is only approxi- mate, the degree of accuracy depending upon the accuracy of the drawing and the scale used. The point of intersection of two straight lines represents graph- ically the solution obtained by solving the two equations as simultaneous. Graphs Intersections of y = 3 x, and x* + y 2 = 25. The graphical presentation shows very plainly that the solution is, approximately, a; = 1.6 and y = 4.8. 82 UNIFIED MATHEMATICS -1-0 Intersections of y = x 2 , y = 3 x + 5. The graph shows that there are two solutions ; in the one, a? = -1.2, y = + 1.4, and in the other, a; = 4.2, y = 17.8. These are approximate values. Plot carefully the graphs of the preceding problems, check- ing on the work presented by the graphs. Plot carefully these two lines and verify the statements made: 2 x + 3 y 26 = 0, | Graphically, parallel ; 2x-\-3y 8 = 0. j algebraically, no solution. The point of intersection of two graphs represents graphically the solution of the two equations regarded as simultaneous. 3. Intercepts. Any given line or curve cuts off on the co- ordinate axes distances that are called the intercepts of the line or curve. The ^intercept is obtained analytically by sub- THE LINEAR AND QUADRATIC FUNCTIONS 83 stituting y = and solving, i.e. by solving as simultaneous the equations of the -axis and the given line ; the ^-intercept is obtained by substituting x = 0. The ^-intercept of 2 x + 3 y 26 = is 13, obtaining by substituting y = in 2 x + 3y 26 = 0; the ^-intercept is + - 2 ^ 6 ; of a 2 + y 2 = 25, the cc-intercepts are 5, the y-inter- cepts are also 5. Note that the problem of finding the intercepts of a given graph is a special case of the problem to find the intersections of two given curves ; the OS-intercept designates the inter- section of the given curve with the a-axis, y = 0, and similarly the ^-intercept refers to the intersection with x = 0. Rule. To find the ^intercept, put y = 0, and solve ; similarly for the ^-intercept. 4. Pencil of lines. The straight lines which pass through a common point constitute what is termed a pencil of lines. If the common point is determined as the intersection of two given lines, we may write the equation of the pencil of lines in terms of the two expressions which put equal to zero repre- sent the given lines. The pencil of lines through the intersection of y-3x-5=0 (IJ 3y + 2aj + 7 = (7 2 ) is given by the linear equation, A; being assumed constant, (3) y-3x-5 + k(3y + 2x + 7) = 0. (/ 3 ) Evidently any point on the first line, ? makes y 3x 5 = 0, and any point on the second line, Z 2 , makes 3y-{-2x-\-7 = 0; the point of intersection substituted in our equation (3) gives + k or 0, hence the point of intersection of l t and 1 2 satisfies equation (3) for all values of k. By giving k successive values 1 3 can be made to pass through any point of the plane. Thus to pass through (1, 5) sub- stitute (1, 5) in 1 3 and solve for k, giving 5 _ 3 - 5 + k(15 +2 + 7) = 0, 84 UNIFIED MATHEMATICS or 24 k = 3, fc = . The line y 3ar 5 + |(3y+2* + 7) = 0, or 8#-24 a-40+ 3 y +2 x+7 = reduces to lly-22x 33=0, or y 2 x 3 = when simplified. In solving as simultaneous the two equations y 3x 5 = 0, and 3y + 2a; + 7 = 0, the particular lines parallel to the axes of reference and passing through the point of intersection of /! and Z 2 are sought. Thus, after multiplying the upper expression by 3 and adding, you get the line / 3 with k = ^. gives 11 x 22 = 0, or x = 2. To eliminate x we multiply the upper expression by 2 and the lower by 3 and add ; this gives, 11 y + 11 = 0, or y = 1. The same line given by 11 y + 11 = is obtained from line 1$ with k = | ; i.e. : y-3x-5 + %(3y + 2x + 7) = gives 11 y + 11 = 0, or y + 1 = 0. The point of intersection of the two lines, (2, 1), is given as the intersection of x = 2 and y = 1. PROBLEMS 1. Solve y-3x 5 = 0, 3y + 2x 7 = 0, both graphically and algebraically. 2. Plot the graphs of x + y 2 = 0. Do these three lines appear to meet in one point on your diagram? Have these three equations a common solution? Substitute the solution of the first pair (obtained in problem 1) in the third equation. Later it will be shown that a point whose coordinates when substituted in a first-degree expression give a small numerical value is near the straight line repre- THE LINEAR AND QUADRATIC FUNCTIONS 85 sented by the equation formed by putting that expression equal to zero. 3. Plot 15 points whose coordinates satisfy the equation 4. Plot the lines x = 3 and y = 4 ; what point is repre- sented by these equations? Note that the Cartesian system (x, y) of representing points implies each point as the inter- section of two lines. 5. Solve both graphically and algebraically. 6. The weight of a cylindrical vessel of water when filled to a height of 10 inches is 6.8 pounds, when filled to a height of 6 inches it is 4.4 pounds ; plot the two points (6, 4.4) and (10, 6.8). The straight line joining these two points gives the weight of the vessel when filled to any height from to 10. The equation may be written iv = k h -f c, where w and h are the variable weight and height, k and c are constants. This equation is the simple statement of the fact that the weight of the water and the container for any height h is the weight of the vessel (c) plus h, the height, times the weight of the water which fills the container to a height of one inch. Note the sig- nificance of the intercepts. pn \ 7. The equation, w = ^ v, may be used to express the relation between the volume in cubic inches and the weight in pounds of a given mass of water.. Plot this carefully and find approximately the weights of 100 cubic inches, 500 cubic inches, and 700 cubic inches of water. Find the volume of 15 pounds of water ; the volume of 25 pounds ; of 30 pounds. 8. The volume of mercury at any temperature between and 40 C. is given by the equation V= fc(l -f- at), wherein a = .00018 ; for A; =1000 cu. cm. this becomes V= 1000 + .18 t. 86 UNIFIED MATHEMATICS Plot this equation taking the horizontal axis as at 1000. This is equivalent to plotting the increase in volume, /=.18<. Plot for to 40 C. and find the increase in volume when 1000 cu. cm. of mercury at C. are heated to 27 C. 9. Find the equation of the straight line through (3, 5) and through the intersection of 3x y 7 = and 5 x + 12 y - 17=0. 10. Plot degrees Fahrenheit as abscissas and degrees Centi- grade as ordinates, connecting (32 F., C.) to (212 F., 100 C.), by a straight line. Find the equation of this straight line. Find the Centigrade reading corresponding to Fahrenheit, to 100 F. Discuss the meaning of the slope of the line. 11. Find the intercepts of the line 9 y 5 x = 160. Com- pare with your result in the preceding problem. 12. Plot the graph of s = 16 t z , for values of t from t = to t = 5, using one inch for 1 second on the horizontal axis, and 1 inch for 100 feet on the vertical axis. Find value of s when t = 4.3 from the graph. Check by computation. 13. Plot carefully x 2 +y z = 64, and y = 3 x 5. From the graph get the approximate solution. 14. Show graphically how to change a system of marks from a scale of 100 to a scale of 75 ; from 75 to 100. 15. Sound travels at the rate of 1089 feet per second in air at 32 F. (or C.) ; at the rate of 1130 feet per second in air 13 t at 70 F. The formula, v = 1054 +^ gives very closely the \a velocity in feet per second at temperature t Fahrenheit. Plot the graph of the function, plotting the excess above 1000 feet as ordinates and temperature Fahrenheit up to 80 F. as abscissas. At what temperature is the velocity 1100 feet per second? How would you adapt these figures to the Centigrade scale for temperature beginning C.? v = 1089 + 2 t is the resulting equation. THE LINEAR AND QUADRATIC FUNCTIONS 87 16. The velocity after t seconds of a bullet shot straight upwards at 800 feet per second is given by the equation v = 800 - 32 t. Plot the graph, taking 100 feet as | inch on the vertical axis, and 5 seconds as inch on the horizontal axis ; negative values of v mean that the bullet is descending. 17. Plot v = 600 -f 32 1, and interpret as downward velocity of an object thrown downwards from a height. 18. Time yourself on plotting the following 10 lines ; five may be plotted with respect to one set of axes : a. __ b. 3y = 2x-5. ' 3 5 c . X -y-S = Q. h. -axis, y = 0, in the two points whose abscissas satisfy the equation, 2 # 2 -f- 8 #+ 7=0. y=2.T 2 +8#+8 represents a curve which is tangent to the o^axis, correspond- ing to the fact that the roots of the equation, 2x* + 8o; + 8 = 0, are equal to each other. The equation y = 2a^ + 8x + ll repre- sents a curve which does not cut the cc-axis, corresponding to the fact that the quadratic 2x 2 + 8x-|-ll = has for solu- tions, x = , values corresponding to no points on Z the x-axis, i. e., to imaginary values of x. Plot the graphs in- dicated. THE LINEAR AND QUADRATIC FUNCTIONS 89 The quadratic equation is solved algebraically by reducing the problem to the solution of two first-degree equations : b -f- V& 2 4 ac 2~a~ 2a ' . 2a 2a The quantity & 2 4 ac which appears under the radical sign is called the discriminant of the quadratic. The nature of the roots of the quadratic equation is determined by this dis- criminant, when a, b, c represent real quantities, i.e., a, b, and c having values which can be represented by points upon a scalar line. When b 2 4 ac > 0, i.e. positive, the two roots are real and un- equal, when 6 2 4 ac = 0, the roots are real and equal, and when b- 4 ac < 0, i.e. negative, the roots are imaginary. Further, the condition that the roots of the quadratic should be equal given by b- 4 ac = 0, may be obtained by inspection, or by actually setting the two roots equal to each other and simplifying; ax 2 + bx + c may then be written a(x+- ) V 2 a,/ Graphically these conditions correspond to the fact that the curve y = ax 2 + bx + c cuts the co-axis in two points, is tan- gent to the ;K-axis, or does not intersect it at all, according as b 2 4 ac is greater than, equal to, or less than 0. Frequently the two roots of the quadratic ax 2 + bx + c = are designated by x t and x 2 . Thus x, = -6+V6 2 2a , - b V6 2 4 ac and x 2 = . 2a The sum and the product of the roots, x l + x 2 and XiX 2 , are b c given, respectively, by and + - . The expressions o^ -f x. 2 a a and x&z are representative symmetric functions of the roots of 90 UNIFIED MATHEMATICS a quadratic function of one variable, being expressions which re- main unchanged when a^ and x 2 are interchanged. 6. Historical note. The solution of linear equations was known four thousand years ago to ancient Egyptians. The equation x + - = 19, was proposed and solved in the work of an Egyptian writer named Ahrues ; the problem reads, with " ahau " representing " heap " or " unknown," " ahau and its seventh, it makes 19." In other ancient Egyptian documents problems leading to pure quadratics are found. The Greeks were able to give as early as 450 B.C. a geometrical solution of any quad- ratic having positive roots ; the numerical application appears in Greece somewhat later. In India numerical quadratics were solved in the fifth and sixth centuries A.D. The first systematic treatise combining clearly analytical statement with geometrical illustration is given by an Arab, Mohammed ibn Musa al-Khowarizmi, about 825 A.D. His work continued in use for centuries. The complete quadratic with general, literal coefficients, did not come, of course, until after the in- troduction of literal coefficients by Viete late in the sixteenth century. 7. Graphical solution of the general quadratic equation. The general quadratic equation ax 2 + bx + c = can be solved graphically by means of one fixed curved line, y =x 2 , and a variable straight line. The intersection of y = x* and ay -f bx + c = gives the solution of the equation ax* + bx + c = 0, for the solution is obtained algebraically by substituting for y its value x 2 in ay + bx + c = 0, giving ax 2 -j- bx 4- c = 0. The graphical solution of the quadratics, 2 # 2 6x 5 = 0, 2x z -6x = 0, 2x z - 6x + f = 0, and 2x 2 6x+ 10 = 0, is presented upon the diagram ; the student is urged to solve these equations algebraically and to trace the correspond- THE LINEAR AND QUADRATIC FUNCTIONS 91 Graphical solution of quadratics 2x 2 6x-5 = 0; 2x 2 6x = 0; 2x 2 -6x + f = 0; 2x 2 -6x + 10 = 0. Two real solutions. 2y-6z + f = 0; Two coincident solutions. Two real solutions. 4y- 6x +10 = 0. Two imaginary solutions. 92 UNIFIED MATHEMATICS ence between the algebraic and graphical solutions. Two sets of real and different roots are indicated by two of these straight lines on our diagram ; one set of equal roots is in- dicated ; one pair of imaginary roots is indicated by the line which does not meet the curve. The graphs of the corresponding functions, y = 2x z 6x 5, y = 2 & 6 x, etc. should also be drawn. PROBLEMS 1. Plot the graph of y = 3x 7; give to x the integral values from 2 to -f- 5 and find the values corresponding of y. 2. A freely falling body falls from rest in t seconds a dis- tance s, given by s = 16 1 2 ; plot points given by corresponding values, using horizontal axis as -axis, and vertical axis for distance. Take values of t from to 10, and as s will vary from to 1600 take 1 cm. to represent 100 on the s-axis. 3. The simple interest on $ 100 for n years at 5 % is given by 1= 5n ; plot n on the horizontal axis and / on the vertical. On the same axes plot A = 100 + 5 n where A is the amount at the end of n years, plotting A on the vertical axis. On the same axes plot A= 100(1 -f .05)" for = 1 to 10, finding the values of (1.05) 2 , (1.05) 3 ... by logarithms ; this gives the amount at compound interest, compounded annually. Check by the table at the end of the book. 4. Take problem 3, using 4 % as the interest rate and 6 % as the interest rate ; take the values of (1.04) n and (1.06)" from the tables. 5. Take the data of problem 3, using 6 % simple interest paid semiannually, and 6 f interest compounded semi- annually. 6. The velocity of a freely falling body is given by the formula, v = 32 1, when falling from rest ; or v = 32 t + k, where k represents the velocity at the instant when t = 0, or k THE LINEAR AND QUADRATIC FUNCTIONS 93 is the velocity at the instant when you begin to measure the time. Plot for values of t from to 10. 7. A bullet shot straight up into the air at a velocity of 1000 feet per second, has its height above the earth given by the equation h = 1000 1 16 1*. Plot this equation for values of t increasing by intervals of 5 seconds from t = to t = 100. If the bullet is shot at an angle in such a way that the vertical velocity when leaving the gun is 1000 feet per second, the given equation continues to hold for the height of the bullet above the earth. The resistance of the air (considerable at the velocity mentioned) is neglected in these equations. NOTE ox NUMERICAL APPROXIMATIONS. In several of the problems below, as well as in some of the preceding problems like the fifteenth of the preceding set, numerical approximations are given involving simple fractional expressions as substitutes for decimal values. The method of making this kind of substitution which is frequently of use involves simply the application of addition and subtraction to aliquot parts of 100, rejecting, with discretion, places which are not necessary to attain the degree of accuracy warranted by the data of the problem. | = .33333, | of ^ = .033333, etc. I = .25, J of ^ = .025, etc. = .16666, I of ^ = .016666, etc. j = .50, Jof j^=.05, etc. | = .125, | = .875, J of any number is usually obtained by subtracting J of the number from the number. \ = .666666 ; f = .75. Thus, .365 = | + ^ - ,$ ff ; .1918 = J- + jfr + T^nr 5 the error & about 'As of * % .1492 = ^ + A TtfW 5 tne error is at)0ut } of 1 %. These numbers are chosen at random. 8. The volume in gallons of a cylindrical container, meas- ured in inches, is given by V = ^ . Replace - 7r ' 4 X AOL 4 X 2ioL by its approximate value, .003399, or \ of 1 % of cPh +2% of this result, or (1.02) x ^ % of cPh ; i.e. use the formula F=1.02 x(.00)d 2 /i. Plot values from d = 1 to d = 20, for h =10. 94 UNIFIED MATHEMATICS 9. The volume in barrels per foot of height of a cylindrical cistern is given by V= X , in which 7.48 is the number 4 31.5 of gallons per cubic foot and 31.5 is the number of gallons per barrel ; d is to be measured in feet. Use for - - -^ - the 4 x 31.5 value + A plotting then F = a+ T fo)d*, for d=l to 10. Plot the number of barrels per foot of height of a square tank using the formula F = ^ , computing the 7 48 quotient - - by logarithms and using this multiplier for 31.5 values of d from 1 to 20. Choose appropriate scale to get the data on paper. 11. Given h = 800 1 16 P, find t when h = 100, 1000, 10,000, 12,000 respectively. This equation represents the height to which a bullet would rise when shot vertically up- wards at a velocity of 800 feet per second, neglecting air- resistance. Interpret your results. This bullet has a velocity at time t, v = 800 32 1 ; find the velocity at the various heights mentioned. 12. Solve 16 1- 800 1 + 7i = for t, regarding h as a con- stant. See preceding problem and find maximum value h can have. 13. In solving 16 1 2 800 1 + h = 0, two roots are obtained ; find the sum of these roots and the product. Interpret the sum, i.e. give the physical meaning. 14. Find the nature of the roots, without completely solving, in the following equations : a. xi + 3x-5 = 0. c. x 2 + 3x - 40= 0. 5. x * _|_ 3 x - 8 = 0. d x* - 3 x + 40 = 0. e. x 2 -3z + f = 0. /. 4 x 2 -12 x +9 = 0. THE LINEAR AND QUADRATIC FUNCTIONS 95 15. Determine the nature of the roots : a. 4* 2 - 16* -160 = 0. c. 3 y 2 + 16u + 20 = 0. b. 7P + 16- 160=0. d. 3v 2 + 16v + 25 = 0. 16. Plot the graphs of the functions in 15. 17. Solve the equations of 15 graphically, using the inter- section with y = t 2 or y = y 2 (one half-inch may be taken for 10 units on the vertical axis). 18. Find the sum and the product of the roots in the problems of 14 and 15. f x 2 -\- w 2 = 36 19. Solve { ' by substitution. g __ 20. Solve I _ 'by substitution. Draw graphs. I s == ot o, 21. Solve .1 t 2 50 1 30 = 0, to 2 places of decimals. 22. Solve t 2 50 1 .0001 = to 2 places of decimals. 23. Time yourself in solving the following 10 quadratics, writing the roots in simplest form but not approximating the square root. a. 2 x 2 + 3 x- 5 = 0. /. 9 x 2 = 20 x + 10. b. 3z 2 -2z + 7 = 0. gr. c. 5 ; y 2 + 12z + 3 = 0. h. d. ?/ 2 -3 y- 7 = 0. t. e. 2v 2 24. Time yourself in finding to one decimal place the roots in the above 10 equations. 8. Equations reducible to quadratics. The solution of ax 2 + bx + c = is a value of the variable x, which when it is substituted in cu 2 -\-bx-\-c, makes the expression ; similarly this solution gives a value of the variable v, or t, or p, or 2 , or f 2 , or 3 t z 1, or 7 t 2 + 2 t 3, which makes the expression of the same form 96 UNIFIED MATHEMATICS in that variable zero ; viz., a value which makes av* -f bv 4- c equal zero when the value is put for v, or a(t 2 ) 2 + bt 2 -+- c, equal zero when the value is put for t 2 , or a(3 t z I) 2 + b(3 P 1) + c equal zero when the value is put for 3 1 2 1. Equations which can be put in the form cue 2 + bx + c = are called equations in quadratic form, the term being applied, in gen- eral, to expressions which are not quadratics directly in the principal variable. Thus, in any expression involving x, x 2 , x 3 , x*, x b , or x 6 , the value of the expression depends primarily upon the principal variable, x ; an expression like 3 x 4 2 x 2 7, involving the variable x 2 , its square, and constants as coeffi- cients, is said to be in quadratic form, and it is a quadratic in the variable x 2 , but a quartic in x. 9. Illustrative exercises. 1. Solve 3 t* - 5 t 2 - 7 = 0. As a quadratic in t 2 , the formula for the solution of a quadratic gives : = 5V109, whence 6 6 There are four values represented here, of which two are imaginary. 2. Solve x 3 = 1, or x 3 1 = 0, and x 3 8 = ; these illus- trate a type of equation reducible to a quadratic by factoring. x - 1 = (z - 1) (a? + x + 1) = 0. x - 1 = 0, x - 1 3 2 These values ^ , ^ and 1 are called the cube 2 2 roots of unity ; note that V 3 is denned as a quantity whose square is 3 ; the systematic discussion of such numbers is deferred until a later chapter. Squaring either of the two imaginary cube roots of unity gives the other ; these roots may then be designated as 1, w, w 2 . The cube roots of 8 are 2, 2 w, and 2 w 2 ; of 7 are 7*, 7^ w?, and 7^ w> 2 , wherein 7* denotes the real cube root of 7. THE LINEAR AND QUADRATIC FUNCTIONS 97 PROBLEMS 1. Solve for t\ and then for t. t 6 1 ? 8 = 0. 2. Solve and check by substitution : or 4 + 3 x~ 2 - 5 = 0. 3. 2x*-7x*-5 = 0. 4. Note that this expression when cleared of fractions gives x 4 + xs + x 2 + a; + 1 = 0, a factor of x 5 1 = ; the imaginary roots of x which are obtained by solving are the other four fifth-roots of unity. 5. (3 x2 _ S) 2 + 2(3 x 2 - 5) - 7 = 0. 6. v + v* = 10. 7 ' 8. Find the value of x?, and of x in a? + 3 x* - 7 = 0. 9. Find the value of x 2 , in x _ 3 ar* - 7 = 0, and compare with the preceding. The real test of a value found as a root is obtained by substituting the value in the given expressions. Squaring may introduce a new root ; thus squaring x = 2, gives x 2 = 4, or is equivalent to multiplying x 2 = 0, member by member by x -f 2. 10. 3 x + Vx + 5 = 7. 11. 3 x - Vx + 5 = 7. 10. Limiting values of a, b, c. As c approaches more and more nearly to zero as compared with, a and b, it is evident that some value of x also near to zero will satisfy the equation ax 2 + bx + c =0 ; this value will be of the same sign as c if 6 is negative, and opposite in sign to c if b is positive. Thus 98 UNIFIED MATHEMATICS 3x* 2x .000001 = is an equation which will be satisfied by a value of x very near to .0000005 ; substituting the value, -.0000005 we have .00000000000075 +.000001 - .000001 which is surely near to zero ; even if a were very large compared with b, this expression has one root near to zero. Solving 3 a; 2 -2 a -.000001 = 0, _ 2 V4 + .000012 ~6~ 2 2.000003 4.000003 .000003 = = or 666 = | or - .0000005. Similarly in 1000 x 2 3000 x 1 = 0, one solution will be small, approximately 3 -^ . When c = 0, the roots of ace 2 + bx+ c = are the roots of ax 2 + bx = 0, giving a(acc + 6)=0; whence #=0 and x = . When both b and c approach zero, both a roots of the quadratic ax 2 + bx + c = approach zero. When a approaches zero in comparison with b and c, one root of the quadratic becomes very large and the other ap- proaches -. Thus in the quadratic b x>- 1000 x -3000 = 0, x = 1000 VlOOOOOO + 12000 = 1000 1006 = 1003 or _ 3 2 2 /1 000 1005.982 - - are the more exact values, giving 1002.991 V 2 or - 2.991.") As a approaches nearer and nearer to zero one root becomes larger and larger without limit. Thus if above we had .001 x 2 - 1000 x -3000 = _ 1000 VlOOOOOO + 12 _ 1000 1000.006 .002 .002 = 1000001.5 or - 3 (more exactly 2.991 as before). THE LINEAR AND QUADRATIC FUNCTIONS 99 Both roots become large if both a and b become small as com- pared with c. PROBLEMS Find first approximate values, and verify by solving the quadratic : 1. 3x 2 - 7x -.0001=0. 2. 5x z 7x .1 =0. 3. 5 x 2 .007 x - .001 = 0. 4. 4000 = 3000 1 - 16 1 2 ; one root is the number of seconds for a bullet to rise 4000 feet, initial velocity 3000 feet per second, air resistance neglected ; what does the other root represent ? 5. .01 x*- - 300 x - 500 = 0. 6. .003 1 2 + 2 1 42 = ; this gives a more exact equation for the temperature at which the velocity of sound in air becomes 40 feet greater than it is at C. 7. 1000000 x 2 - 3000000 x - 5 = 0. REVIEW PROBLEMS 1. Plot the graph of y = 3 x 5. 2. Plot the graph of the following functions : a. y = x 2 4 x + 5. b. y = x 2 4 x + 4. c. y = x z 4 x. d. ?/ = a; 2 4 a; 2. 3. For what values of x is y equal to in the four functions of the preceding question ? The graphical solution is desired. 4. Plot 15 points from x = .5 to x= +8 and join by a smooth curve representing for what values of x is y equal to zero ? 100 UNIFIED MATHEMATICS 5. s = 20 t,+ 50 16 1 2 . This equation represents the mo- tion of a body thrown from a height of 50 feet straight up into the air with a velocity of 20 feet per second. Plot the graph and locate the position of the body at the end of 1 second ; at the end of 5 seconds. 6. Plot the graph of s = 800 1 16 1 2 , for values of t from to 50 ; note that it is desirable to get the values of s first for intermediate values and to choose the y-scale accordingly. This equation represents approximately the height after t sec- onds of a bullet shot straight into the air with a velocity of 800 feet per second. 7. Plot the graph of x-axis and 7/-axis off the paper -2t -140 v= between d = 12 and Shifted lines of reference d = 20, taking the scales so as to enable you to read vol- umes as correctly as possible within these limits. Plot only values above 100 on the y-scale, and to the right of 12 on the x-scale. This gives the volume in cubic units per unit of height of cylindrical containers which have radii varying from 12 to 20 units. Apply this to cans and to silos. 8. Plot the graph of t? = ^ ; this gives the time of beat of a pendulum I centimeters long where gravity is 980 cm. per sec. per sec. 9. Plot the graph of y = as*, for values of x from to 8. 10. Plot the graphs of the following linear functions : a. y = 3 x 5. c. v = 10 + 8 1. e. s = 100 40 1. b. y = 3 x. d. s = 6 - 3 1. f. y = 2 x + 10. .CHAPTER VI STRAIGHT LINE AND TWO-POINT FORMULAS 1. Slope-intercept formula : y = mx + k. The equation y = mx + k, into which form the equation of any straight line can be put, is called the slope-intercept form of the equation of a line ; m represents the slope of the line and k is the intercept on the y-axis. The equation of a line parallel to the ?/-axis, x = k, cannot be placed precisely in this form, as the ^-intercept is infinite. 2. Point-slope formula : y y\ = m(x x^. As it is frequently desired to find the equation of a line of given slope and passing through a given point, a separate equation in terms of the slope and coordinates of the given point is desirable. Let the equation of the line be conceived as in the form, y mx + k ; since (x l} y^) is on the line, y l = mx! -f- k ; subtracting gives y y v = m(x o^), the equa- tion of the straight line in terms of m, the given slope, and (#!> y\) the coordinates of the given point. 3. Two-point formula: = y2 ~ Vl X X Xn X\ The equation of the straight line through (x 1} yi)(x 2 , ?/ 2 ) is also easily derived from the slope-intercept form. As before ?/! = mo^ + k, yi = mx 2 + k, whence y 2 y 1 = m(x 2 o^), and m = ^ 2 ~ -^ , giving m, the slope X 2 -Xi of the line, in terms of a/ 1} y 1} x 2 , and y 2 . 101 102 UNIFIED MATHEMATICS Hence, y y l = ^ 2 ~ ^ 1 (x a^) is the equation of the line x% Xi in a form involving only the given constants. The expression, m = ** ~~ * l , represents the slope of a line joining (xi, y^) to (x z , y z ). Similarly - represents the slope of the line joining any point (a;, y) to (x t , yi). The preceding equation of the line in the form - = & is an equality of two slopes. l The formula m = ^ 2 ~ ^ 1 is frequently used ; it should be memorized with the aid of the diagram as placed in quadrant I. This formula gives the rate of increase of y in the interval from (a^, y) to (a^, t/ 2 ) as compared with the increase of x in the same interval ; it compares the change in y in the interval with the change in x in the same interval. PROBLEMS 1. Find the equation of the line of slope 3 and y-intercept 5 ; with m = 3, k = 5 ; m = 3, fc = 8 ; m = 0, fc = 4 ; ra = 5, 2. Put the following equations into slope-intercept form : a. 3y 2 a; + 5 = 0. d. y 3x 7 = 0. b. 3x + 2y-7 = Q. e. y + 5 = 0. c. x + 2 y = 0. /. x + 3 = 0. 3. Write the equation of the straight line through ( 2, 5) and (1, 4) ; through (3, 5) and (2, 1). Find intercepts on both axes and the slope in each case. 4. Write the equation of the straight line through (1, 5) having the slope 3. Find the x and y intercepts. 5. Find the equation of the straight line through (a, 0) and (0, 6), i.e. the line having intercepts a and 6, respectively, STRAIGHT LINE AND TWO-POINT FORMULAS 103 /j AI and put this equation into the form - + " = 1. This is called a b the intercept form of the equation of a straight line. 6. Given 9 C = 5 F 160, the formula connecting centi- grade and Fahrenheit readings of temperature, find the slope and the x and y intercepts. Find the slope of the line join- ing (32, 0) to (212, 100). What is the rate of change of C in the interval as compared with the change in F? What physi- cal meaning have the intercepts ? 7. Given that 1000 cu. cm. of mercury at C. increases to 1007.2 cu. cm. at 40 C., find the rate of change of volume per degree of temperature, and finally per cu. cm. Note that it is not necessarily true that this rate found for an interval of 40 C. should be the uniform rate everywhere in the interval. Write the equation representing the volume in terms of temperature, assuming that the relation is linear, i.e. that the increase in volume is proportional to the temperature. Mercury expands differently at different temperatures, but the variation is slight in the interval from to 40, not varying by more than | of 1 % from .00018 cu. cm. per degree for 1 cu. cm. 8. Join (0, 0) to (100, 39.37) and interpret for converting centimeters to inches and inches to centimeters ; what is the meaning of the slope? Find the value in inches of 18 cm., 39 cm., 47 cm. Note that 100 cm. = 39.37 inches. 9. 59.8 pints of water weigh approximately 62.4 Ib. Draw the graph connecting (0, 0) to (59.8, 62.4) which will give the approximate weight of any given number of pints of water. How could you read the weight of quarts or gallons? Use 1 inch for 10 units on both scales, in plotting. 10. Find the equations of the straight lines joining the following pairs of points, timing yourself : a. (3, 5) to (- 2, 7). e. (0, 8) to (0, 5). 6. (3, 5) to (2, -7). /. (1, - 3) to (- 1, - 5). c. (0, 8) to (7, 0). g. (1, - 3) to (1, 6). d. (0, 8) to (7, - 6). h. (-1, -3) to (-3, -5). 104 UNIFIED MATHEMATICS i. (-1, -3) to (3, 3). j. (8, - 3) to (- 3, 2). k. (-3,5) to (7,0). I (-3, 5) to (-7, -2). m. (- 3, 5) to (7, - 2). (2, 2) to (-2, -2). o. (100, 60) to (0, 0). n. 11. Find the equations of the following lines : a. of slope 3, y-intercept 5. b. of slope 3, ^-intercept 5. c. of slope 2, y-intercept 0. d. of slope 1, through (2, 2). e. of slope + 4, through (2, 2). /. having intercepts of 7 and 5 on the x- and r/-axes respectively. o = y z y\, 4. Distance between two points : d = V(x z j^) 2 + (y 2 Since Whatever the positions of P l and P 2 , parallels to the x- and y-axes through P l and P 2 form a rectangle (a straight line if *i/i == 3/*> or Vj ^s v->) WIIOSG sides are in absolute value | x 2 x l | and | y 2 y l \ ; the bars indicate that only the numerical value is considered. As a posi- tive distance P\M^ if Xi > x 2 , would have to be written x v x 2 But since the numerical value of the expression (x 2 o^) 2 is the same as Distance between two points d* = (zj - a*)' + (y* - 2/i) 2 - the value of (a^ a; 2 ) 2 we may use in every case (x 2 x^) for in the above expression for d wherein only the square of enters. d = STRAIGHT LINE AND TWO-POINT FORMULAS 105 The distance from any point (x l} y t ) to any point (0%, y 2 ) is given by this formula ; this distance is taken in general as a positive quantity. d This formula may be used to derive the equation of the straight line joining PI(XI, y\~) to P 2 (x. 2 , y. 2 ) for any point P(x, y) on the line is such that PP V + P\Pz PPz 5 and for no point not on the line is this relation true. 5. Point of division formula : _ *iX 2 4- Mi . . . _ *u 5/2 + fc^ 4 ~~ til, ' y ' A ~ K!' + Kn i fcl+*2 A A A A II > II P 3 r* 1*3 A 1 > For any three points P,, P 2 , and P 3 on a directed line we have P X P 2 4- P 2 P 3 = PiP 9 5 if P lies between P l and P 3 , all three segments have the same algebraic sign but otherwise positive and negative segments are involved. OPi + PiP-2 = OP 2 is then, similarly, the fundamental rela- tion true for any three points on a directed line, whence P X P 2 = OP 2 - OP! = x 2 - *!. In words the distance on the avaxis (or any other line parallel to the as-axis) from any point whose abscissa is a^ to any point whose abscissa is x 2 , is given by x z x t . Similarly with respect to points on the y-axis, or two points on a line parallel to the t/-axis, the distance from the point whose ordinate is y l to the point whose ordi- nate is y% is ?/ 2 y^. To find the coordinates of the point P 3 which divides the line joining P]P 2 into two segments which bear to each other T. the ratio -*, note that -1! -"A Point of division formula drawing lines through P,, P 2 , and P 3 parallel 106 P /-* k 1 i f 3_H. -43X2 A'o to the axes, similar triangles are formed, or the proposition of plane geometry that a series of parallels cut off on trans- versals proportional parts may be directly used. Whence oc 3 a?, _ A 3 A 2 fc, Similarly, 3 = ; whence ' 2/2-2/3 , -, H W herein r = l. If P 3 (a^, 2/ 3 ) divides the line PiP 2 externally in the ratio Jc , or r, the segments must be regarded as of opposite signs and k * j. consequently, the ratio J , or r, is negative. Either k\ or A* 2 can 2 be regarded as negative ; shifting the sign from A" 2 to ^ is equivalent to changing the sign of the numerator and denom- inator in the value of x% and 2/3, no change is necessary in our above derivation of the values of a^ and y 3 . By eliminating k t and Jc 2 between the two equations, + k the equation tained. of the straight line joining P 1 and P 2 is ob- STRAIGHT LINE AND TWO-POINT FORMULAS 107 Mid-Point : Place k { = Jc 2 , or place r = 1, _ This mid-point formula is of such frequent use that it should be separately memorized ; the truth of % it is obvious from the figure. PROBLEMS 1. Plot the locus of each of the following equations : x - y - 8 = 0. Plot the two graphs on one diagram with reference to the same system of axes. Locate the point of intersection, graph- ically and analytically. 2. Plot the graph ofy = x 3 3x i 8x 2. Discuss. 3. Plot the graph of5y+2z-5 = 0. 4. Plot p = 51 + 50. 5. Find the equation of the straight line joining A( 2, 5) to B(3, 7). Find slope of this line. Find length of AB. Find the point of trisection nearest A. Find a point on BA extended that divides the segment BA externally in the ratio 1:2. 6. Given that the velocity of sound at C. is 1090 feet per second, and at 30 C. is 1150 feet per second, find the velocity at 20 C., assuming that the relation is linear ; the point dividing the line joining (0, 1090) and (30, 1150) in the ratio 2 : 1 will give the velocity as the ordinate. At what temperature will the velocity be 1100 ft. per second ? What are the velocity and temperature at the middle point of the range given? 108 UNIFIED MATHEMATICS 7. The resistance of wire increases uniformly with the temperature, r = r (l + at), the rate of increase depending upon the material of the wire ; r is the resistance at C. and a is a constant. If a given piece of wire has a resistance of 200 ohms at 10 C. and of 208.4 at 30 C., find the resistance at the middle point [of (10, 200) and (30, 208.4)]. Find the equa- tion for r in terms of L Find the value of r when t = ; interpret ; find the value of t when r = 0. The theory is that at a temperature of absolute zero ( 273 C. or thereabouts) the resistance would be zero. Ans. r = 195.8 -|- .42 t. 8. The resistance of copper wire of fixed diameter varies with the length. If the resistance 9f 1450 feet of a given spool is 184 ohms, and the resistance of feet is ohms, find the equation for r in terms of 1. Plot (0, 0) and (1450, 184). What would be the resistance of 5280 feet of this wire ? 9. Between (1, 5) and (8, 37) insert 9 points dividing the line into ten equal parts, using the formulas x = kte + frga?! and J " r~] 2344 114 UNIFIED MATHEMATICS 4. Give the value in radians of 1 revolution ; 180 ; 45 ; 135 ; 60 ; 120 ; 225 ; 3 right angles ; 390 ; 765. 5. What is the percentage error in using 57.3 as the value of 1 radian ? 6. What error in seconds is introduced by using 57.3 for 1 radian in finding the value of 3 radians ? 3 r = 171.9 ; an error of 1 / would be approximately 1.7. 7. A bicycle rider pedals at the rate of 20 miles per hour ; how many revolutions does the rear wheel, diameter 28 inches, make per minute ? The rear sprocket wheel, diameter 4 inches, makes the same number of revolutions as the rear wheel ; how many revolutions does the front sprocket wheel, diameter 10 inches, make ? Changing gear shifts the chain to a smaller rear sprocket ; what speed will be attained at the same rate of pedaling by shifting to a 3-inch rear sprocket ? 8. Place the following angles in their proper quadrants : 150, 240, 760, - 840, ^-, , - ~ if. Give the cor- o 4 3 responding positive angles less than 2 TT T . 9. In the circle of radius 10 what is the length of the arc of an angle at the center of 60? What is the difference between an arc of 60 and an angle of 60? W T hat is the length of the arc of 30, 45, -, ? 6 6 10. What is the angle at the center in radians and degrees, in a circle of radius 100, subtended by an arc of length 100 ? 50 ? 30 ? 100 TT ? Find the areas of the corresponding sectors of the circle. 11. In the artillery service angles are measured in "mils"; a " mil " is defined as 6 ^ d of a complete revolution. Com- pute the value in radians of one mil. 12. On the " mariner's compass " the complete revolution is divided into 32 parts, called " points " of the compass ; com- pare the " points," with degrees, " mils/' and radians. TRIGONOMETRIC FUNCTIONS 115 13. Compute the value of the " mil " in minutes and give approximate formulas for converting " mils " into minutes and conversely. 14. At what rate per second in degrees, and in radians, do the hands of a clock turn ? 15. A grindstone of diameter 18 inches is turning 246 times per minute. Compute the linear velocity of a point on the rim. 16. In grinding certain tools the linear velocity of the grinding surface should not exceed 6000 feet per second. Find the maximum number of revolutions per second of a 10-inch (diameter) emery wheel and of a 5-inch wheel. 17. Find the angular velocity in revolutions and in radians of an Ohio grindstone, 2 feet in diameter, which should have a circumferential speed of 2500 feet per minute. 18. The path of the earth is approximately a circle with radius 93,000,000 miles ; find the distance traveled in 1 day. What percentage of error would be introduced by using 365 instead of 3,65i days ? Show that the fact that we give r as 93,000,000 implies that the position of the point on the earth would not affect our computation. 3. Polar coordinates and angular variables. Any point P in the plane may be located by giving its distance from a fixed point 0, called the pole, and the angle which a line from the pole to the point P makes with a fixed line OR, called the polar axis. In general terms the polar coordinates of any point, of a variable point, are designated by r and 6, radius vector and vectorial angle. (See p. 116.) r will be assumed to be a positive quantity, and may be assumed as the angle generated by the rotation of the vector OP from an initial position on OR. A negative angle is generated with the polar axis by a line which turns from the polar axis, about 0, in the clockwise direction. Thus the /- R OP is taken as +30; this same figure may also be con- 116 UNIFIED MATHEMATICS Polar coordinate paper Location f (1 ' 3 0) ' (10 ' 150&) ' (10 ' - 150 )' and ( 10 ' - <" U 10 - f)- ceived as representing 330. Angles which differ by multi- ples of 360, generated by lines rotating from an initial posi- tion upon the polar axis, are represented by the same diagram ; two such angles are commonly called " congruent " angles. Each rotation of 360 brings a line back to its starting place. PROBLEMS 1. Locate the points (3, 30), (6, 90), (4, 45), (8, 135), (3, 270), (6, - 90), (5, 180), and (2, 390). ^ Locate the points (&, \ /4, \ (6, 0), (3, -\ ), and (3, 3^). V V 3. What is common to all points on OR ? 4. What curve is represented by r = 10 ? 5. What curve is represented by 6 = 30 or 6 = ? 6 TRIGONOMETRIC FUNCTIONS 117 "/r 4. Trigonometric functions sine and cosine. Assume an x-axis to coincide with the polar axis, and a y-axis to be drawn perpendicular to the polar axis at the pole. When is any fixed angle, the coordi- nates (x, y) in rectangular coor- dinates and (r, ff) in polar coordi- nates, of points upon the ray making the angle 6 with OX, are connected by the following relations : = = = - , for points upon the ray, TI r 2 r s r -Va and ^=^ = ^ = 2?. 1 2 3 x 2 + y 2 = r 2 , for any point (x, y) in the plane. We may say that - is a constant for any given angle 0; r this constant changes as changes. It is evidently a function of 0. Since r remains positive, this function is positive for all angles represented in the upper quadrants ; negative for angles in quadrants III and IV. This constant is ^ f or = 30 or .707 for = 45, \ V3 or .866 for 6 = 60, 1 for or f) 2 = 90, .866 for 6 = 120, .707 for = 135, \ for = 150, and for 6 = 180, all by elementary geometry. When is an angle which lies in quadrant III or IV, i.e. values of between + 180 118 UNIFIED MATHEMATICS and -f 360, this function of becomes negative, of is called the sine of 6, or sin 0. This function 1 sin 9 = " T cos* = -- r Polar coordinates, r and Rectangular coordinates, x and COS 6 = Similarly the ratio - " is a constant, whose value depends entirely upon the position of the- moving ray ; this func- tion of we define as cosine 0. sin0 = The consideration of the changes in value of these functions of 0, sin 0, and cos 0, as 6 changes, is facilitated by thinking of the moving ray as fixed in length. For positive values of less than 90, 6 in I or, in symbolic language, < < 90, it is evident that the complementary angle to any angle gives a tri- angle similar to the triangle in- volving 0. In this second tri- angle the ordinate and abscissa correspond respectively to the ab- scissa and ordinate in the original triangle, whence ?L _ r r Now y ~= sin (90 -0), and ?=cos0; r r hence cos = sin (90 0), or, in words, the cosine of any angle 6 (0 < < 90) is the sine of the complement of 0. This explains the name, cosine 0, which is simply the " complement's sine." Complementary angles, 9 + 9' = 90 Further, = - , whence cos (90 0)= sin 0. r r TRIGONOMETRIC FUNCTIONS 119 Either one of the triangles may be regarded as the origi- nal, the complementary angle will be found in the other; the demonstration, as given, applies in either case. The above figure serves, then, to demonstrate the two formulas, sin (90 0) = cos 6 and cos (90 0) = sin 0, for any positive acute angle 0. Later these formulas will be shown to hold for all angles 6, without restriction as to magnitude or sign. The formula cos (90 $) = sin may be derived from ' sin (90 - 0) = cos by substituting for the value 90 - 0', and finally replacing 0' by 0. Since may vary from to 90, 90 6 varies between the same limits. sin (90 -6)= cose, cos (90 8) = sin 6. 5. Historical note. The function sin 6 is Hindu in its origin, dating back probably to the fourth century A.D. The Hindus called the sine " ardha-jiva," meaning half-chord. In the eighth century A.D. the Arabs becoming familiar with Hindu astronomy and trigonometry, as used in astronomical work, transliterated the word "jiva" or "jiba" into "geib"; the word in Arabic means curve and in the twelfth century Euro- pean translators into Latin of Arabic works of science trans- lated this word as " sinus." Into English the word comes by transliteration again, the sound and not the sense being pre- served. Plane trigonometry is possible using the chords instead of the half-chords ; this system was developed by the Greeks, but it leads to much more complicated formulas and methods. 6. Tangent and the reciprocal functions. The quotient cos varies as varies ; this is then a function of 0. This function is called the tangent. By definition, tan 8 = cos Y xj 120 UNIFIED MATHEMATICS The reciprocals of sin 6, cos 0, and tan 6 are also functions of ; to these the names cosecant 0, secant 0, and cotangent have been given. By definition, cogecant ^ ^ e = _L_ sin6 4 secant 0, or sec 8 = cotangent 6, or cot 6 = cos 6' 1 tan 6 PROBLEMS 1. Given sin = .29, find cos using the formula sin 2 + cos 2 = 1. The negative value has a meaning. 2. In what quadrants is sin 9 positive? in what quadrants is cos 6 positive ? 3. Given sin = .29, in what quadrants may 6 lie ? 4. In what quadrants is tan positive ? 5. As a rotating arm of length 10, moving about from OX, turns through 90, discuss the changes in value of the y of the end of the moving arm ; consider r as 10 and discuss the change in value of sin as the angle generated increases from to 90 to 180. What change in sin as increases beyond 180? 6. Discuss similarly the changes in values of cos 6 as 6 varies from to 90 ; from 90 to 180. 7. Discuss the possible values of tan 0. Take x = 1, |-, .1, .01, .001 and compute y in a circle of radius 10. Discuss the values of tan 0. When x = .000001, y = 9.99999999999995 what is the approximate value of tan 6 ? 8. Given tan 0=3, find sec 6 from the formula sec 2 = 1 + tan 2 0. Compute both the positive and the negative values of cos 6. 9. Express in terms of the sine of the complementary angle : cos 48, cos 84, cos 56, cos 48 10', cos 90. TRIGONOMETRIC FUNCTIONS 121 10. Express in terms of the cosine of the complementary angle sin 48, sin 84, sin 56, sin 48 10', sin 90. 11. Complete the following table : cos 45 = .7071 = sin 45 cos 46 = .6947 = sin 44 cos 47 = .6820 = sin cos 48 = .6691 = sin cos 49 = .6561 = sin cos 50 = .6428 = sin Reverse the table, beginning sin 40 = sin 41 = 12. Complete the following table : sin 35 = .5736 = cos sin 35 10' = .5760 = cos sin 35 20' = .5783 = cos sin 35 30' = .5807 = cos sin 35 40' = .5831 = cos sin 35 50' = .5854 = cos sin 36 = .5878 = cos Notice that the sines of 35 + some minutes are cosines of angles 54 + some minutes ; the cosines of 35 + minutes are sines of the complements, 54 + minutes. In our tables you have written at the left of the table 35 and 54 at the right ; sin at the top and cos at the bottom. 35 1 sin cos .5736 .8192 60 10 .5760 .8175 50 20 .5783 .8158 40 30 .5807 .8141 30 40 .5831 .8124 20 50 .5854 .8107 10 60 .5878 .8090 cos sin 54 122 UNIFIED MATHEMATICS 7. Fundamental formulas. Since x* + y z = r 2 , for any point on this circle of radius r, 1 Note that although x or ?/ or both may be negative, the rela- tion continues to hold since ( x) z =.x z , and ( y) z = y 2 . Whence, by substitution, cos 2 6 + sin 2 6 = 1, for all values of 6. By division by cos 2 0, sin 2 6 1 1 + cos 2 cos 2 0' or 1 + tan 2 = sec 2 0, for all values of 0. Similarly, 1 + cot 2 = esc 2 0. sin 2 6 + cos 2 6 = 1. 1 + tan 2 6 = sec 2 0. TRIGONOMETRIC FUNCTIONS 123 These formulas are of fundamental importance. They should be memorized. 8. Functions oi plane geometry t determined for t structed with ru most important 45% 60, and 72 for and 90 (as ] sin 45 = cos 45 = tan 45 = 0, 30, 45, 60, and r le values of these functic he angles which can be ler and compass. The of these angles are 30, ; the values are evident imits). 1 v 2 7ft7 el m 8 ited angles. By i can be precisely eometrically con- / / s / z / A/2 2 1 A/2 ~^r~v =:i 7 ' I = cot 45. / i Functions of 45 One half a unit square. qin fif> * . S / cos 30 2 f S *N ^ 1 4 ._ /JAO 1 s ' / JO \ cos oO = % X ^ JO' 2/ ' \ y 2 = sin 30. S s^ / \ tan 60 A/3 X ^ 1 - / "t V V 2" ^ \ = cot oO . S x^ 1 ] tin 30 Functions of 60 Equilateral triangle. A/3 Functions of 30 = cot 60. Equilateral triangle. V3 V3 These diagrams should be memorized as half of a unit square for 45, and half of an equilateral triangle placed ver- tically for the functions of 60 and directly related angles, and the same placed horizontally for the functions of 30 and related angles (-30, 150, 210). sinO = 0. sin 90 = 1. cos = 1. cos 90 = 0. tanO = 0. tan 90 = 00. 124 The meaning of the expression tan 90 = oo (infinity) is that as the angle approaches nearer to 90 the tangent becomes larger than any quantity we may assign, however large; strictly at 90 the tangent function has no meaning, as a divi- sion by zero is involved. The expression tan 90 = x is not, then, an equality, like tan 60 = V3. Construction of the regular decagon OM divides OA in " extreme and mean " ratio. Algebraical method by solving, 3? 10 (10 x). The method of constructing a decagon combined with the solution of a quadratic equation enables us to find the sine of 18. The radius of the circle is divided in extreme and mean ratio to obtain the side of the inscribed decagon : 10(10 x)= x*, in a circle of radius 10. Whence, z 2 + 10 x - 100 = 0, x = - 5 V125 = - 5 11.1803 = - 16.180 or + 6.1803, of which we take the positive value. One half of this value is the value of y in the triangle of reference for 18 when r = 10. Hence the sine of 18 is 3.090 10 = .3090. TRIGONOMETRIC FUNCTIONS 125 9. The Greek method, using chords. By the methods of plane geometry, using chords instead of half-chords, the sine of half an angle and the sine of the sum and the difference of the two given angles can be computed. One theorem involved, in addition to the Pythagorean theorem, is not given in many geometries. It is called Ptolemy's the- orem, as it is fundamental in , the method of computing chords de- veloped by Ptolemy, a Greek writer of the second century A.D., whose textbook on astronomy, the Almagest, continued in active use for fifteen hundred years. The theorem is that in an inscribed quadrilateral the product of the diagonals is equal to the sum of the products of the oppo- site sides. From the chord of 60 one can compute the chord of 30 ; thus the sine of 15 is obtained. From 36 and 30 the sine of 3 can be obtained by using hah' the chord of the difference of two given arcs ; from this the sine of lj, f, |, ^ 6 , 3 V, ,y, T |j, 5 | s , ... can be computed. The sine of 1 cannot be obtained by this process, nor can the sine of ; these are found by other methods, giving approximations as accurate as desired for any practical purposes. 10. Origin of the tangent and cotangent functions. In the study of astronomy the angle of inclination to the horizon of ftiiis H v/) Arabic shadow function The shadow varies as the cotangent of the angle of inclination of the sun. the sun and of other heavenly bodies is important. The ratio of the length of the shadow to the length of the vertical object 126 UNIFIED MATHEMATICS casting the shadow gives the cotangent of the angle of in- clination of the sun. This function of the angle appeared before the tangent function in the works of the Arabic as- tronomer, Al-Battani, of the tenth century A.D., and it was called the shadow and later, right shadow or second shadow. The tangent function, being the ratio of the length of the shadow cast on a vertical wall to the length of a stick placed horizontally out from the wall, was called later the first shadow. The Arabs took the length of the stick as 12. Variation of sine and cosine as 9 varies. 11. Variation. As varies the trigonometric functions also vary ; it is desirable to fix in mind the changes of the three principal functions, viz. sin 0, cos 0, and tan 0, as changes by rotation of the moving arm. Taking r = 10, it is an easy matter to follow on the graph the changes in the x and y of the end of the moving ray. As the moving ray starts from OX, an angle of 0, the y or ordi- nate is zero. So we have that the sine, -, begins at zero for r ; as increases the y increases, reaching a maximum of 10 when 6 is 90 and the maximum value then of sin is \$ or 1. As increases beyond 90, the ordinate begins to decrease, arriving finally at when the moving arm is on OX'. For angles greater than 180 up to 270 the ordinate decreases, finally reaching a minimum or lowest value of 10 ; the cor- responding minimum of sin is 1 ; from 270 on to 360, completing a revolution, the sine increases from 1 up to 0. For angles greater than 360, or for negative angles, the moving ray would move through no new positions ; for any such angle the trigonometric functions are equal to the func- tions of the corresponding positive angle having the same position. The limits + 1 and 1 of the sine function and cosine function are evident, of course, in the figure. In any position of the moving ray x and y are the sides of a right triangle of which r is the hypotenuse, except that on the axes x or y equals r ; hence the quotients - and - are either numerically less than 1 or at most equal to 1. Note particularly on the diagram the sines of 30, 45, and 60, as A, approximately 111, and ^; the values, .500, 0.707, and 0.866 may well be memorized. On the diagram it is a simple matter to read the sines of 10, 20, 30, 40, 50, 60, 70, 80, and 90 as the corresponding ordinates divided by 10, correct to two decimal places. The cosines of these angles are read as the corresponding abscissas divided by 10. The tangent as - is not in a form to give the numerical value without computation ; however, by drawing the tangent line to the circle at A and producing r to cut the tangent line at 128 UNIFIED MATHEMATICS A T v AT T, you have = = *- ; whence - = tan a, and so the value of OA x 10 the tangent of the angle can be read as the ordinate at A di- vided by 10. The tangent read as a length A T = 10 tan . tan 30= -^ = ^=.58; V3 10 tan 45 = ^ = !; -./Q -17 q tan 60 = ^ = ^ = 1.73. When the angle increases beyond 90 the position of the terminal arm fixes the sign of each function ; the sine is positive when the arm is in the upper quadrants, I and II, and negative in the lower, and the cosine positive to the right, I and IV, and negative in II and III. The tangent is positive in I and III, and negative in II and IV ; when posi- tive the corresponding vertical lengths are cut off above A on the tangent at A, and when negative, in II and IV, the corre- sponding vertical lengths are cut off below A on the tangent. If the radius is taken as unity, the ordinate, abscissa, and tangent length represent numerically and in algebraic sign the sine, cosine, and tangent values of the corresponding angle. However it is usually more convenient to take a radius of 10, 25, 50, or 100 and to interpret the trigonometric functions as ratios, as indeed they are. 12. Related angles. From our definitions it is evident that sin has the same value for two angles, symmetrically placed with reference to the y-axis, 6 and 180 6 ; cos has the same value for two angles symmetrically placed with respect to the x-axis, 6 and 0, or and 360 ; tan has the same TRIGONOMETRIC FUNCTIONS 129 value for two angles which differ by 180, 6 and 180 + 6. All functions are the same for angles which differ by 360, or by any integral (positive or negative) multiple of 360, for the terminal arms of such angles will coincide when the angles are placed in position to determine the trigonometric functions. The trigonometric functions of 360 - 6, 180 - 6, 180 + 0, 90 + 6, 90 6, and 0, in terms of the functions of are of particular importance in later work. In the figure the vectors I a f v 0, -9, 180 -0, 180 + 0. Related angles Read the corresponding functions on the diagram. OPi, OP 2 , OP 3 , and OP 4 are the terminal arms of related angles in quadrants I,. II, III, and IV. The vector OPi de- termines, we may say, a positive acute angle XOPi, and, further, any angles which differ from the positive acute angle by any integral multiple of 360 ; OP 2 represents the terminal arm of 180- 0, OP 3 of 180+ 6, and OP 4 of 360 -0, or of - 9. If is the angle represented in quadrant 1, 180 d is the angle here represented in II ; and conversely, if is in II, 180- $ is in I ; if is the angle in III, 180- is the angle in IV, and conversely. Evidently if d in I is 30, 180 - 6 is 150, represented in II, and if = 150, 180 - 6 = 30 ; further, if Q is - 330 in I, differing from 30 by - 360, 180 - will be 180 -(- 330) or 510 which is in II, 360 + 150, differing from 180- 30 by 360. The ordinates in I and II are equal 130 UNIFIED MATHEMATICS and of the same sign, and similarly the ordinates in III and IV are algebraically equal ; r is the same in all. Hence for all angles 0, sin (180" - 9 1 = sin 9. The abscissas in I and II are numerically equal but opposite in sign, similarly in III and IV : the vectors are the same and positive. Hence cos (180 - 6) = - cos 6. By definition, tan (180 - 6) = , for all angles. cos (18v 6) Bv substitution, this formula also holds for all angles 0, since every formula in- volved has been shown to hold for all angles 6. If is an angle in I, 180 4- 6 is in III ; the corresponding positions are represented for any such angles by our figure. If is represented by the vector in II, 180 + 6 is represented by the vector in IV. The ordinate in III equals numerically the ordinate in I, but is opposite in sign ; similarly the ab- scissas of I and III ; similarly the ordinates and abscissas, respectively, of II and IV are equal in value and opposite in sign. Hence sin <180" + 9)= sin 6. and cos < 180 + 0)= cos 6. These equalities hold for all angles 0. By definition, tan (180+g) = o = = tan 0, cos(180+0) costf which holds for all angles B. In precisely the same way sin ( 9) = sin 9. cos (- 8)= cos 9, and tan ( 6) = tan 9, for all values of 0. TRIGONOMETRIC FUNCTIONS 131 e , 90 + 6. Related angles which differ by 90 Our second figure can be used to show that sin (90 + 0)= cos 0, cos (90 + 0)=- sin0, for all values of 6. If 6 is the angle represented in I, 90 + 6 is the angle here represented in II ; if is in II, 90 4- is rep- resented in III ; if d is in any quadrant, 90+ 6 ism the quadrant following in the counter-clockwise sense. Now the ordinate in any quadrant here equals numerically and algebraically the preceding abscissa ; thus y 2 = x 1 ; y 3 = x. 2 ; y^ = x 3 ; y^ = x. For any angle 6, sin (90 + 6) = cos 8. Similarly cos (90 + 6) = sin 0, tan (90 + cos (90 + 0) - sin 6 = cot 6. 132 UNIFIED MATHEMATICS The following relations have now been established for all angles 6: sin (180 - 6) = sin 9, sin (180 + 9) = - sin 6, cos (180 - 6) = - cos 6, cos (180 + 6) = - cos 6, tan (180 - 6) = - tan 9, tan (180 + 6) = tan 9, sin (9)= sin 9, cos (9)= cos 9, tan (9) = tan 9. The formulas, sin (90- 9)= cos 9, cos (90 - 9) = sin 9, and tan (90 - 9) = cot 9, have been established for acute angles. However, the preced- ing formulas for 90 + which we have established for all values of 0, positive and negative, can be used to prove that these formulas for 90 B hold for all values of 6. Thus sin (90 6) can be considered as sin (90 + ( 0)), and as the formulas for 90 + hold for all values of 6, we have: sin (90 + ( 0))= cos (- 0)= cos 6, and cos (90-f (- 0)) = sin (_#) = _ (_ sin0)= + sin0. Hence sin (90 - 6} = cos 0, and cos (90 0) = sin 0, for all values of 0. T? . , /nno /IN sin (90 0} cos 6 Further tan (90 6) = - - 1 = = cot 8, cos (90- 0) sin again for all values of 0. The student will do well to remember the diagrams and to connect the formulas with these. It is necessary to recollect only the representation for an acute angle ; it is more desir- able to connect the formulas with the diagrams than merely to memorize the formulas. TRIGONOMETRIC FUNCTIONS 133 PROBLEMS 1. Find the sine, cosine, and tangent of 210. The triangle is the same as that used for the functions of 30, but it is placed in III. 2. sin 150 = ; cos 150 = ; tan 150 = 3. sin 315 = ; cos - 45 = ; tan - 45 = 4. sin 225 = ; sin 495 = ; tan 750 = 5. Express the following in terms of functions of positive angles less than 45 : a. sin 170= b. sin 130 = c. cos 170 = d. cos 130 = e. sin 220 = /. tan - 40 = 6. Express in terms of functions of x : . a. sin (x- 90) = HINT. Use first sin ( 0) = sin 6. b. sin (270 + a) = HINT. Express first as (180 + 6) ; i.e. sin (180 + 90 + z) = - sin (90 + z) = .... c. cos (x- 270)= d. tan (360- a;) = 7. Draw one quadrant of a circle of radius 10 half-inches ; construct the angles of 30, 45, and 60 and read their values. Bisect the angle of 30 and so obtain the values of the func- tions of 15. Make a table of values of sines, cosines, and tan- gents, advancing by 15. Note that the chord of 30 may readily be computed ; one half of this chord divided by the radius gives the sine of 15. Find the sine of 7^ similarly. From the table of sines of acute angles from to 90 by 15 intervals, give the sines of the related obtuse angles up to 180. 8. Given tan = 3, find sec and cos ; what is the signifi- cance of the double sign in the answer ? 134 UNIFIED MATHEMATICS 45 9. Express in terms of functions of positive angles less than a. sin 100. b. cos 100. c. tan 100, d. sin 200. e. tan 200. /. sin 300. g. cos (60). h. cos (-160). t. tan (-420). j. sin 750 50'. k. cos 1030 40'. I tan 218 10'. 13. Angles constructed from given functions. Given sin = 4, construct both values of (in I and II). The problem is in geometrical language to construct a right triangle with the hypotenuse and one side given, since sin0 = ^. r Take r as 7 and y as 4 ; since sin = - , 7 is to be one side of r our angle ; with 7 as a radius and as center describe a semi- Graphical solution of sin 6 = - 7 circle above the cc-axis ; y = 4 is a line parallel to the axis of x, cutting the circle, x- + y 2 = 49, in two points. Find the inter- sections a x and a 2 ; geometrically the angle is found. Using the Pythagorean theorem a 2 -f 4 2 = 7 2 , and x z = 33, x = 5.75. The positive value of x is to the right and the negative to the TRIGONOMETRIC FUNCTIONS 135 left; to the first corresponds the acute angle 0, and to the other with the terminal arm in II. In I, sin 6 = ^ = .57. In II, sin = |. 5.75 cos e = = .82. 7 tan B -'= .70. 5.75 cos = - = - tan 6 = 7 4 5.75 = - .70. This problem should also be solved using the formula sin 2 6 + cos 2 = 1. Given tan = 1.4, find the other functions of and discuss the two solutions. cos 6 x can be in I or III. Take y as 1.4, a; as 1 (or y as 14, x = 10, or other values as con- venient) ; evidently x as 1 and y as - 1.4 gives in III. Graphical solution of tan t = 1.4 Since x 2 + y* = r 2 , ?- 2 = 1 -f 1.96 = 2.96. r = 1.72+ In I, sin0 = ^ - or -M_ = M. V2T96 = .81 (or .814 to l.<2 V2.96 2.96 three places). cos0 = - =.581 + . 1.72 In III, sin = - .814, cos = .581. This problem should be solved also by using the formula 1 -f tan 2 6 = sec 2 0. 136 UNIFIED MATHEMATICS EXERCISES 1. Use 10 of the larger units on a sheet of cross-section paper and find by construction the sines of the angles 0, 10, 20, 30, . 180. Compare with the table. Note that the values for 10 and 170, 20 and 160, 30 and 150 corre- spond. Use the formula cos = sin (90 0) to find the cosines of 0, 10, 20, 90. Note that in the second quadrant the cosines become negative. 2. Construct an angle of 60, using 10 as side of the equi- lateral triangle used. Find cos 60, sin 60, tan 60 to 2 places. 3. Find the sine of 150, 210, 330. Use half the equilateral triangle, placed horizontally with vertex of 30 angle placed at the origin. 4. Find the sine and cosine of 120, 135, 225, - 30. 5. Find the tangent of 120, 135, 225, - 30, from the data of the preceding problem ; find tan 120, tan 135, tan 225, tan ( 30) from the geometrical figure. 6. By construction of a square, side 10 units, find approxi- mate values of the functions of 45. Find the values using the Pythagorean theorem. 7. Construct an angle of 30, and find values of the functions. 8. Construct angles of 15 and 7|- , and find the values of the functions. 9. Given sin = -fa, find cos and tan ; indicate both solutions. 10. Given cos 6 = .432, compute sin 6 and tan to 3 places, 6 in I. 11. Given tan = 4.32, compute sin and cos for in III. 12. Given tan = \ , construct geometrically. 13. Construct 9 geometrically, given sin = T ^-. 14. Given sin = 1 A S -, find cos 6, in IV. 15. Given sin .43, find cos 0, in III. TRIGONOMETRIC FUNCTIONS 137 16. Given tan 9 = .43, construct in II, and find values of sin and cos from the figure. 14. The inverse functions. If we are given the sine s of an angle and desire to speak of the angle we can say "the angle whose sine is s" and we can abbreviate this expression in writing to arc sin s or to sin" 1 s. Similarly the angle whose cosine is ra is written arc cos m, or cos" 1 m. Note that in sin" 1 s, cos" 1 m, and tan" 1 k, the 1 is not at all a negative expo- nent ; these expressions for angles are read anti-sine s, anti- cosine m, and anti-tangent k, or sometimes, inverse sine s, etc., respectively. In what follows we shall use mainly the symbols arc sin, arc cos, arc tan, arc esc, arc sec, and arc cot, although the other symbols are also in common use. Whether the " arc " or " 1 " symbols are used the student is strongly advised to read " arc tan t " or " tan" 1 1 " always as " the angle whose tan- gent is t" and similarly expressions like arc cos cc, arc sin 1, and sin" 1 k. A given angle has only one sine, but a given number is the sine of many different angles. A similar remark applies to the other five functions. To illustrate : sin 30 is 0.5 and no other value. But arc sin .05, the angle whose sine is 0.5, may be 30 or - 330 or 390 or 750 or 150 or 510 or 870 or any angle differing from 30 or 150 by an integral multiple of 360. The sine of any one of these various angles is 0.5 ; sin (k 360 + 30) = .5 and sin (k 360 -f 150) = .5, where k is any integer. PROBLEMS 1. Given arc cos =6, construct both in the first and in the fourth quadrant. Note that the problem is precisely the same -as though the requirement were to construct when given that cos = -J ; . or to construct arc cos ^. 2. Between what values must k lie to have any solution for = arc cos k ? 138 UNIFIED MATHEMATICS 3. Given that the angle, arc sin ^, is obtuse, construct the angle. 4. Construct the following angles of the first quadrant : arc sin^, arc tan (+ 2), arc cos y^, arc sin .43. Give the approxi- mate value of the other two principal functions in each case. 5. Give five solutions of each of the following : arc cos ^ = ; arc tan 1 = ; arc sin = ; arc cos 1 = 6. If arc sin .438 = 26, what is arc cos .438 ? 7. What is the value of arc sin .438? Give four answers. Give the general formula representing angles 9 which satisfy B = arc sin (- .438). What is arc cos - .438 ? 8. On the following diagram, regarding the circle as having a radius of 100, read the numerical value to two decimal places, of the sine, cosine, and tangent of each angle repre- sented. Each minor division represents 4 units. CHAPTER VIII TABLES AND APPLICATIONS 1. Tables. The tables of the trigonometric functions are computed by processes dependent upon formulas derived in the higher mathematics. We have shown the graphical method of finding sine, cosine, and tangent, which serves also to bring out the fact that the sines of angles from to 45 are at the same time cosines of the complementary angles ; simi- larly since tan (90 x) = cot x, it follows that the tangents of angles from to 45 are cotangents of the complementary angles, from 90 down to 45. Since tables are given of both sine and cosine it is necessary to give values of both functions only up to 45, and similarly with tangent and cotangent. Thus sin 26 10' is found in the table, of sines which reads down with 26 at the left, and below 10' as given at the top ; if the cos 63 50' is sought we look for 63 at the right of the table of sines with the minutes to be read below ; and we find that the cosine table is the same as the table of sines, but reading up ; this brings us to precisely the same place in the tables as sin 26 10', the complementary angle ; similarly sin 63 50' is sought in the row marked 63 at the left of the table and leads to the value which read as a cosine represents cos 26 10'. For angles greater than 90 the formulas which we have given for related angles are applied. Probably the simplest formulas to apply to obtain the functions of obtuse angles are the formulas, sin (90 + x) = cos x, and cos (90 + a;) = sin x. 139 \ 140 UNIFIED MATHEMATICS Thus sin 128 35' = cos 38 35' ; cos 128 35' = - sin 38 35'. It is well to note that subtracting 90 from angles greater than 100 and less than 200 simply increases the tens' digit of the angular measure by one, dropping the hundreds' digit. The formulas for 180 + a, and for 360 or , are used for angles in III or IV. Since computation is largely effected by means of logarithms, it becomes desirable to have separate tables of the logarithms of the trigonometric functions. The sines and cosines of all angles are numerically less than 1 and so are tangents of angles less than 4t> ; hence the logarithms of these numbers will have negative characteristics. In the logarithms of the trigonometric functions, 10 is to be annexed to the logarithm as given in the table for sines, cosines, and tangents up to 45. Thus log sin 30 is 9.6990 - 10 ; log sin 56 10' = 9.9194 - 10 ; log tan 34 20' = 9.8317 - 10 ; but log tan 56 10' = .1737. 2. Interpolation. The insertion, by interpolation, of the natural and logarithmic functions of angles lying between those expressly given in the tables follows precisely the same lines as in the corresponding problem in the logarithms of numbers. Our tables give these functions for angles increas- ing by multiples of 10 minutes ; interpolation enables us to compute the functions of angles to minutes ; in using tables giving the functions to minutes interpolation enables us to compute to tenths of a minute. Note that the assumption is always that if angles are read to minutes you compute only to minutes ; the tables used should correspond to the precision of measurement of the given data. Four-place tables are, in general, sufficiently accurate for measurements which are made to four places in numbers, and to minutes in angular measurement. Illustrative problems. 1. Find by interpolation (a) sin 36 15', (6) log sin 36 16', (c) log cos 36 18', and (d) log tan 36 14'. TABLES AND APPLICATIONS 141 TABULAR VALUES Compare with your tables angle sin log sin COS log cos log tan 36 10' .5901 97710 .8073 9.9070 9.8639 53 50' 36 20' .5925 9.7727 .8056 9.9061 9.8666 53 40' cos log cos sin log sin log cot angle The values to four decimal places of the functions of angles between 36 10' and 36 20' evidently lie between the values which are here given. Thus sin 36 10' is .5901 and sin 36 20' is .5925, an increase of 24 units of the fourth place ; this 24 is called the tabular difference. To the ten equal steps of in- crease from 36 10' to 36 20', by minutes, correspond ten in- creases approximately equal to each other, in the sines of these angles, making a total increase in ten steps of 24 units of the fourth place. The tenths of 24 are respectively, .1 .2 .3 .4 .5 .6 .7 .8 .9 2.4 4.8 7.2 9.6 12 14.4 16.8 19.2 21.6 In adding, as our logarithms are given only to four places, we add rejecting tenths, and retaining in the last place the nearest unit. Thus for tenths of 24 we use always 2, 5, 7, 10, 12, 14, 17, 19, and 22. The interpolation does not always give the correct result to four places, although in the values of the sine the error is always less than 1 unit of the fourth place. In the above values of sin 36 11' to 36 19' as given by the addition 2, 5, 7, 10, 12, 14, 17, 19, and 22 units of the fourth place to .5701 the first value .5703 should be, to 4 places, .5704 ; the error is less here than -^ of 1 % of the value taken. 1. a. sin 36 15' = .5701 + T % of .0024 = .5713. Method : Tabular difference is 24 ; to .5701 add .5 of 24 units of the fourth place. 6. log sin 36 16' = 9.7710 - 10 + T %(.0017)= 9.7720 10. Tabular difference is 17 ; 10.2 is replaced by 10, and this is added in the third and fourth decimal places to 9.7710. 142 UNIFIED MATHEMATICS 0. log cos 36 18' = 9.9070 - 10 - ^(.0009) = 9.9063 - 10. Tabular difference is 9 ; cosine and log cosine are decreasing functions ; 7 units of the fourth place must be subtracted. d. log tan 36 14' = 9.8639 - 10 + T 4 ff (.0027) = 9.8650 - 10. Tabular difference is 27 ; 10.8 is replaced by 11. 2. Find to minutes, by interpolating, the angle when given (a) sin a = .5919, (6) log sin a = 9.7717, and (c) log cot = 9.8650 ; find a in each case. 3. (a) sin a = .5919 ; tabular difference is 24 ; given differ- ence .5901 to .5919 is 18 units of the fourth place. Among the tenths of 24 find the nearest to 18 ; 16.8 and 19.2, respectively .7 and .8 of 24, are equally near and the even number of tenths is commonly taken, in such cases, by computers. sin a = .5919 ; a = 36 18'. (6) log sin a = 9.7717 ; a = 36 10' + T ? r of 10' (to minutes). a = 36 14'. Tabular difference is 17 ; 7 is nearest to .4 of 17. (c) log cot a = 9.8650 ; a = 53 40' + |f of 10'. Tabular difference is 27, a decrease; given decrease is 16; among the tenths of 27 the nearest to 16 is 6 ; hence a = 53 46'. Had log cot been given as 9.8651 - 10 or 9.8649 10, the angle a would again be given as 53 46'. PROBLEMS 1. Find the 20 natural trigonometric functions following, without interpolation ; time yourself ; limit 6 minutes. a. sin 36 10'. g. tan 70 30'. b. tan 63 20'. h. sin 28 50'. c. cos 34 10'. t. tan 16 20'. d. cot 80 00'. j. cos 8 40'. e. sin 59 30'. k. sin 157 10'. /. cos 48 50'. I cos 214 10'. TABLES AND APPLICATIONS 143 m. cot 141 00'. n. tan 329 30'. o. cos 136 50'. p. cos -28 10'. q. tan-6420'. r. sin 384 00'. s. cot 756 00'. t. sin 242 40'. 2. Find the logarithms of the above 20 trigonometric func- tions, timing yourself. Limit 7 minutes. 3. Find the following 20 yourself. Limit 12 minutes. logarithms, interpolating ; time a. log sin 36 14'. / log cos 48 57'. Jc. log sin 152 15'. b. log tan 63 29'. 9- log tan 70 33'. I log cos 214 26'. c. log cos 34 14'. h. log sin 28 51'. m. log cot 141 05'. d. log cot 80 06'. i. log tan 16 22'. n. log tan 329 33'. e. log sin 59 32'. j- log cos 8 48'. o. log cos 136 57'. P- log cos - 28 11'. Q- log tan - 64 26' r. log sin 384 03'. s. log cot 756 08'. t. log sin 242 44'. 4. Find the angles less than 90 corresponding to the follow- ing 20 logarithms ; no interpolation ; a. log sin a = 9.6878 - 10 Jc. b. log cos a = 9.9954 - 10 I c. log tan a = 9.4858 - 10 m. d. log cot a = .5102 n. e. log cos a = 9.8241 - 10 o. f. log tan a = 9.7873 - 10 p. g. log sin a = 9.3179 - 10 q. h. log tan a = .2155 r. i. log cos a = 8.9816 - 10 s. > log cot a = 9.9341 - 10 t. time 6 minutes. log sin a = 9.9499 - 10 log cos a = 9.8081 - 10 log cot a = .8904 log tan a = 8.9420 - 10 log cos a = 9.9640 - 10 log cos a = 9.9757 - 10 log tan a = .5720 log sin a = 8.9403 - 10 log cot a = .0152 log sin a = 9.9977 - 10 5. Give in each case another angle which would satisfy the above relationship, in problem 4 ; e. q. if log sin a = 9.699010, a = 30 or 150. 144 UNIFIED MATHEMATICS 6. Find the following 20 angles interpolate ; a. log sin a 6. log cos a c. log tan d. log cot a e. log cos /. log tan a g. log sin a h. log tan a i. log cos a .;. log cot a A 1 , log sin a /. log cos a m. log cot a w. log tan a o. log cos a />. log cos a g. log tan a r. log sin s. log cot a <. log sin a time yourself. = 9.6881 - 10 = 9.9956 - 10 = 9.4861 - 10 = .5104 = 9.8228 - 10 = 9.7879 - 10 = 9.3200 - 10 = .2144 = 8.9912 - 10 = 9.9358 - 10 = 9.9502 - 10 = 9.8092 - 10 = .8955 = 8.9492 - 10 = 9.9645 - 10 = 9.9753 - 10 = .5699 = 8.9404 - 10 = .0137 = 9.9978 - 10 Angle 10 in a circle of radius 5 inthes PM = .868 in. ; arc PA = .873 in.; AT- .882 in. 3. Angles near and 90. For angles near zero, from to 2, the cosines vary only from 1.0000 to .9994; the cosine function to 4 places cannot then be used for determination of the angle to minutes. Simi- larly, of course, the sines of angles from 88 to 90 vary between the same limits. For ordinary pur- poses it will suffice to avoid the use of the cosine TABLES AND APPLICATIONS 145 in the interval from to 2 or 3 or 4 ; the method of avoid- ance is explained below. In computing graphically the values of sin 6 and tan even with a radius of 10 cm., or of 5 inches, the difference between tan 6 and sin becomes too small to read accurately when 6 is less than (i.e. 7|; .131'). For 10 which is .1745 radian, sin .1745 r is .1736 and tan is .1763 ; for 5 or .0873 r , sin #=.0872 and tan0 is .0875 r ; for 1 or .Ol745 r , sin 0=. 01 745 and tan is .01746 or 5 places are necessary to exhibit any difference between 0, sin 6, and tan 0. sin < < tan 8 for small acute angles 0, is measured in radians Evidently, triangular area GAP < sector GAP < area OAT, but the area of the triangle OAP=GAx MP = \ r x r sin Q = r 2 sin B. The area of the sector GAP = ^ r 2 0, since is measured in radians, and the area GAT=^ r 2 tan 0. Whence, by substituting, i r 2 sin 6 < l^B < ^f* tan 0. e i 1< sin cos 6 8 Whence, as 6 diminishes, - T , lying between 1 and a num- ber approaching .1, can be made as near to 1 as we please. By methods of plane geometry, using 30, 15, 7, 3|, together 146 UNIFIED MATHEMATICS with 72, 60, 12, 6, and 3 it can be established that cos f differs from 1 by less than ^ of 1 % ; cos = .99991 ; for any angle 0, less than f , will exceed sin by less than T ^ of 1 f and tan will exceed by less than y^ of 1 %. Similarly the discrepancy between sin 6 and tan 6 for 0, any angle less than 3|, is less than ^ of 1 %, and for any angle up to 8 the difference is less than 1 % of either value. On the earth's surface ordinary dis- tances are regarded as straight lines. However for many purposes the deviation from a straight line is of importance : thus particularly with projectiles of long range, the deviation is of vital importance. In the figure given if PA represents an arc on the earth's surface, PT may be regarded as the altitude of a balloon, aeroplane, or top of a mountain, and TA gives the distance of the horizon. /. TO A is equal to the dip of the horizon. AM is the drop in the distance twice PA, i.e. from T an observer would note, on the ocean, the complete disappearance of a ship of height AM when the ship is at Q. By algebraic process, A T = V2 rh + h z ; when h is measured in feet, r in miles, and TA in miles, this gives for values of Jo 7 "1 h less than 15 miles, AT = A/ -, correct to - of 1 %. Check using 3960 miles as r. Arc PAQ on the earth's surface TA, horizon dis- tance. PT, height of ob- server. PROBLEMS 1. Given that an observer is at a height of 1000 feet, com- pute the distance to the horizon, r = 3960 miles. What is the dip of the horizon ? Note that the tangent of the dip- angle is the horizon distance divided by the radius. TABLES AND APPLICATIONS 147 2. Find the angle subtended at the center of the earth by an arc of length 1 mile, 10 miles, 20 miles. 3. What is 1 of latitude in miles ? 4. Degrees of longitude vary in length from degrees on a great circle of the earth at the equator to at the poles. Find the radius of the small circles on which degrees of longi- tude are measured, for 40 north latitude. Where else on the ^- L -- ~, X ^ ^ X / / ^. s /' X --" >cc .s4 ) ^^ \A / ^~-, ^_ ^~ 0-- s ; .. / \ / 1 I n > J / \ t -y s \ / s \ > L 0' \ g / i 1 Circle of 40 N. latitude earth's surface would degrees of longitude be the same ? From 35 to 45 N. latitude discuss the percentage variation in degrees of longitude, as compared with degrees of longitude at 40 N. latitude. 5. How far below the arc of 1 mile on the earth does the corresponding chord fall at the lowest point ? Find the same distance in inches for arcs of 2 miles, 8 miles, 10 miles, 16 miles, 20 miles. 6. What part of the height of a mountain, measured on the altitude, is not visible from a point 20 miles distant ? 7. From what distance can the top of a mountain 10,000 feet high be seen ? 8. What distance from shore is a ship whose masts, 55 feet high, are just disappearing from view? 148 UNIFIED MATHEMATICS 9. Using the figure in the text, find an approximation for TA in miles when h is small and measured in feet. =v'- 2h-r f h \_ /2 - h 60 / h V 5280 \5280y ~"V 80 \52Wj 16, , / h V \/-/t4-[ - ] = *4 V5280; AT for values of h less than 5 x 5280, the ( ] can be neglected. 10. Find the " dip " of the horizon and the distance from the balloon for h = 100, 500, and 1000 feet. 11. Find the distance from the point below the balloon on the earth's surface to the points on the horizon viewed by the observer in the balloon. 12. According to the approximate formula of Huyghens the length of a circular arc, a, is connected with the chord, c, of the arc and the chord, h, of half the arc, by the formula a = . Compute the actual length. CHAPTER IX APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 1. Parallel and perpendicular lines. The slope of the line joining fa, yd to (x 2 , y. 2 ~), x. 2 x l evidently represents the tangent of the angle which the line joining these two points makes with the positive ray of the -axis, i.e. the angle from the oj-axis to this line. We have taken P^fa, y 2 ) to the right of Pfa, y^, but obviously interchanging Pzfa, 2/2) and P r fa, yd simply changes sign of both numerator and de- nominator of the fraction representing the slope m ; Q is in each figure the point fa, yd, and P^Q and QP 2 have like signs if PiP 2 or P^Pi makes a positive acute angle with the positive ray of the or-axis ; and P } Q and QP 2 have unlike signs in the contrary case when P^P Z or P z Pi makes a negative acute angle with OX. It is to be noted that shifting the y-axis, parallel to itself, either to the right or to the left does not affect the value of x^ x 1} since 149 150 UNIFIED MATHEMATICS whatever the position of 0, A^A^ = P 1 Q = OA 2 OA l = x 2 x\ ; similarly no change is made in the value of the slope by shifting the <-axis parallel to itself, up or down. Given y = m^ + k, any straight line, raj represents the tangent of the angle which this line makes with the positive ray of the #-axis. Any parallel line has the same slope ; ra 2 = ?% for two parallel lines. Any perpendicular line has the slope angle , = 90 -f ! ; tan <% = tan (90 + ttl )= cot ^ = tan ax whence m 2 = . Of two parallel lines the slopes are equal, m x and of two perpendicular lines the slope of the one is the negative reciprocal of the slope of the other, i.e. ra 2 = -- or, by solving, m l = -- m l m 2 Given y = mx + b, any family of parallel lines of slope m. y = x + k represents the family of perpendicular lines. m. Illustrative problem. Given _3x + 4y 1 = 0, find the slope, the parallel line through the origin, the family of perpendicular lines, and the perpendicular line through (1, 5). = f, 0=-36 52'. y = f x is the parallel line through the origin. Derive this both from y = mx + b and y y l = m(x x\). 1 4 The perpendicular line has the slope, m 2 = -- - = + - MI y = | x + k is the family of perpendicular lines. y 5 = |(x + 1) is the perpendicular line through ( 1, 5). EXERCISES 1. Write the equations of the sides of the triangles used in finding the functions of 30, 45, and 60. APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 151 Acceleration down a plane, g sin . 2. Gravity imparts to a falling body a vertical velocity of 32 1 feet per second, with t seconds as time during which the body has fallen ; on a smooth in- clined plane gravity imparts a ve- locity of 32 t sin a where a is the angle of inclination of the plane. Find the velocity imparted at the end of 1 second to a body sliding (without friction, assumed) on an inclined plane of slope 10, 20, 30, 40, 50, ... to 90. 3. In a freely falling body s = 16 t z ; while on a plane s = 16 1 2 sin a ; find s for t = 10, a = 30, 45, and 60. 4. To pull the body up the plane requires a force of TFsin a + kW- cos a, where A; is a constant dependent upon the friction. Find the force to pull a weight of 1000 Ib. up an incline of 30, Jc = ^. 5. Find the slope of the line joining (3, 7) to (5, 9); find the middle point of this line ; find the equation of the perpendicular bisector of the segment. 6. Write the equation of the line through (3, 5) making an angle tan" 1 T 5 ^- (m = ^ 2 -) with the a-axis, and write the equation of the perpendicular from (1, 8) to this line. 7. Find the foot of the perpendicular line found in prob- lem 6 and then find the distance between (1, 8) and the origi- nal line, using the distance formula. 8. Find the slope angles in degrees and minutes of the following lines : (a) 5 y 12 x 7 = 0, (6) 127/ + 5x-3 = 0, (c) x - y - 5 = 0, (d) 3x-y- 8.= 0. 9. Find lines through (1, 5) parallel and perpendicular to each of the lines in the preceding exercise. 152 UNIFIED MATHEMATICS -O 10. Find the perpendicular bisectors of the sides of the triangle formed by the three lines given by the equations, 5y 12 # 7=0, 12y+5o; 3=0, and x+y 5=0. Find the area of this triangle graphically and analytically. 2. Projections of vectors. OP has been designated by r, for radius vector of the point P. The line OP has magnitude, given by r, and direction, given by the angle 6. We may use this system of representation to repre- sent velocities, forces, and other physical quantities. As a velocity this vector may be resolved into two component velocities, repre- sented by OA and OB. OA repre- sents the velocity in the x direction, x = r cos 6 ; OB represents the y velocity, r sin 6, the vertical com- ponent of the velocity of a body moving with velocity represented by OP. The projection of any vector upon a directed line is defined as the directed distance between the perpendiculars dropt from the extremities of the given vector upon the line ; it is given by v cos wherein v represents the vector and a is the angle between the positive rays of the two lines. Since cos ( )= cos a, we do not need to distinguish between the two lines, i.e. the angle can be taken as obtained by ro- tation from the given line to the given vector, or vice versa. It is a fundamental as- sumption that any two vector quantities which- may be repre- sented acting together at the same point may be replaced by Components of a vector OA represents the x com- ponent of OP. OB represents the y com- ponent of OP. OP is the resultant of OA and OB. Vector parallelogram APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 153 a single vector which is the diagonal of the parallelogram formed by the two given vectors. The process is called vector addition. . This assumes that in space, for example, an imparted velocity S. E. of 50 miles per hour increased by a velocity N. E. of 30 miles per hour produces the same displace- ment whether the two forces which produce the velocities act together for one hour, or whether both act in succession each for an hour. The projection of a broken directed line upon a given directed line is the same as the projection of the straight line joining the ends of the broken line. This follows from the fact that on a directed line 1 whatever the relative po- sitions of MI, M 2 , and M 3 . The directed length is the projection of MM Z is the pro- jection of P 2 P 3 , MiMz is the projection of PiP 3 . The physical interpreta- tion is simply that the total component in the x direction (or any other) imparted by two (or more) vectors is the algebraic sum of the two (or more) x components of these vectors, taken separately. When the velocity is given as v, v z and v y are commonly used to designate the x and y components of the velocity; evidently, also Projection of a broken line on a directed line v x = v cos 6, v,, = v sin 0. 154 UNIFIED MATHEMATICS PROBLEMS 1. A bullet, muzzle velocity of 3000 feet per second, leaves the gun elevated at an angle of 10. The position, neglecting air resistance, is determined at the end of t seconds by the two equations : y = 3m t gin 1Q o _ 16 ^ x = 3000 1 cos 10. Find t when y = ; when y = 5 ; explain the two values in each case. Find x for both values of t which make y =. 0. 2. The velocity is a vector resolved into components v x = v cos and v y = v sin a. Find v x and v v when = 10, 20, 30, 45, 60. i 3. A ship sails S. E. for 2 hours at 8 miles per hour and E. N. E. (22f off East) for 2 hours at 6 miles per hour. Find the x and y of the resultant position. 4. The propeller imparts to a steamer a velocity of 8 miles per hour S. E. ( 45) and the wind imparts a velocity of E. N. E. (+ 221) of 6 miles per hour. Find the position at the end of 1 hour. 5. A boy runs east on the deck of a steamer at the rate of 20 feet per second ; the steamer moves south at the rate of 15 miles per hour. Find the actual direction in which the boy is moving and his total velocity. 6. Find the velocity in miles per hour of a point on the earth's surface due to the rotation of the earth on its axis ; find the velocity per second due to the revolution about the sun ; compare, and note that the resultant can never be greater than the sum nor less than the difference of the two. Take values only to 3 significant figures ; 3960 mi. = r ; 93,000,000 miles as distance from sun. 7. The United States rifle, model 1917, has a muzzle velocity of 2700 feet per second. Find the horizontal velocity of the bullet when the angle of elevation is 1, 10, 20, 30, and 45 respectively. APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 155 3. Normal form of a linear equation. The slope-intercept, point-slope, and two-point formulas correspond to the fact that a straight line is determined when one point on the line [(0, A;) or (x l} y^) respectively] and the direction of the line are given, or when two points are given. A straight line may be determined in many other ways ; one method which gives a further useful form of the equation of the straight line deter- mines the line in terms of the length and direc- tion of the perpendicular from the origin upon the line. Thus if a perpendicu- lar from the origin upon a given line is 5 units long, and makes an angle of 120 with the x-axis (positive ray) geometri- cally we construct the line by constructing the ray 6f 120 and upon it taking a length of 5 units. At the extremity of this line of 5 units length a perpendicular is drawn which is the required line. The point N is readily found to be (5 cos 120, 5 sin 120) and the slope is therefore the equation of the line to be found is A line determined by the normal to it from the origin Normal length, 5 ; a = 120. tan 120' y- 5 sin 120 = -1 tan 120' , (a? -5 cos 120). tan 120 = Sm ; substituting, clearing of fractions, trans- posing cos 120 x cos 120 + y sin 120 - 5 (sin* 120+ cos 2 120)= 0, x cos 120 + y sin 120 -5=0, since sin 2 a + cos 2 a = 1, 5 = 0. 156 UNIFIED MATHEMATICS In general, given the normal ON to the line from the origin, of length p, and making angle a with OX, the extremity N is (p cos a, p sin a) ; the slope is , and the equation becomes sin a x cos a. + y sin a p = 0, ^> is taken as a positive quantity just as r has been taken. Evi- dently if p = 0, x cos + y sin = represents a parallel line through the origin. Evidently also for parallel lines on opposite sides of the origin the angles a and a' differ by 180 ; i.e. a' = 180 + a, whence _ _ cos a' = cos a. x cos + y sin a p = Normal form. A The projection on ON } ofOM + MP ] the projection on ON ofOP 4. Normal form derived by projection. We have shown that the projection of any broken line upon any given line is the APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 157 same as the projection upon the given line of the vector join- ing the ends of the broken line. Let P(x, y) be any point on the line whose equation is sought; drop PM the perpen- dicular from P to the avaxis; the projection of the broken line OM + MP on the normal OjVis equal to the projection of OP on ON. Now OM = x makes the angle a, by hypothesis, with ON, and MP makes the angle a 90 ; hence the projec- tion of OM on ON is x cos a (OA, negative in the figure since a is obtuse) and of MP on ON (AN in the figure) is y cos (a 90) ; the projection of OP on ON is ON itself, or p; further y cos (a 90) = y cos (90 a) = y sin a. Then, since projecting on the line ON, projection of OM + projection of MP= projection of OP, we have x cos a + y sin a = p, whence x cos a + y sin a p = 0. 5. To put the equation of a straight line in normal form. Let the given equation be 3 a; 4 y -f 7 = 0, and let x cos a + y sin a p = be the same equation in normal form. If these two equations represent the same line, these lines must have the same slope and the same y (or x) intercept. cos a __ 3 1 _ p sin 44 sin a cos a = ^ sin a. 4 cos 2 a = T 9 ^ sin 2 a. But cos 2 a = 1 sin 2 a, whence 1 sin 2 a = T 9 g sin 2 a ; T | sin 2 a = 1 ; sin a = |. p = + % sin a, whence since p is to be positive, sin a must be taken as positive. Hence sin a = + , p = - ; cos a = | ; and thus the normal form is fx + fy | = 0. This equa- tion is obtained by dividing each member of the original equation by 5. 158 UNIFIED MATHEMATICS 27) In general to put Ax -f By + C = in the normal form, a; cos a -f-t/ sin a p = 0, one must multiply through by some quantity k, so that fc^l = cos, A,-JB = sin, and kC = p-, kC=p shows that k must be chosen opposite in sign to C; squaring both members of the first two equations and adding gives k z (A- + B 2 } 1, whence k = , of which the sign is taken as opposite to C. vA 2 + B 2 RULE. To put an equation Ax + By + C= in normal form divide through by \ f A* + B 2 , with the sign taken opposite to that of the constant term. 6. To find the perpendicular distance from a point to a line. In solving this problem one considers the various forms of the straight line which may be employed. Evi- dently the normal form is most hopeful for use, since it involves the perpendicular distance of the given line from the origin. Through the point PI(X I} yO draw a line paral- lel to the given line ; evidently the difference be- tween the nor- mals to the two lines gives the distance. Three possibilities must be considered : 1. P 1} on the opposite side of the given line from the origin; 2. P lt '0* Distance of a point from a line APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 159 on the same side of line as 0, the origin, but such that the normal angle is the same, i.e. so that the parallel line through PI(XI, y t ) falls on the same side of as the given line, P\ 011 the figure ; 3. P 1} on the same side as the origin, but the normal angle increased (or diminished) by 180, designated by P"i on the figure. Let x cos a + y sin a p = be the equation of the line. 1. x cos a -f- y sin a (x^ cos a -|- y sin a) = is the parallel line through PI(X } , y^), since this equation is evidently in normal form and the line passes through (x ly y^). ON Z = Xi cos a + y! sin . d = ON% ON = x t cos a -\-yi sin p. The perpendicular distance is obtained then by writing the equation in normal form and substituting for (a/!, y\) the coordinates of the given point. Evidently if PI (x l} y^) is on the line, this gives also the correct distance, which is then zero. 2. x cos a+y sin a (x } cos a-\-y l sin a) is the equation of the parallel line ; again, ON' 2 = x cos a -f- y sin a. d=ON- ON' 2 : whence d= ON' 2 - ON = Xi cos a + yi sin a p. The same rule holds, but the distance in this case is negative. Evidently the rule holds if ON 1 ^ is 0. 3. ' = 180 -f a ; cos ' = cos a, sin a'= siu a. To write the equation of the parallel line in normal form, the coefficients of x and y must both be the negatives of the coeffi- cients of x and y in the given equation. x( cos a) + y( sin ) ( a^cos a y^ sin)=0 is the equation of the parallel line in normal form. = (x { cos a + /! sin a). d = ON" 2 + ON= a?! cos y^ sin a +p, or d = #1 cos a + y t sin a p. 160 UNIFIED MATHEMATICS RULE. To obtain tin- iHxtmice from a point to a ////< the. equation in normal form, substituting therein for # and y tlie coordinates of the given point. The resulting number gives the distance as positive if the point and the origin lie upon opposite sides of the given line, as negative if P l and are upon the same side of the given line. x cos + y sin p represents the perpendicular distance then from P(x, y) to the line x cos + y sin p 0. For all points on one side of this line the expression is positive, and on the other side, crossing the line to the origin side, the expression is negative. A line which passes through the origin, p = 0, will be said to have its equation in normal form when sin a is taken as positive, i.e. when the coefficient of y is made positive. Points on this line make x cos a + y sin a = ; points above the line make the expression x cos a -f y sin a positive, and points be- low the line make it negative. Thus 3x 4 y = is written ^ x -f - y = 0, or 5 o -3a + 47/_ to be in normal form. The perpendicular distance from any point to such a line will be positive for points above the line, and negative for points below the line. 7. Bisector of the angle between two lines. Geometrically the bisector of an angle is the locus of the points equidistant from the two sides of the angle ; analytically we express the condition that two distances should be equal to each other. Let the equations be given in normal form, as x cos t + y sin ! p l = and x cos a^ + y sin a, Pz = 0. Let P(x, y) represent any point on either bisector of the given angle ; analytically x cos ! + y sin a^ p L = (x cos a, + y sin a p~). APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 161 Bisector A, in the opening which includes the origin, is ob- tained by taking the + sign since both perpendiculars are of the same sign for points on A. Bisector B is obtained by taking the negative sign since any point on B is on the same side as the origin with respect to one of the lines, and on the opposite side with respect to the other ; hence if PM l comes out negative, PM 2 will be positive (by the formula) and the equality will be obtained by putting PMi = - PM 2 . Just as the two axes divide the plane into 4 quadrants in Bisectors of ^ angles betwe en two which the distances to these lines given normal form axes are + +, +, , and + respectively, so any two lines in the plane divide the plane into 4 sections in which the perpendicular distances, as given by our formula, to these lines are ++,+ ,- > and + respectively. The + + and sections are separated by the -f and + sections respectively, as it is evident that you pass from + + to + by crossing the second line. The bisector of the -f- + and opening is given by equating the left-hand members of the equations of the two lines in normal form; the bisector of the -f- , - + opening is obtained by equating the one to the negative of the other left-hand member. If one of the lines passes through the origin, or if both do, then the above-mentioned convention is necessary to es- tablish the part of the plane in which the left-hand member of the equation of the line is positive. It is customary to make sin a positive, which makes the portion of the plane above the line the positive side, i.e. the coordinates of any point above 162 UNIFIED MATHEMATICS the line when substituted in the given equation give a positive value, and of any point below the line give a negative value. PROBLEMS 1. If a line makes an angle of 30 with the ar-axis what angle does the normal to the line make with the x-axis ? 2. What is the slope of the line, y = 2 x -+ 5 ? What is the slope angle ? What is the slope of the normal to this line ? AVhat is the angle which this normal makes with the z-axis ? Find from the tangent of the angle made by the normal with the x-axis the sine and cosine of the same angle. Write the equation in normal form and interpret the constants. 3. Given a and p, as below, slope angle of the normal and length of the normal from the origin to the line, find the equations of the lines, and draw the lines : a. a = 30, p = 5. b. = -30, p = 5. c. a= 150, p = 5. d. a= 210, p = 5. e. a= 137, p = 5. /. = 137, p = W. g. a = - 63, p = 10. k. a = 223 15', p = 8. 4. If a remains equal to 40 and p varies, what series of lines will be obtained ? if p remains equal to 5, and a varies, what series of lines will be obtained ? 5. Write the following equations in normal form : a. 3x 4y 5 = 0. 6. 5x - c. x d. 3z-5#-4 = e. y = 2 x 14. 35 APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 163 6. Find the distances of the points (1, 5), (2, 3), (0, 5), (0, 5), (2, 3), and (3, 7) from each of the lines in the preceding problem. 7. What is the distance of the point (x, y) from the line 3x 4y 5 = 0? Under what circumstances does the formula give a negative value for this distance ? What is the distance of any point (x, y) from 5z-fl2?/-f8 = 0? What does equating these two expressions, i.e. the left-hand members of each normal form, give ? Interpret on the diagram. What is obtained by setting one of these expressions equal to the negative of the other? 8. Find the bisectors of the angles between the following pairs of lines : a. 3x + 4y 5=0 and 12 x 5 y 10 = 0. b. y2x5=0 and 2x + y + 7 = 0. c. y 2x 5 = and 3y+x 8=0. d. y -2x = Osmd3y + x-8 = Q. e. y 2 x = and 3 y -f- x = 0. 9. Find the distance of the points (1, 3), (3, 0), (3, - 7), and (0, 8) from each of the lines in the preceding problem. 10. Find the distance between the following pairs of paral- lel lines : a. y = 2 x 7, y=2x + 3. b. 4?/-3a = 5, 4y-3a-16 = 0. c. 4 y 3 x = 0, 4y-3z 16 = 0. d. x + 2y-7 = 0, 2a + 4y + 17 = 0. e. 7.2 x + 8.3 y- 15 = 0, - 8 = 0. 164 UNIFIED MATHEMATICS 11. In problem 8 show that each bisector obtained is one of the pencil of lines through the point of intersection of the given two lines. 12. Find the area of the triangle having as vertices the following points : a. (3, 4), (0, 0), and (0, 8). b. (3, 4), (0, 0), and (10, 2). c. (1, 1), (4, 5), and (7, -3). 13. Find the area of the triangle formed by the three lines : 3x + y - 5 = 0, 12 x - 5 y - 10 = 0, and 4x 3y 7 = 0. 14. What is the distance of any point (x, y) from the point (0, 0) ? What is the distance of any point (x, y) from the line x 5 = ? Equate these two expressions for distance and simplify. The resulting equation has for its graph all points which are equally distant from the point (0, 0) and the line x 5 = 0. 15. Find the locus of all points which are equidistant from the point (0, 0) and the line y 8 = 0. Let (x, y) represent any point satisfying the given condition. 16. Find the locus of all points at a distance 10 from the point (0, 0) ; from (1, 3). Find the locus of all points at a distance 10 from the line 3x 4 y 7 = 0; at a distance 10 ; explain graphically. 17. Find the locus of all points equally distant from 3 x y 5=0 and from (1, 5). 18. In problem 12 find the equations of the three bisectors of the angles of the triangle formed ; find the perpendiculars from the vertices to the opposite sides ; find the perpendicu- lar bisectors of the sides ; show that in each instance you have three lines which have a point in common. APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 165 19. What points, when the coordinates are substituted for x and y, make the expression 4 y 3 x 5 positive ? What points make this expression zero? What points make this expression negative ? Locate three points of each type, plot and discuss. 20. Substitute in the expression x 2 + y 1 25 for x and y the coordinates of the points (0, 3), ( 3, 2), ( 1, 4). Plot these points. Substitute (0, 5), ( 3, 4), ( 4, 3), and ( 5, 0). Plot. Substitute also (0, 8), ( 7, 0), (5, 3), and (4, 6). Note that the graph of x 2 + y 2 25 = separates the plane into two parts ; in the one part inside this curve are all points whose coordinates substituted for x and y, respec- tively, make the expression x 2 + y z 25 negative, and in the part outside lie all points which make this expression positive. CHAPTER X ARITHMETICAL SERIES AND ARITHMETICAL INTERPOLATION 1. Definition of an arithmetical series. In the table of natu- ral sines the values of the sines of 21 to 22 are given as follows, sin 21 siu 21 10 7 sin 21 2 / U / d Sv a */ "/ d >a Id i I -^ / d A l \ a\ 1 ! is 34 5 < 17* \ X dinates represent the terms of an arithmetical series with first term a and the common difference . To The ordinates of y = dx + a at x = 0, 1, 2, 3, repre- sent terms of an arithmet- ical series any series of equal increases or increments given to x there correspond a series of equal increments given to the ordinates ; 170 UNIFIED MATHEMATICS this depends upon the theorem of plane geometry that if a series of parallel lines cut off equal parts on one transversal they do on every transversal, and this theorem is equally true for any straight line in the plane. 5. Interpolations in sines and other functions. The value of the sines of the angles are given by the corresponding ordinates in a circle of radius unity, or the or- dinates divided by 100 in a circle of radius 100, or the ordinates di- vided by 1000 in a circle of radius 1000. On our diagram, with radius 100 the straight line joining the end of the ordinate corresponding to 20 to the end of the ordinate at 30 does not differ materially from the circular arc connecting these points. Were we to plot these angles in a circle of radius 1000 the points of intersection would appear as in the second part of the diagram, lettered AB, and constituting a tenfold linear en- 8 = n 2 largement of AB. If the sines of the angles were given by intervals of ten degrees, interpolation by tenths would give the sines by degrees ; the circle with radius 100 mm. (or 100 twentieths of an inch) permits the sine and cosine to be read to two places accurately, and this rather low degree of refinement corresponds to a table of sines given by intervals of ten degrees ; interpolation would give substantially correct values to two decimal places, e.g. for the sines of 21, 22, - 1 2 3 4 n-ln Graphical representation of the sum of an arithmetical series ARITHMETICAL SERIES AND INTERPOLATION 171 Arc of 20 to 30 in circles with radii 50, 100, and 1000 fortieths of an inch The marks on the long chord indicate the points given by interpolating between sin 20 and sin 30, and between cos 20 and cos 30. Even on the arc with 25 inch radius nine interpolated points on the chord and cor- responding points on the arc, between 21aud 22, coincide. 172 UNIFIED MATHEMATICS ... 29 by interpolating between sin 20 = .34 and sin 30 = .50. The arc of 10 on this circle differs slightly but appreciably to the eye from the chord of 10, but the interpolated points on the chord are not easily distinguished from the ten points on the curve. Angles given by degrees permit interpolation by intervals of 6' or by intervals of 10', with substantially correct values to the third place if the values are given only to three places ; values given to four places give by interpolation values substantially correct to the fourth place. On our circle with radius 1000 the sine and cosine can be read to three decimal places ; interpolation between the values of sin 20 and sin 30 give points markedly different from the true points on the curve. These points are indicated by checks on the chord of 10. Interpolating five points (for 10', 20', 30', 40', 50') on the chord from the 20 point to the 21 point gives points not readily to be distinguished from the correct points on the arc. On this diagram it is not possible to distinguish the sub- divisions for minutes on the arcs from the corresponding points on the chords of central angles of 10'. With the proper changes, noting particularly that as in- creases cosine decreases, the argument given holds 'for inter- polated values for cos 0. Interpolation of the tangent values is similar, except in the neighborhood of 90 where the tangent changes very rapidly ; in a separate table are given by minutes, the tangents of angles from 88 to 90. The graph of the function y = log w x, or 10" = x, is a con- tinuous curve which for small arcs approximates a straight line. Similarly the graphs of the functions y = sin x, and of 2/=sin x, log cos x, log tan x and log cot x approximate straight lines within small intervals, and so are subject to our ordinary process of interpolation. 6. Arithmetical means. If two numbers a and b are given, the arithmetical mean between the two is the number x which ARITHMETICAL SERIES AND INTERPOLATION 173 makes a, x, b three consecutive terms of an arithmetical series ; to insert n arithmetical means it is necessary that a, the n means, and b form n + 2 consecutive terms of an arithmetical series. Ordinary interpolation is the insertion between two tabular values of some particular one of 9 arithmetical means. If a, x, b form an arithmetical series, b x = x a, a + b whence x = ~ -- If a, a -f- d, a + 2 d, a -\- 3 d, a -\-(n l)d, a -|- nd, b form an arithmetical series, b = a +(n + l)d; whence d = ~ a . n + 1 The sum of n terms of the series a, a -\- d, a+2d } a + (n l)d, is the average value of these n numbers is n(a + ) _ a -\- l n 2-n 2 ' termed the arithmetical mean of the n numbers ; the sum of an arithmetical series is seen to be the " average value " mul- tiplied by the number of terms. Similarly of any collection whatever of n quantities, the arithmetical mean is regarded as the total sum divided by the number of quantities. In statisti- cal work the latter mean, total sum divided by the number of given quantities, is called the " weighted mean." PROBLEMS 1. Between .3584 and .3746 insert 5 arithmetical means; if .3584 = sin 21 and .3746 = sin 22, what do these means represent? Between .3746 and .3584 insert 5 arithmetical means ; interpret as cosines. 2. Given sin 21 = .3584 and sin 21 10' = .3611, find 9 intermediate values ; interpret. 174 UNIFIED MATHEMATICS 3. Given sin = 0, sin 30 = .5000, what value would arithmetical interpolation give for sin 21? What is the error ? 4. Given sin 20 = .3420 and sin 30 = .5000 ; find to 4 places sin 21. How many terms in the arithmetical series which is implied ? 5. What is the sum of the first ten integers ? 6. If cards are marked 1 to 190, what is the total sum ? What is the average value of the total group of numbers ? 7. How many years of life have been lived by a group of 30 individuals, aged 21, 22, 23, ... 50 years? 8. Falling from rest a body falls approximately 16 feet in the first second, and 48 in the second, and in each succeeding second 32 feet more than in the one which precedes. What distance will the body fall in 10 seconds ? How long will it take such a body to fall 1000 feet ? 9. If it takes a lead ball 8 seconds to fall to the earth from a balloon, what is the height of the balloon ? 10. How long will it take a ball to reach .the earth if dropped from the top of the Washington monument, 550 feet high ? 11. Draw figures to show that between x = 3 and x = 4, the graph of xy = 1 approximates a straight line. 12. Write the equation of a straight line representing for integral values of x, from to 10, the arithmetical series with a = 10, d = 3. Represent the series also by the series of rectangles, each of width 1. In summing the arithmetical series we reversed the series and added ; show the geometrical equivalent on the figure, page 170, with the rectangles. 13. Given that the first term of an arithmetical progression is 8 and the last term 100, what equation must n and d satisfy ? If d = 2, what is n ? If d = 3, what is n ? Interpret. ARITHMETICAL SERIES AND INTERPOLATION 175 14. Show that in an arithmetical series of n terms (a, a + d, ) the average value is ^ the sum of the first and last term. Note that the average value of the terms is the total sum divided by the number of terms. This average value is termed the arithmetical mean of the n terms. 15. Find the average value of the following 10 heights, and the arithmetical mean : 40 6 feet 126 5 feet 9 inches 64 5 feet 11^ inches 138 5 feet 9 inches 86 5 feet 11 inches 120 5 feet 8 inches 92 5 feet 10| inches 112 5 feet 8 inches 142 5 feet 10 inches 80 5 feet 7J inches 16. If there are 1000 men measured and they are grouped in height as above, find the average height of these men. Note that the easiest way to find this average is to take the variation above and below some one of the " middle " values, e.g. with reference to 5 feet 9 inches as origin, 6 feet is re- garded as + 3 inches and this group has a total of 120 inches excess above 5 feet 9 inches per individual ; 5 feet 1\ inches is regarded 1^ inches and the total group of 80 has a total deficiency of 120 inches, or 120 inches ; the two neutralize each other. Whatever total remains is divided by 1000 and added, algebraically, to the 5 feet 9 inches. 17. Draw the graph of I = a +(n l)d; assuming a and d as constants. 18. Historical problem. In the Egyptian manual men- tioned above, occurs the following problem : If 100 loaves of bread are divided according to the terms of an arithmetical series among 5 people so that ^- of what the first three receive equals what the last two receive, find the number received by each person. Solve the problem. The Egyptian reckoner assumes that the last person receives 1 loaf, and without any explanation, that the second receives 6^ loaves, and so on in arithmetical progression ; the sum he finds to be 60, and to arrive at the correct values all numbers are increased in the ratio of 100 to 60. Compare with your solution. CHAPTER XI GEOMETRICAL SERIES AND APPLICATIONS TO ANNUITIES 1 . Geometrical series . A series of terms in which each term is obtained from the preceding by multiplying by a fixed number is called a geometrical series ; by definition the ratio of each term to the preceding is a constant. Designating this ratio by r, and the first term by a, the type series becomes : a, ar, ar 2 , ar 3 , ar"" 1 . l n = ar n ~ l . The sum of such a series to n terms is obtained as follows : Let s n = a + ar + ar 2 + ar 3 + ... + ar n ~ l . rs n = ar + ar 2 + ar 3 + ... -+- ar"" 1 + ar". s n rs n = a ar n . s n (l r) = a ar". _ a ar* _ a ar" _ ar" a 1r 1r 1r r 1 2. Sum to infinity, r< 1. When r is numerically less than 1, the terms of a geometrical series become smaller and smaller without limit. The sum of n terms differs from the fixed quantity, , by the quantity - '- , which decreases 1r 1r C1 Cl ClV n as n increases ; this difference between - and - 1r 1r 1r can be made smaller than any assigned quantity however small by taking n sufficiently large ; the value - - is termed 1 r 176 GEOMETRICAL SERIES AND ANNUITIES 177 the " sum to infinity " of the series or, more strictly, the " limit " of the sum of the series as the number of terms in- creases indefinitely. A simple and familiar illustration of a geometrical " sum to infinity " is found in the recurring decimals of elementary arith- metic. Thus .33333 ..., or .3, is the series T 3 -fyf^ + y-^ + ... with r = y 1 ^, a = T 3 ; .it represents the series T y T + y^y^ + TTlWW + - With * = T*1F a = loV. = * = |j. Another illustration of an infinite geometrical series is found in the total distance traversed in passing from one point to another by passing first through -| of the distance, then through half of the remaining distance, and successively in each sub- sequent movement through half of the distance which remains to the goal. One of the famous paradoxes of Zeno is to the effect that the hare cannot overtake the tortoise since the hare must first cover one half the distance intervening between their original positions, then one half of the remaining distance, then one half the remaining distance, and in each subsequent interval of time one half of the distance which remains ; conse- quently the hare cannot overtake the tortoise as by this I 1 I TT~ A 8 M l 4 M t 1 M 3 \ M^lB Hare Tortoise process there is always distance intervening, and in the mean- time, further, the tortoise has advanced. If each of these intervals of space of the infinite series traversed required the same length of time, or any finite portion of time, the argu- ment would be sound, but you have here two infinite series each with a finite sum. As each space interval becomes smaller, the time required to traverse that space becomes smaller. The total distance evidently is the sum of 8 + 4 + 2 + 1 -f ^ + which has the sum 16, both by the formula and by the figure ; to cover the whole distance requires 1 hr. and this takes the hare up to B. The fallacy in Zeno's argument lies in restricting the discussion to limited intervals of space 178 UNIFIED MATHEMATICS and time preceding the instant and place at which the tortoise is overtaken. 3. Geometrical means. If a, x, b form a geometrical series b x - = -, whence x 2 = ab, x = Vab. x a If a, ar, ar 2 , ar 3 , ar 4 , ar"" 1 , ar n , b form a geometrical series, evidently b = ar n ' r = ar n+l , Vfty* whence r = ( - ] , and the series is \aj 1 2 _n n+1 a, a( - ) , a. - ) , , a(-\ , af - ] or 6. \aj \aj \aj \aj PROBLEMS 1. Sum to twenty terms the series 2, 4, 8, 16, . In the following series give the sum to twenty terms, and where possible give the sum " to infinity." 2. 3, - 6, + 12, - 24, + 48, .... ^^11 i ... d - ^5 2> T2> 72> ' 4. .1717171717 , or .17 (repeating decimal). 5. .01717171717 -, or .017. 6. 3.16161616 .... 7. 5, 15, 45, 135, . 8. 4, 10, 28, 82, 244, ... the wth term being 1 + 3". 9. 3, 7, 11, 15, 19, .... 10. 3, 7, 17, - 27, . 11. v, u 2 , -y 3 , v*, .... Sum to n terms. 12 1 r r 2 r 3 r 4 .. AX<. , /,/,/,/, 13. (1 + i), (1 + i)*> C 1 + O 3 ? (1 + 0*1 '" Sum to n terms. 14. The number of direct ancestors which an individual has is represented by the series, 2, 4, 8, 16, . Find the total num- ber in the preceding 10 and in the preceding 20 generations. GEOMETRICAL SERIES AND ANNUITIES 179 15. According to Galton's law of heredity the parents con- tribute to the hereditary make-up of an individual -|- of what is contributed by all the ancestors ; the grandparents contribute ^ ; the great-grandparents, of whom there are eight, contribute ^ ; and so on ; find the total contribution. Find the individual contribution of a single individual four generations back. 16. Insert three geometric means between 2 and 17 ; com- pute to one decimal place. 17. Insert one, two, and three geometric means between 1 and 2. 18. One of the so-called " three famous problems of an- tiquity " is to construct, using only ruler and compass, a cube which is double the volume of a given cube. This problem was very soon reduced to the insertion of two geometric means be- tween a 3 and 2 a 3 ; insert two geometric means between a 3 and 2 a 3 and show that this gives the algebraic value of the side of a cube of double the volume ; the geometrical solution has been demonstrated to be impossible if ruler and compass are the only instruments of construction. 19. In the population statistics on page 65 find the population of the United States in 1810 and 1910 ; between these two numbers insert 9 geometric means ; find r ; this represents the approximate decennial rate of increase in the population at 1810 which would give the final actual population in 1910. Compare with the actual census figures at the end of each ten years. 20. How would you find the regular annual rate of increase in the population of the United States, in other words the fixed annual percentage increase in population which would change the population from 7.2 millions in 1810 to 101.1 millions in 1910 ? 21. The plunger chamber of an air pump is approximately J T of the total air capacity ; at each stroke ^ of the air in the receiver is removed ; after 10, 15, and 20 strokes find the 180 UNIFIED MATHEMATICS proportionate amount of air ivmiiiiiiiig in the receiver ; approx- imately how many strokes must be made to remove 99% of the original air ? 4. Graphical representation Geometrical series. On the straight line y rx of slope angle , with tan a = r, if x = a, ar, ar 2 , ar 3 , the successive abscissas represent the terms of Graphical summation of the ge&metrical series, r < 1 Note that the " sum to infinity," r < 1, is represented. the geometrical series ; by means of the 45 line through the origin these successive abscissas are readily constructed. However, a more favorable construction for a graphical treat- GEOMETRICAL SERIES AND ANNUITIES 181 ment of the series regards each term of the series a, ar, ar 2 , ar 3 , as an addition to the preceding abscissa; the succes- sive additions to the ordinates, increments of the ordinates, will be ar 2 , ar 3 , a?* 4 , . By drawing the line, y = x a at an angle of 45 with the #-axis, the successive abscissas are readily constructed ; by drawing through the point Pi(a, ar) a line parallel to the or-axis it intersects the 45 line drawn through (a, 0) at a point M 2 such that 1\M Z = PiA ly since Z. PiAiM 2 = 45 ; a parallel to the y-axis through M 2 intersects the line y = rx at a point P 2 such that M 2 P 2 = r P\M* = ar 2 , and A 2 P 2 = ar + a?' 2 . Similarly, if P n represents the wth point found on our line, y = rx, M n P n = ar n , OA n = s n = a + ar -f- ar 2 -f- ar"" 1 , P A = = tan a = r, whence P n A n = rs n ; A n M n = sr n ar" ; AA n = s n a ; but A l A n = A n M H , .since the slope of the line is 1. .. s n a = rs n ar", s n (l ?) = a ar", a ar" = 1-r If r < 1, as in our figure 1, a < 45, and the two lines inter- sect at K to the right of the origin ; and the points of intersec- tion M 2 , M s , M n will fall below K on the line BK. The abscissa of K represents the " sum to infinity " of our series, as it is evident that the series of triangles could be continued indefinitely in the opening OKA. Evidently also, solving y = x a, y = rx, rx = x a ^ 1 r 182 UNIFIED MATHEMATICS Graphical summation of the geometrical series, r > 1 For r > 1, the figure is quite similar ; the two lines, y = rx and y = x a, diverge ; tan = r, a > 45. Evidently, A n M n = A : A n , A n -L n := ^*^ B J A n M n = rs n ar n , OA n = s. A t A n = s a, whence s n a = ?-s n ar". s B (l r) = a ar", as before =."" ar " or art> ~ a . 1-r r-1 GEOMETRICAL SERIES AND ANNUITIES 183 5. Historical note. Arithmetical and geometrical series are found in the oldest mathematical documents known, both in the remains of ancient Egypt and of ancient Babylon. The system of numbers used by the Babylonians as early as 2000- 3000 B.C. was sexagesimal, increasing by powers of 60 in geometrical series. Further an early Babylonian clay tablet gives the portion of the moon's surface illuminated on each of fifteen successive nights, from new moon to full moon by a geometric and an arithmetical series. The moon's surface is conceived as divided into 240 parts ; on the first five nights 5, 10, 20, 40, and 80 parts, respectively, are illuminated and 011 the following ten nights, 96, 112, 128, 144, and on in arithmetical progression to 240. The Egyptian manual of mathematics of 1700 B.C. (or there- abouts), includes two rather complicated problems on arith- metical series, involving also the insertion of means, and one problem involving the summation of a geometric series. The equivalent of a general formula for summation of a geometrical series was first established rigorously by the Greeks, and appears in Euclid's Elements, Book IX, prop. 36. The first summation " to infinity " of a decreasing geometrical progression was effected by the great Archimedes (287-212 B.C.) who employed the formula in finding the area of a seg- ment of a parabola. In a great part of the later development of mathematics such series have played a prominent role, in some measure because of their own intrinsic importance and in some measure as fundamental in the discussion of other types of series. 6. Annuity formulas, a. Accumulated value of an annuity. The geometrical series plays a large role in the theory of investments and insurance. We have shown (p. 54) that at rate i per year, compounded annually, 1 will amount in n years to (1-f /)" ; if 1 is invested at rate i at the end of each year, i.e. annually, for n years, these payments constitute an annuity of 1 for ?2 intervals, at i per annum. The total accumulated 184 UNIFIED MATHEMATICS value SJPJ, at the end of n years of such an annuity, is the sum of the geometrical series (i + t)-i (i + o- 2 > (i + 0-' (i + O- 4 , - (i + ! since the first payment made at the end of the first year ac- cumulates for (n 1) years, the second for (n 2) years, -, and the w.th payment of 1 is made at the end of the n years. The sum, called the amount of the annuity, represents the accumulated value of an annuity of 1 per inter- val for n years or intervals at a rate of i per year or interval. b. The annuity ichich will accumulate to 1. Very evidently K per annum will produce at the end of n years /| _j_ j'Nn ^ Ks n = K * ; the annuity which in n years will i amount to 1 is evidently the value of K which makes Ks n = 1 ; this value is = c. Present value of an annuity of 1. The accumulated value of 1 to be paid at the end of n years at i per year is (1 + i) n ; K will accumulate at the rate t in n years to K(l + ?')"; this means that K dollars (or units) in hand ac- cumulates to JT(1 + i) n dollars, and that K(l + )" dollars to be paid n years hence is worth K dollars now, money at rate i per year. Hence the present value of 1 to be paid n years hence is a value of K which makes /i"(l + i) n = 1, or K = = y", wherein v = The present value of (i + 0" i + * an annuity of 1 per annum for n years when money is worth i per annum is the sum of the geometrical series, v, v 2 , v 3 , v 4 , v 5 , i , i: 7 , v n . The first term v is the present value of the first payment of 1 which is to be made 1 year from date ; the second term is the GEOMETRICAL SERIES AND ANNUITIES 185 present value of the second payment of 1 to be made in 2 years ; ; v n is the present value of the final payment of 1 to be made n years hence. v v n+l 1 v n 1 V d. TJie annuity which 1 will purchase. The present value of K per annum for K years is Ka^\ = K . ; the annuity v which is worth 1 at the present time is evidently a value K which makes Ka tTl = 1, whence K= = This is the annuity which 1 will purchase. e. Summary of interest functions. These six functions, r n (1 -f )*, accumulation of 1, v n = (1 4- j)~", discount value, s^n = ^ ' ~ , accumulated annuity value, i 1 (1 4- rr n 1 v n a^i = * = , present value of annuity, i i , annuity to accumulate to 1, ti (1 + 0" 1 i and = , the annuity which 1 will purchase, a, r | 1 (1 4- " are of fundamental importance in the valuation of bonds, and in all problems where stipulated payments are to be made at stipulated intervals, and also in the theory of interest. If interest is to be compounded semiannually and pay- ments are made semiannually, the interval can be considered as 6 months and the rate of interest as one half the stated rate ; similarly the interval can be considered as 3 months and the rate of interest as one fourth the stated interest if interest 186 UNIFIED MATHEMATICS is compounded quarterly and payments made quarterly. Other types of problems with payments falling between in- terest periods are beyond the scope of this work. PROBLEMS 1. Compute the value of s f or &%, 5%, 4%, using loga- rithms to obtain (1 + .06) 20 , (i + .05) 20 , and (1 + ,04) w . What percentage of error is introduced using four-place tables ? Dis- cuss the effect in finding the accumulated value of an annuity of $ 100 per year for 20 years. Check by the tables given at the back of this book. 2. Compute a,, and discuss as in 1. 3. Find by logarithms from your values of s^ and a^-, the annuity which will accumulate to 1 in 20 years, and the annuity which 1 in hand will purchase. 4. If payments of $50 per year are made semiannually and interest is compounded semiannually, find the accumu- lated value of this annuity at the end of 20 years for a nominal interest rate of 6 % , 5 % , and 4 % respectively, per annum (3 %, 2^ %, and 2 % per interval). Use the tables. 5. What annual payments continued for 10 years are equal to $1000 cash in hand? 6. What annual payments continued for 10 years are equiv- alent to $ 1000 to be paid at the end of 10 years? to $ 1000 to be paid at the end of 20 years? 7. Prove 8^= (1 + i)"^; and a^sw"- Sg-]. Discuss. 8. Show algebraically that = i. a,7] s \ NOTE. The difference between 1 in hand and 1 to be paid in n years is simply the earning power of the 1 in hand for this period of n years ; if money is worth 6 % per year, 1 in hand will earn every year for n years .06 in addition to preserving itself ; 1 in n years is worth simply 1 then ; is the annuity for n years which 1 in hand purchases, and is GEOMETRICAL SERIES AND ANNUITIES 187 the annuity equivalent to 1 to be paid in n years ; the difference is the annual earning of the 1 in hand, i. 9. Give the arithmetical series represented by the ordinates of y = 3 x + 7, for integral values of x from to 10. What is the sum of this series ? Between 1 and 2 interpolate 9 values, at equal intervals, and state corresponding series ; what corre- sponds to the tabular difference ? 10. Discuss as in problem 9 the corresponding ordinates of 11. Sum to 20 terms the series 7, 5, 3, 1, 1, . 12. Write the 20th term of 7, 5, 3, 1, 1, .... 13. Write the tenth term of 7, 5, ^ 5 -, -L 2 /, .... 14. Find the arithmetical and the geometrical means be- tween 7 and 5. 15. Historical problem. It is related that an Indian prince who wished to reward the inventor of the game of chess suggested to the inventor that he should name the reward he desired. The scholar replied that he would take 1 grain of wheat for the first square of the board, 2 for the second, 4 for the third, 8 for the fourth, 16 for the fifth, and so on in geometrical progression to cover the 64 squares. The prince agreed but found, on the computation, that the value exceeded that of his realm. Taking 10,000 grains as approximately a pint, make a rough calculation of the amount involved. 16. Historical problem. In textbooks of the sixteenth century the following problem frequently appears. A black- smith being asked his price for shoeing a horse replied that for the first nail he would charge one fourth of one cent (use this in place of farthing, or pfennig), cent for the second nail, 1 cent for the third, 2 for the fourth, and so on for the thirty-two nails. Compute the price. 188 UNIFIED MATHEMATICS 7. Annuity applications. Brief tables of the annuity func- tions are given at the back of this book ; somewhat larger tables will be found in the Bulletin No. 136 of the U. S. Bureau of Agriculture, which includes also a more extensive treatment of the subject of bonds and annuities by Professor James W. Glover. a. Common annuity. To find the purchase price of an annuity of k dollars per interval for 71 intervals when the current rate is i per interval is obviously a direct application of the a;n table, as the purchase price is simply the present value of the series of payments. If the first payment of the annuity is to be made r intervals hence, a deferred annuity for n intervals, the price may be considered as the difference between an annuity for r 1 intervals and an annuity for n + r 1 inter- vals, fln+r-i! gives a payment every interval for n + r 1 intervals, the first made at one interval from the present time ; a f=ri gives a payment every interval for (r 1) intervals, the first as before, and the last (r 1) intervals from the present time ; the difference is the value of the deferred annuity of n payments, first payment to be made at the end of r intervals. Some large banks, trust companies, and insurance companies do this type of business. Frequently a purchaser desires an annuity to be paid annually terminating with the death of the purchaser ; this involves then a life contingency, and the dis- cussion and solution of this problem require new methods and new tables. b. Farm loans. To extinguish or amortize a debt by n annual payments of fixed amount is the type of problem which arises under the recent Farm Loan Act. Thus a farmer borrowing $ 10,000 at the bank at 5 % interest may desire to make such a payment as to extinguish the debt in 30 years, money being worth 5 % ' annually. The problem may be solved by considering the annuity which $ 10,000 will purchase at 5 % interest for 30 years or $ 10,000 x .06505, giving $ 650.50. The problem may also be solved by considering the interest as paid each year, $ 500, in addition to which an GEOMETRICAL SERIES AND ANNUITIES 189 annual payment must be made to accumulate at 5 % to $ 10,000 at the end of 30 years. This annual payment is found to be $ 150.50. The value found, $ 650.50, may be checked, as below, by using the $$ table. Or another check is to find the value at the end of 30 years of the $ 10,000 or $ 10,000 (1 + .05) 30 and compare this with the value of $ 650.50 x %,. Suppose, on the other hand, that the borrower desires to pay approximately $ 600 per year, applying the extra amount each year to the debt. The annuity which 1 will purchase, 5 %f! is the function involved. The question here may then be put, For what period of years at 5 % interest will $ 10,000 purchase an annuity of $ 600 per annum ? We will consider only ap- proximate solutions, taken from the tables. The tables show that at 5 % interest $ 10,000 will purchase an annuity of $ 10,000 X = $ 10,000 X .06043 for 36 years, 36l or $ 604.30 annually for 36 years ; $ 10,000 will purchase an annuity of $ 598.40 for 37 years. 36 years would be taken, and this period of 36 years is provided for, as an amortization term, by the government. By paying every year $ 100 more than the interest, at the end of 36 years the accumulated value of this annuity, the excess $ 100 over the interest, would be worth at 5 % : $ 100 x SMI = $ 100 x 95.8363 = $ 9,583.63, lea,ving $416.37 due at the end of 36 years. This amount with interest at 5 % should be the next and final payment. c. Sinking funds. If a city issues bonds to be redeemed 20 or 30 or 40 or n years hence, it is commonly desirable to pro- vide for the repayment of the bonds by an annual (interval) payment allowed to accumulate at i per annum (per interval). Similarly in business a manufacturing concern using an ex- pensive piece of machinery which has a probable lifetime of 20, 30, or 40 years must provide for the eventual replacement of this machine by an annual payment, out of earnings, into a sinking fund. In this type of problem the function involved is the annuity which will accumulate to 1 in n years. Thus 190 UNIFIED MATHEMATICS to provide for replacement of a $ 10,000 piece of machinery in 30 years money at 5 % requires an annual payment of $ 10,000 x or S 10,000 x - - which equals $ 150.50 5g6] (1 + I') 30 - 1 per annum. d. Bonds at premium and discount. If a city issues bonds at 5 % when money is worth in the money market 4 %, the bonds will sell at higher than face value, since they pay on each $ 100 an annuity of S 5.00 per year for the term of the bond, when investors are demanding only S 4.00 per annum with security of capital. This higher price is, on the basis of money at 4 %, the present value of an annuity of $ 1.00 per annum for the term of the bond or 1 x a^ at 4 % . The dif- ference between the par value or face value of a bond and the price offered by investors is called the premium (or discount, when the price offered is less) on the bond. If a city issues bonds at 5 % when investors are demanding 6 %, a bond for $ 100 will sell at a discount of 1 X a^ at 6 %, since the investor receives from these bonds not S 6.00 per annum but only $ 5.00 per annum. Evidently the longer the bond has to run the greater would be the discount. In general terms the premium on a bond of face value C, paying a dividend rate g, bought to yield j per annum is P = C(g j)a^ at j per annum. PROBLEMS 1. If a father sets aside annually $ 100 per year as a fund for his son when the latter becomes of age, to what will the fund amount at the end of 21 years, the money accumulating at 4 % interest ? 2. If a farm mortgage of SI 0,000 draws 5 % interest and the farmer pays annually S 600, what is the accumulated value at the end of 21 years of the excess payments of S 100 per annum, accumulated at 5 % ? GEOMETRICAL SERIES AND ANNUITIES 191 3. What annual payment will accumulate at 5 % in 30 years to $ 10,000 ? What annual payment would have to be made on a $10,000 mortgage, to extinguish the debt in 30 years, money worth 6 % ? 4. Find in the tables the annuity for 30 years which $ 10,000 will purchase. 5. What semiannual payment will accumulate in 30 years (60 payments) to $ 10,000, interest being 5 % com- pounded semiannually ? 6. Find the cost of an annuity of $ 10,000 per year to run for 10 years, 20 years, and 30 years, respectively, money being worth 4 %. 7. If a city issues $ 10,000 in bonds, what amount must be set aside annually to accumulate at 4 J interest to redeem t^e bonds at the end of 20 years ? 8. Find the cost of an annuity of $ 500 per annum for 10 years, the first payment to be made 10 years hence, 20 years hence, and 30 years hence, respectively. 9. W T hat premium can you afford to pay on a $ 10,000 bond drawing 5 % to run 20 years, if money is worth 4 / ? What discount should you receive if money is worth 6 / ? 10. What is the present value of $10,000 to be paid 20 years hence, money at 5 % ? 11. W r hich is the better offer for a piece of property, money being worth 5 %, a rental of $600 per year for 20 years, or a price of $10,000? A rental of $600 per year for 10 years, and $700 per year for the following 10 years, or $12,500? Assume no change in the price of the -real estate in 20 years. 12. What sum at 4 % interest should a railroad set aside each year to replace engines worth $35,000, which have an estimated life of 25 years? to replace buildings worth $ 1,000,000 which have an estimated life of 100 years ? 192 UNIFIED MATHEMATICS 13. If a man invests 6100 each year for 20 years, what annuity, for 20 years, can he purchase at the end of the first 20 years, money at 5 % interest ? 14. If a man agrees to take $ 1000 a year for five years for a house originally offered at S 5000, what is the discount when money is worth 5 % ? 15. At 5 % interest what annual payment for five years is equivalent to $ 5000 cash in hand ? 16. Answer questions 14 and 15, assuming that the first $ 1000 is to be paid immediately. 1. Binomial series. The expressions (1 + i) 4 , (1 -f i) 20 , can be developed in powers of i by means of the binomial expansion, ( In particular, if a = 1, -- - 1 1 ^ i Z o n(n l)(n 2)(n 3) to (r 1) factors r _ 1 1.2.3.4 ... (r - 1) This formula expresses a rule for the formation of successive terms of the expansion of (a + x) n or (1 + x) n ; the first term contains a with the exponent n of the binomial ; the second term has as coefficient the exponent, or index, of the binomial, x appears to the first power, and the exponent of a decreases by 1 ; the coefficient of the third term has two factors in numerator and denominator, in the numerator n(n 1) and in the denominator 1 2 ; a appears with exponent 1 less than in the preceding term, and x with exponent 1 greater ; each following term can be obtained from the preceding by introducing one further factor in numerator and denominator and at the same time decreasing the power of a by one and increasing that of x by one ; the further factor in the new numerator is one less than the last one introduced there, and 193 194 UNIFIED MATHEMATICS the further factor in the new denominator is one greater than the last one of the preceding denominator ; the coefficient of the term in x"~ l contains in the numerator (r 1) integral factors from n down, and in the denominator (r 1) integral factors from 1 up ; x appears with exponent r 1 and a with the exponent which added to r 1 makes n, i.e. n r + 1. Illustrations of the binomial expansion. (\o Oit5 n , O , O ^ J. Q a + x) 3 = a 3 + - a*x + ax* + x 3 = a 3 + 3 a 2 x + 3 ax 2 + x 3 . IST 2o 8n 6. (a + a) 15 = a 15 + a u x + ^ ^ lOrif TERM 15 14 . 13 . 12 . 11 . 10 . 9 8 7 1.2.3.4.5.6.7.8.9 Note that it is well in writing the tenth term to begin with x g ; then a enters to the sixth power as in every term of this expansion the exponents of a and x together make 15 ; the denominator contains 9 factors, 1, 2, 3, 9; the numerator contains 9 factors, which should be counted as they are writ- ten ; finally cancellation should be made, giving 5 7 11 13 aW. 15 - 14 . 13 IOTH TERM _ 5 7 11 13 a s x 9 + .... 10TH Note that the powers of a in each term can be dropped, as every power of one equals one. BINOMIAL SERIES AND APPLICATIONS 195 e. Compute to 4 decimal places (1 + .04) 15 . IK. 1A 1 K 1A 1Q . + .04) 16 = 1 + 15(.04) 1-2 v 1-2-3 15-14.13.12 15.14.13-12-11. Q 1.2.3.4 l ' 1.2.3.4.5 .60 second .0291 1.00000 first term _1_ .12 .60000 second term 4.20 5).OQ349 fifth .16800 third term .04 .00070 .02912 fourth term 3).168Q third term 11 .00349 fifth term .0560 .0077 .00031 sixth term .04 .04 1.80092 Ans. .002240 .00031 sixth .02240 10 times 672 3 times .02912 fourth term, to be multiplied by .12 Note here the method of computation given at the left ; each term is obtained from the preceding term ; three new factors of which one is .04 enter into each succeeding term, two in the numerator and one in the denominator ; these three factors after the second term (.60) are 14 X ' 4 or 7 x .04 ; then 13 X - 04 ; 2 o then 12 x - 04 or .12: then U x - 04 which might well be 4 5 treated as .008 x 11 ; then the factor would give the 6 seventh term from the sixth, making about 2 in the fifth deci- mal place. The expansion of may be written as follows : 196 UNIFIED MATHEMATICS in which T 2 , T 3 , T t , T b , designate the second, third, fourth, terms respectively. This type of representation, in which each term is obtained from the preceding, is frequently of use in statistical work and in computation. /. Write the sixth and sixteenth terms of (1 -f- #)". 6-rii TERM 16 TEEM 17-16.15.14.13 17-16... 1.2.3.4.5 1.2... g. What decimal place is affected by the sixth term of (1.06) 17 ? 17 . 16 15 . 14 13 1.2.3.4.5 (.06) 5 ; (.06) 3 = .000216 ; (.06) 4 = .00001296 ; (.06) 5 = .00000078- ; multiply .00000078 by 4, this by 7, this by 13, and then by 17, rejecting any beyond 3 significant fig- ures ; this gives .00000312, .0000218, .000286, and finally .00476. Note that the computation of six terms of (1 + .06) 17 involves not very much more numerical labor than this determination. h. Write six terms of (a - 12 a" (3 x) + ^~ 10 (3 *) - ^'.S* ^ ^ 12.11,1019 _ 12.11.10-9.8 1.2.3.4 1.2.3.4.5 a i2 _ 36 a a? + 2 - 3 3 . 11 a 10 * 2 - 2* 3 3 5 . 11 a?tf + 5 . 3 8 11 a 8 ar - 23 - 3 7 It is not necessary or desirable to perform the multiplica- tion in such an expression as 2 2 3 3 5 - 11 ; such terms, if desired numerically, are usually obtained progressively from preceding terms as in example (e) above. PROBLEMS 1. Expand to 6 terms, (a + x) 6 , (a + a) 14 , (a + 2 x) 10 ' (a - 3 x) 9 . BINOMIAL SERIES AND APPLICATIONS 197 2. Write 5 terms in simplest form (prime factors) of (1 + *) 6 , (1 + a) 14 , (1 + 2 x-) 10 , (1-3 a-) 9 . 3. Compute to 4 decimal places (1 -f .05) 6 , (1 -f- .05)", (1 + -05) 10 , (1 - .05) 9 . 4. Compute to 2 decimal places the value at the end of 10 years of $ 100 placed at interest at 6 % compounded annu- ally ; use 100 X (1 + .06) 10 . How could you use the result obtained to find the value at the end of 20 years ? 5. From problem 3 give the amount at the end of 6, 14, and 10 years, respectively, of $ 256 at 5 % interest, com- pounded annually. 6. Find the value at the end of 6, 14, and 10 years respec- tively of an annuity of 1 per annum, paid at the end of each year, interest at 5 %. Use the formula s^ = ^ -*-* and the preceding results. 7. Compute the values in 3, 4, and 5 by logarithms and compare. 8. Given 2 10 = 1024, find to 1 decimal place (2 + .Ol) 10 . Ans. 1077.7. Find also (2.1) 10 to one decimal place, and check by logs. 9. Find (12.3) 3 to five significant figures. 10. Find the amount at the end of 20 years of $ 100 placed at interest, 3%, compounded semiannually. 11. Write the 8th term of (1 - 3x)" and of (1 -f a?). 12. How many terms in (13 a;) 17 ? Write the middle terms. 13. Write in simplest form the coefficient of & in (1 2 #) 28 . 14. Write the series for (1 + a;) 5 , (1 -f- a;) 6 , (1 -f- a;) 7 , and (1 + x) s . What is the sum of the coefficients ? (NOTK. Substitute 1 for x.) 15. What is the sum of the coefficients in (1 + x) 19 ? 16. Write 7 terms of (1 +Va;) 10 and of (1 198 UNIFIED MATHEMATICS 2. Proof of (a + *)" = a" 4- a"~ 1 jr + ~ a"- 2 * 2 + 1 1 2 mi TKRSI TO A TOTAL OF (r 1) V\> -ioi:s n(n-l)(n-2) , . n(n-l)(n - 2) - r+1 1.2-3 1-234 (r-1) The proof of this expansion for positive integral values of n is effected by the process called mathematical induction. Evidently (a + x) 2 = a 2 + 2 ax + x z , follows the rule ; also (a + a) 3 = a 3 + 3 a?x + 3 ax 2 + X s , follows the rule. By the rule, ,4, ,4-3, ,4.3-2 ,4.3-2.1, ( -f a;)4 = a ^ + _ a3a . + __ a2a .2 + __ __ aa ,3 + j-^- X* = a 4 + 4 a 3 + 6a 2 a; 2 + 4 aic 3 + a; 4 ; by actual multiplication we find the same series, showing that the rule as given holds for n = 4. 1.2-3-(r-2) mi TERM 4. n(n-l)(n-2)-.(w-r + 2) _ r+1 , . 1 - 2 - 3 - (r - 1) multiply by a + x = a + x (a + *)" +1 = a n THE NEW mi TERM n(n - l)(?i - 2) ... (n - r + 3)(n - r + 2) 1.2-3- (r 2)(r 1) n(M _ i )(n _ 2 ) ... (n - r + 3)\ -^j^ BINOMIAL SERIES AND APPLICATIONS 199 Note that these two coefficients of the new rth term have the first (r 2) factors of numerator and denominator the same ; multiply the denominator and numerator of the second term by r 1 and then add the numerators, taking out the common factors, giving NEW mi TERM COMMON REMAINING REMAINING FACTORS FACTOR or FACTOR OP FIRST TERM SECOND TERM I 1.2.3.4.5... (r-1) The new rth term may be written, then, (n + l)(n)(n - l)(n - 2) ... (n - r + 3) (I1+1) _ r+1 x . 1.2- 3. 4. (r-1) the rth term of (a + #) B+1 is formed according to the rule with (>i + 1) substituted throughout for n ; the numerator contains (r 1) factors beginning with (n -\- 1), and the denominator contains the same number of factors beginning with 1. Hence if this expansion assumed for (a + a;)" is correct for any value n, it is correct for a value one greater, n + 1. The theorem is true, by trial, for n = 4 ; hence it is true for n = 5 ; since it is true for n = 5 it is also true for n = 6 ; ... and so for every integral value of n. 3. Binomial series ; any exponent. The equation, n(n-l)(n.-2)(n-8) 4 "1.2.3.4 can be shown by methods of the higher mathematics to hold for all values of n when 1 < x < 1. This means that a series for (1 + x)* can be obtained by substituting n ^ in the above formula. Similarly (l+x)~ 3 and (1 + #)~ 7 , (1 + cc)~^ or (1 -}- )V? can be developed in powers of x, when |a;| < 1, by 200 UNIFIED MATHEMATICS substituting for n the values 3, and 7, and the like in the above formula. These series are of frequent use in statistical work. Thus if money is worth 6% per annum the interest for one half year is not 3 % of 1, since this rate continued for the full year would give at the end of one year in addition to the 6 Jo of 1, the interest on the 3 % of 1 for one half year ; the interest on 1 for one half year is taken to be j=(l + . 06)* -1, and this rate of interest per half year accumulates at the end of two half years a principal of one to (1.06), or is equivalent to 6 % per year. Similarly the effective rate of 6 % per annum means that the interest for one fourth of a year will not be .015 times the principal, but rather (1 + .06)* 1, since this is the rate of interest for one quarter of a year which continued for four quarters will accumulate a principal of 1 to 1.06, since [(1 + .06)*]*= 1.06. In our illustrative problems we will assume that the terms that follow any given term in the expansion of an ex- pression like (1 + x)* are together less than the last term given ; the general proof of the convergence of these series is reserved for the calculus ; however, it is evident here that each new factor, when x is less than 1, diminishes in value and finally the terms are in turn respectively less than the terms of a geometrical series with ratio x ; thus, below in the ex- pansion of (1 -}- .06)*, beyond any given term the terms are respectively less, term by term, than the terms of a geometri- cal series with ratio .06 ; the sum of all terms beyond the fourth term is certainly less than , wherein a is the fourth .94 term, since the corresponding geometrical series even to an infinite number of terms has only this sum, - = Note particularly that the expansion of (1 + x) n for values of n other than positive integers leads to a series which has no termination, i.e. to an infinite series ; this series is valid BINOMIAL SERIES AND APPLICATIONS 201 and has meaning only when | x \ < 1 ; similarly any infinite series given by (a + x) n has meaning only when -- < 1. In our further discussion this limitation will be consistently assumed. As an illustration of the possible absurdity from the point of view of finite series of the infinite series given by (1 + x) n , let us take the fraction - - which may be written (1 a?)" 1 , and developed by the binomial theorem as, l + a3 + # 2 + ar 5 + :E 4 + 2r i -|- .... For values of x numerically less than unity this series is valid, but if you put x equal to 3, or 5, or other value greater than unity, you obtain an absurdity. 4. Illustrative problems. a. Compute (1 + .06)^ to 5 deci- mal places. (1 + .06)* = 1 + i(.06) - 1 (.06)2 + ^ (.06)3 + ... = 1 + .03 - .00045 + .0000135 + ... = 1.02956. If the value of (1.0ft)* were desired to eight decimal places, progressive computation would be desirable, appearing as follows : Ti = 1 4) .0000135 T 2 = i(.OO) TI = .03 .000003375 = -. 00045 2 \ 2 / ' .000016875 ; = +. 0000135 - 03 2 3 .00000050625 .042 1012 2024 .00000002125 T 5 = - ^ . 'I* T 4 = - .000000506 1 -030013521 - .000450507 T 6 = _ I . M T& = + .000000021 1.029563014 r 7 = -? . ^ T 6 =- .000000001 w 202 UNIFIED MATHEMATICS The letters TI, T 2 , T 3 , T 4 , represent the first, second, third, fourth, and succeeding terms ; each term is obtained from the preceding, intro- ducing the new factors. 6. "Write six terras of the series for (1 + x)^ and for (1 as)*. 1 - 1 3 i -i -3 -5 1 riJ.^S.jzJi -7 2' 2 ' 2 ' 2 ' 2 1.2.3-4.5 - = l + |x- Jx- r 2 -lx. T,-Jx. T 4 -^x- T, ---- . Alternate terms after the second are positive and negative. I. =1.^3 1-1-3 -5 2 ' 2 ' 2 2 ' 2 ' 2 " 2 1/-J\ 1 2\ 2 / 1.2.3.4 1-1-3 -5 -7 2 ' 2 ' 2 ' 2 2 1.2.8.4.6 = l-ix + Jx- Every term after the first is negative. c. Compute V.98 to 5 places. - .02) = 1 - J(-02) + ~ (.02)2 _ ~- (.02)3 + ... 1*3 1*8*3 = l _ .01 - .00005 - .0000005 - = .9899495 or .98995 to 5 places. d. Compute the cube root of (1012) to 6 significant figures. (1012)5 =(1000)3(1 + .012)^ = 10(1 + .012)i (1 + .012)* = 1 + K-012) + * ( \ ~ ^ (.012)2 + Ki - !)(^ ~ 2 ) (.012)3 + ... 1 2t 1 o = 1 + .004 - .000016 + .0000001 ... = 1.003984. (1012)' = 10 x 1.003984 = 10.03984. e. Compute (1.05) 5 to 4 places ; treat this as 1.05 x (1.05)^ ; .05 may be taken as ^. Check by logarithms. BINOMIAL SERIES AND APPLICATIONS 203 /. Compute the square roots of 26 and 30 to 4 places. V26 = 26^ = (25 + 1)2 = 5(1 + '&)$ = 5(1 + .04)* = 5(1 + .02 - .0002 + .000004) = 5(1.019804) =5.09902. 30^ =(25 + 5)2 = 5(1 + .2)* = 6(1 + .1 - .005 + .0005 - .0000625) = 5(1.0954375)= 5.47718. Check roughly, using logarithms. g. Compute (1.05)~ 7 to 5 significant figures. (1.05)-* = 1 - 7(.05)+- 7 (- 8 )(.06) + ~ 7( 1 ~^ ( 3 ~ 9) (.05)3 I ---- Y _---- r y , 1-2-3-4 1-2-3.4-5 = l _ .35 + .07 - .0105 + .0013125 - .0001444 + .0000144 = .71068. 5. Historical note. The binomial theorem as applied to (a + b) 2 and (a + 6) 3 was well known to Euclid (320 B.C.) and other early Greek mathematicians. The great Arabic mathe- matician and poet Omar al Khayyam (died 1123 A.D.) extended the rule to other positive integers. In China there appeared in 1303 a work containing the binomial coefficients arranged in triangular form, the so-called Pascal (1623-1662) triangle of coefficients. 11111111 1234567 13 6 10 15 21 1 4 10 20 35 1 5 15 35 6 21 1 1 1 121 1331 14641 15 10 10 51 1 6 15 20 15 6 1 1 1 7 1 PASCAL'S TRIANGLE CHINESE FORM OF TRI- OF COEFFICIENTS, ANGLE OF COEFFI- PRINTED 1665. CIENTS. m 121 1331 14641 1 5 10 10 51 1 6 15 20 15 6 1 STIFEL'S (1486-1567) TRIANGLE OF COEFFI- CIENTS, 1544. The general rule for any exponent -- was first discovered n by Sir Isaac Newton, and made known in a letter of date Oct. 204 UNIFIED MATHEMATICS 24, 1676, to a friend named Oldenburg. It is of interest to note that Newton wrote each coefficient in terras of the coeffi- cient immediately preceding, following the lines indicated in our numerical problems above. The complete proof for the general case, any real or complex imaginary number, was finally effected by a brilliant Norwegian mathematician, Abel (1802- 1829), in 1826. PROBLEMS 1. Find the coefficients of the first five terms of (1 + x)* and (1 x)* and use them in the next two problems. 2. Compute to 3 significant figures the square roots of 27, 28, and 29 as 5(1 + .08)*, 5(1 + .12)*, and 5(1 + .16)*, respec- tively. 3. Compute the square roots of the first eleven integers, to 5 places, taking \/2 as ^(196 + 4)* |(i + ^i, V3 as 2(1 -)*, V5 as 2(_l + i)*,_V6_as (4+ 2)1 = 2(1 + .5)*, V7 as 1(25 + 3)*, V8 as 2V2, VH as (100 - 1)*. Check the first four significant figures by logarithms. 4. Write 5 terms of (1 -+- ), and of (1 #)*. 5. ITse these to compute the values of A/7 and ^9, as 6. From the cube root of 9 find the cube root of 3. 7. Find the cube root of 6. 8. Using the cube roots of 6 and 3, find the cube root of 2. 9. Compute (1 + .05)* to 5 places. 10. Write 5 terms of (2 - 3 *)~*. 11. Write the 6th term of (2 - 12. Find the value to 4 significant figures of expanding (1 - .02)~*. ^ ~ - 02 BINOMIAL SERIES AND APPLICATIONS 205 13. Expand in powers of i to 4 terms - and - What is the approximate percentage error in using (1 i) and (1 -f i) as multipliers, respectively, instead of - : and - ., J- *T~ 1 JL ~~* 1 when i = .01, .05, .005, and .5, respectively ? 14. Find the fifth root of 35 correct to 2 decimal places. What is the shortest way ? Compute the root to 5 places. What method can you use ? 15. Write the term containing x 6 in the expansion of (1-3*)-*. 16. Write the first five terms of (10 + .3) 8 ; of (10.3) 6 ; of (10.3)~ 6 and give the value to 4 significant figures. 17. Time yourself on writing and simplifying 10 terms of the following five expansions : (1 + x) *, (1 x)' 5 , (1 2 )"*, 18. Compute correctly to 4 significant figures, using the formulas for - and -- Time yourself on the exercise. 1+i 1 - i 18 27 54 62 68 .98' 1.02' 1.03' .99' 1.05 19. Compute the following to three significant figures, tim- ing yourself : (1.06)*, (1.06)T2, (1.05)*, (1.06)*, (1.10)*, (1.06)-*. CHAPTER XIII RIGHT TRIANGLES 1. Right triangles. To apply our trigonometric work to the numerical solution of right triangles place the triangle under consideration in quadrant I in proper position to be able to read the trigonometric functions of one acute angle. x = rcos a = rsin p y = r sin a = r cos p tan a = - = cot p --c Fundamental formulas of the right triangle cot a = - = tan B y a + B = 90. The equation x = r cos a may be written cos x a = - r or cos a f as occasion demands ; similar transformations are to be effected upon the other equations given. Given a with x, y, or r; or (3 with x, y, or ?; or two of the lengths ; these formulas enable us to solve the right triangle completely for the remaining three parts. The student is advised to draw the figure to scale on coordi- nate paper, using a protractor to lay off correctly to degrees the angles given, before attempting to apply any formulas ; then write the required equations directly from a consideration of the figure, and not by attempting to memorize t;he solutions for the different types of problems. The lengths as given graphically serve as a check upon the values obtained. 206 RIGHT TRIANGLES 207 2. Right triangles. TYPE I. Given hypotenuse and one angle, i.e. a and r, or /3 and r. A guy wire 168 feet long reaches to the top of a tall chimney, making an angle of 37 with the ground. Find the height of the chimney and the distance from the supporting peg to the foot of the chimney. \J! / i - ^ __5 i ft? :"" Gt ,/fl + ' slid X , a OT- - "jffS82 I Jf * -1 r =m ~\ m so^po-ifto-iji^q sills Solution. First draw the fig- ure, as indicated, using ^ inch to represent 30 units. y - 168 sin 37 x = 168 cos 37 Check, y = x tan 37 Hypotenuse and one angle given Using natural functions there are here three problems in multiplication. The logarithmic solution is as follows : log 168 = 2.2253 log 168 = 2.2253 log sin 37 = 9.7795-10 log cos 37 = 9.9023 - 10 logj/ = logx= 2.1276 x = 134.2. 2.0048 y = 101.1. Check, log x = 2.1276 log tan 37 = 9.8771 - 10 log y = 2.0047. Compare with preceding value of log y ; a dis- agreement in the fourth place is permissible. It would not affect the fourth significant figure of y in this case ; nor would the measurements of height of chimney and length of guy wire be made with greater accuracy than to one tenth of a foot, x 2 + y 2 168 2 could be used as a check. .Ill Angle and side given 3. Right triangles. TYPE II. Given one leg and an angle. If a telegraph pole is 34 feet high, and the supporting wire makes an angle of 62 with the ground, find the length of the wire and the distance from the foot of the supporting peg to the foot of the pole. 208 UNIFIED MATHEMATICS The corresponding formulas are as follows : x = 34 cot 62 34 Check, x-r cos 62 or 342 = r 2 - x 2 sin 62 log 34= 1.5315 log cot 62 = 9.7257 - 10 logx = 1.2572 x = 18.08. log 34 = 11.6315- 10 - log sin 62 = 9.9459 - 10 logr= 1.5856 r = 38.51. Check, logr = 1.5856 + log cos 62 = 9.6716 10 logx = 1.2572, checks. 4. Right triangles. TYPE III. Given a leg and the hy- potenuse. Let the side a = 341 and c, the hypotenuse, equal 725. Evidently, sin a = f f^, b = 725 cos a. log 341 = 12.5328 - 10 log 725 = 2.8603 log sin = 9.6725 10 = 28 4'. Check, b = 341 cot a. log 725 = log cos a = Hypotenuse and one side given Graphical solution gives a rough check to two significant figures. 2.8603 9.9457 - 10 = 2.8060 6 = 639.7. Check, log 341 = 2.5328 log cot a = .2731 log b = 2.8059. Compare with above value log b =2.8060 as a check; the error of 1 here is in- evitable with 4-place log- arithms. Note that on the small graph only two places are accurately representable. 5. Right triangles. TYPE IV. Given the two legs. Entirely similar except that the initial formula is for tan a instead of sin . Note that commonly in lettering right triangles x and y or a and b are RIGHT TRIANGLES 209 used for the legs, r or c for the hypotenuse, a and p for the angles at A and .B, opposite a and b respectively. In Type III if a and c are nearly equal we may avoid the use of the sine of the angle near to 90 by computing the other side using the formula & 2 = c 2 a 2 = (c )(c + a). Thus if a = 718, c = 725, 6 2 = (725 - 718)(725 + 718) = 7(1443) =10101. b = 100.5, either by inspection as in this case, or from a table of squares, or by logarithms. PROBLEMS Solve the following right triangles by logarithms : 1. Given r = 240, = 30 10'. 2. Given a = 368, a = 30 14'. 3. Given a = 368, r = 579. 4. Given a = 368, 6 = 275. 5. Solve for the missing parts the following ten problems, using logarithms ; time yourself ; the exercise should be com- pleted within 30 minutes. a. Given r = 186, a = 84.3. e. Given a = 930, a = 24. b. Given a = .394, b = .654. /. Given 6 == 184, a =55 15'. c. Given a = 2.89, = 68 24'. g. Given r = .0936, b = .0418. d. Given b = 706, a = 70 10'. h. Given 6 = 3.24, ft = 86 14'. i. Given b = 878, a = 48 19'. j. Given r = 8.4 x 10 6 , ft = 34 16'. 6. Area. In computations of functions involving measured and computed values, measured values are taken, as far as possible, in preference to computed values. The computed value involves not only the inaccuracies or errors of measure- ment, but also the errors of computation, the inevitable errors of computation with approximate numbers as well as the avoidable errors. Among the following formulas for the area of a right triangle the student should select, in accordance with 210 UNIFIED MATHEMATICS the principle mentioned, the formula to be used in each prob- lem involving a right triangle. A = % ab = \ a 2 cot a = \ 6 2 tan a = \ c 2 sin a cos a. 7. Applications. In the application of the solution of right triangles to practical problems we find that the difficulty is frequently a matter of ter- minology rather than of prin- ciple. The student is urged to acquire some real famili- arity with the industrial and scientific application of the -rA 1? Common terms relating to angles Dip depression elevation bias departure. principles explained. The terms " elevation," " depression," " dip," " de- parture," and " bearing," all refer to angles. Thus in the figure ABC, if A C is in the direction of the sun or if C represents the top of a mountain viewed from A, then angle BAG is termed the angle of " elevation " of C as viewed from A : if the observer is at C, on a mountain or in an airship, HCA is the angle of " depression " ; if A represents the horizon as viewed from C, then HCA is called the " dip " of the horizon. If CA repre- sents a vertical section of a vein of coal, the angle HCA or BAG is called the " dip " of the vein ; in navigation if AB represents east, then angle BAG represents the " departure " north of the line AC, whereas in surveying the angular de- flection from north or south is given as the " bearing." Frequently some function of an angle is given from which the angle must be de- termined. The pitch of a roof is given as the height divided by the span, whence the corresponding slope angle of the roof is 6, given by Pitch = - 2s RIGHT TRIANGLES 211 tan = -, s where h is the height and 2 s is the total span. The slope of a railroad is commonly given as so many (K) feet of rise in 100 feet horizontally ; this gives the slope angle from tan 6 = = h %. 1UO A spiral thread winds about a cylinder advancing a height h, called the " lead," in one complete turn ; the circumference , . I , Full size representation of a one-inch cylindrical screw The " lead " is 2 % of an inch. of the cylinder is the base and the "lead" is the altitude of a right triangle which may be regarded as wrapt about the cylinder to give the spiral. The angle a made with h by the spiral line is called the angle of the spiral ; evidently tan a = -^- h Circumference, AC, and length of one spiral, AB, of the above one-inch cylindrical screw 212 UNIFIED MATHEMATICS PROBLEMS 1. A standpipe subtends an angle of 4 at the eye of an observer ; if its height is 280 feet above the level of the eye, find its distance from the observer. 2. If the diameter of the standpipe in 1 subtends an angle of 15' at the eye, what is the diameter in feet ? Suppose that the angle at the eye lies between 10' and 20', what range of diameter would these values give ? 3. The shadow of a flagstaff 60 feet high is 48 feet long. Find the angular elevation of the sun. 4. Using trigonometric functions, find the height of a building which at the same time casts a shadow 87 feet long. 5. Find the lengths of the circle of latitude and the circle of longitude through your home city. 6. When the sun is directly over the equator, the latitude of any place on the earth's surface from which the sun is visible is the angle between the zenith line (the vertical) and the line to the sun, when the sun is on the me- ridian. Find the shortest length of the shadow of a pole 100 feet high, at noon, latitude 40 N., and also for latitude 42 18' N. Q Zenith distance representing latitude Sun on celestial equator, di- rection OE. OP is zenith direction of P. 7. Compute the diameter of a circle circumscribed about an equilateral triangle of side 40. 8. Find in a circle of radius 486 cm. the chords of angles of 30, 60, 45, 90, 120, 72, 68. 9. For any angle a, find the chord and the chord of half the angle in terms of the radius r. Apply the latter formula to obtain the results of problem 8. RIGHT TRIANGLES 213 10. A pendulum of length 34 inches swings between two points 10 inches apart ; compute the arc of the swing. If this is a seconds pendulum, passing the vertical once every sec- ond, what is the velocity of the pendulum bob? Find the chord of this arc, and the difference between the chord and the arc. 11. A circular arch over a doorway is to be 4 feet wide and 20 inches high ; compute the radius. Compute for heights of 10 to 24 inches by 2-inch intervals. 12. Given the radius 10 feet and the span 4 feet of a circular arch. Compute the height. Compute for spans of 2 feet to 20 feet, by 2-foot intervals. 13. Adapt the preceding results to a radius of 8 feet. How closely would interpolation give correct results ? discuss by considering the problem graphically, 14. In a circle of radius 100 inches, compute to one decimal place the lengths of sides and the perimeters of regular in- scribed polygons of 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 sides. Time yourself on the exercise. State the general formulas involved. 15. Compute perimeters of regular circumscribed polygons of 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 sides in a circle of radius 100. Time yourself on the numerical work ; 30 minutes is ample time. 16. Compute the perimeter of a regular inscribed polygon of 96 sides, and of a regular circumscribed polygon of 96 sides, radius 100. How does the circumference compare with these two values ? Archimedes computed these lengths by plane geometry methods and so found ?r to lie between 3| and 3|^. Check his result. 17. Frequently arches of bridges are circular segments; find the radius of the circular arch of the famous Rialto in Venice (see illustration, page 225). The width of the arch is 95 feet and the height is 25 feet. Draw the graph of the arch to scale. 214 UNIFIED MATHEMATICS 18. One of the largest masonry bridges in the U. S. is the Rocky River bridge at Cleveland ; the height of the circular arch is 80 feet and the span is 280 feet. Find the correspond- ing radius. 19. Find the angle of the spiral represented in the above diagram of the one-inch cylindrical screw. 20. Find the angle of a cylindrical screw of diameter \ inch which advances ^ of an inch in one complete turn. 21. A twelve-inch gun has a muzzle velocity of approxi- mately 2500 feet per second (f.s.). The velocity is tested by Determination of velocity of a projectile One screen is at the muzzle of the gun. electrical means ; screens are placed at a known distance apart and the projectile in passing through the screens breaks successively two electrical circuits which serve to give the time of flight of the projectile to thousandths of a second in passing through the known distance. The apparatus may also be used to determine the angle of elevation of the larger guns. In the figure TAT represents the axis of trunnions of a twelve-inch gun ; AM along the axis of the barrel is 25 feet ; MS is 180 feet ; the one screen is over the muzzle and the other screen is at a height of 94 feet above the axis of trun- nions. Determine the angle of elevation of the gun and reduce to " mils." Find the horizontal distance MQ between RIGHT TRIANGLES 215 muzzle and screen and the vertical distance between muzzle and screen, QS ; find the time of flight of the projectile, assuming 2500 f.s. as velocity ; find " horizontal velocity," v t , and " vertical velocity," v v , by dividing M Q and QS re- spectively by this time of flight. The vertical velocity v v divided by 32.2 gives approximately the time in seconds that the projectile will continue to rise ; find this time ; the position of the projectile after this interval of time is given approxi- mately by the product of horizontal velocity, v x , multiplied by the time, as horizontal distance from the gun, and by vertical component of velocity multiplied by the same value of t less 16.1 multiplied by t 2 , as ordinate. The equations are x = v r t, y = v v t- 16.1Z 2 . What error is possible in the angle measured if the height of S is given only within one foot ? The aim is directed at a point two feet below the top of the screen, as, in general, there is a slight " jump " due apparently to the explosion. Estimate the jump in degrees and minutes, and in " mils " if the pro- jectile hits the top of the screen. 22. When two screens are used with a large gun the dis- tance between screens is sometimes measured by taking equal Distances of screens from muzzle, M, determined by right triangles Two screens spaced 100 feet apart (horizontally). distances MH and ME at right angles to the line MS^Sz and measuring the angles MHS^ MHS 2 , MESi, and MES* Note that the screens are 20 to 100 feet in the air on tall standards, 216 UNIFIED MATHEMATICS making it inconvenient to measure the distance with a steel tape. Assuming that the distances ME and MH are taken as 20 feet and that the angles MHS l = MES^ = 70 10', and that angle MHS Z MES 2 = 82 34', compute the distance MS lt MSz, and S^S*. If the screen Si is at an elevation of 34 feet and the screen S 2 is at an elevation of 83 feet, compute the angle of elevation of the gun, and the height of the muzzle above the plane of its axis if the muzzle is 25 feet long from the axis. Note that the relative positions of the screens are usually determined by two observers in towers whose distance apart is fixed ; these observers record positions of muzzle and each of the screens. 23. If a stick of length 12 units casts shadows of lengths 4, 6, 8, 10, 12, 15, 18, 30, and 40 units respectively, determine the angle of inclination of the sun. For angles of inclination of 10 to 20 by degrees, determine the corresponding shadow length to tenths of one unit. This type of table was the first appearance of the cotangent function as direct shadow ; it appeared as early as 900 A.D. in the works of the great Arabic astronomer Al-Battani. 24. The pitch of a roof is given by the vertical height li, from the point C to M, on the diagram, divided by the span, 2 s ; thus ^ pitch is a 45 slope. Find the slope angle of a roof of % pitch, of \ pitch. If 2 s is given as 48 feet, find the length of the rafters in each of the roofs mentioned. 25. In a roof of span 62 feet find to the tenth of an inch the lengths of the rafters if the roof is inclined at 30, 40, 42, 45, 53, and 60. In each case determine the effect upon the length of the rafter of an error of one degree. 26. Find the pitch and the angle of inclination of the roof represented in the diagram above. Pitch equals when span is 2 s 2s RIGHT TRIANGLES 217 8. Railroad curves. In so far as possible the track of a rail- road is laid out in straight lines. Wherever the direction of the track is changed a curved line of track is- introduced lead- Simple curve at a turnout on a railroad track ing from the one straight track to a second; these straight portions of track must be tangent to the curve which joins them, and so they are commonly designated simply as tangents. Let AV and VB in the figure represent two such tangents, meeting at a point V, called the vertex ; the exterior angle XVB is called the deflection angle, and is usually designated by /. A single circular arc, radius It, which joins two tan- gents is called a simple curve and is designated in American rail- Chord 100 feet; arc approxi . road practice by the number of de- mately 100 feet grees D at the center of the circle subtended by a chord whose length is 100 feet, the length of one chain used in surveying. On a simple curve the lengths 218 UNIFIED MATHEMATICS of two consecutive tangents, from intersection point to the circle, i.e. AV and VB, are equal ; this length is called T. The angle D is commonly given only in degrees and half- degrees. Relation between D and R. Let PB on the figure represent a chord of length 100 feet ; drop the perpendicular from A, the center of the circular arc of radius R, bisecting PB. Evi- dently sin = , whence R = . Now for any angle 2 R D sin 2 up to 4 the sine differs numerically from the angle expressed in radians by less than .1 of 1% of itself; hence you may replace sin by ^- , the value of in radians, with an 2 2 180 2 error of less than .1 of 1% when D is any angle up to 8. Note that the error is less than o feet in 5000 ; the circular measure of the angle is larger than the sine so that the error will be a deficiency. 18000 i\ ^ - TTD gives the radius. Relation between I, R, and T. On our figure in the right triangle (MFthe angle AVO is 90 - |, and the angle AOB is evidently equal to the deflection angle /. - , 2J AV T' whence R = T cot , and T = R tan f Evidently the radius R can be expressed in terms of the "degree" D of the curve, giving new formulas involving D, T, and 7. Elevation of outer rail. In turning a curve a railroad train tends to leave the track, due to the tendency of any moving body to continue its motion in a straight line. To keep the train on the track the flanges alone are not sufficient, but the RIGHT TRIANGLES 219 outer edge must be elevated. The formula for ordinary speeds, giving number of inches of elevation, is e = -3- , wherein g is 32 R the gage of the track in feet, v the velocity of the train in feet per second, and R the radius of curvature in feet. PROBLEMS 1. What radii have railroad curves of 8, 7, 6, 5, 4, 3, 2, and 1, respectively ? 2. If a railroad curve is built with the radius of 2640 feet, compute D in degrees. 3. On a circular track of 100 miles' circumference what would be the number of degrees ? 4. On English and continental railroads the curvature is usually given by the length of the radius ; find the number of degrees, American D, corresponding to radii of 8000, 5000, 4000, 3000, 2000, 1000, 800, 600, and 400 feet, respectively. Do not compute beyond minutes. Find D for radii of 300 meters, 1000 meters. 5. Compute e, elevation of outer rail, for g = 4 feet 8.5 inches, standard gage on American railroads, when v = 60 miles per hour, and R = 800, 1000, 2000, 4000, and 5000 re- spectively. Compute for a one-degree and for a two-degree curve. 6. Given that two portions of straight track diverge at 22 14', and that the tangent distance is to be 300 feet, com- pute R ; find R for T, the tangent distance, equal to 200, 250, and 350. Find the corresponding values of D. How could you determine the length of T, approximately 300, so that D will come out in degrees and half-degrees ? 7. Compute R when T = 400, 500, and 600, respectively, the deflection angle being 60 ; similarly when I = 30. 8. Compute e for g = 4 feet 8.5 inches (4.71 feet), standard gage, v = 60 miles per hour, and curves of 1, 2, 5, 6, and 8> respectively. CHAPTER XIV THE CIRCLE 1. Formulas. a- 2 + 2 (x-hy+(y-k == r\ \x h = r cos 6, Parametric equa- = r 2 . | y Jc = r sin $. tfons of the circle. + 2/i 2 = r 2 ; X2 2 + y 2 2 = r 2 ; x 3 2 + y 3 2 = r 2 ; x 4 2 + 2/4 2 = " 2 ; x 2 + y 2 = r 2 . For any point P (x, y) on a circle of radius 10, center the origin, we have the relation, a;2 + 2/ 2 = 100, 220 THE CIRCLE 221 which is the equation then of a circle of radius 10 and center at the origin (0, 0). This equation is obtained directly from the distance formula ; x 2 + y 2 = 100 expresses the fact that the distance of the point (x, y) from the point (0, 0) is 10 ; any point (x, y) which satisfies this equation is at a distance 10 from (0, 0) and any point at a distance of 10 units from O (0, 0) satisfies this equation. The locus of this equation, then, is the circle of radius 10 and center (0, 0). The formula may readily be verified on the figure ; take P any point on the circle, drop PM a perpendicular to the a?-axis. Then OM 2 + MP 2 = OP 2 , in any one of the four tri- angles, representing any possible position of P. Herein OM and MP must be regarded initially as positive quantities, since the formulas of plane geometry were applied only to positive lengths. However OM, as a positive length = #, where the negative sign is taken for points in II and III and MP=y, where the negative sign is taken in III and IV, whence sub- stituting in OM* + MP 2 = OP 2 you have x z + y* = 10 2 . For a circle of radius r the equation x 2 + y 2 = r 1 is satisfied by any point P(x, y) which is upon the circle, for OP will equal r, and every point which satisfies the equation evidently lies on the circle. Hence, by definition, the locus of x 2 + y 2 = r 2 is the circle of center (0, 0) and of radius r, for every point on the circle satisfies this equation and every point which satis- fies the equation lies upon the circle. These two conditions must be fulfilled in order that any given curve may be desig- nated as the locus of a given equation. In other words, the given curve must include all points whose coordinates satisfy the equation and must exclude all whose coordinates do not satisfy the given relation. Similarly the two equations : x = 10 cos 6, y = 10 sin 0, give for every value of 0, called a parameter, the coordinates of a point which lies upon the circle. The locus of this pair 222 UNIFIED MATHEMATICS of equations is the circle of radius 10. Thus the ten values of 0=0, 30, 45, 60, 90, 120,_150, 180, 210, and 330, give' the ten points, (10, 0), (5\/3, 5), (oV2, 5\/2), (5, 5V3), (0, 10), (- 5, 5V3), (- 5V3, 5), (- 10, 0), (- 5V3, - 5), and (+ 5V3, 5), which lie upon the circle. Intermediate values Circle of radius 10 ; units are eighths of an inch f X = 10 cos 0. = 100, or y = 10 sin 6. can readily be obtained using the tables of sines and cosines. The two equations together constitute the equations of the circle in parametric form, a type of equation of particular im- portance in applied mathematics. If desired, we may eliminate as follows : squaring and adding gives a- 2 + f = 100 (cos 2 8 + sin 2 6), or x 2 + y>- = 100 (since sin 2 + cos 2 = 1), THE CIRCLE 223 a relation independent of 0. But for many purposes it is more convenient to keep the equations in parametric form. For the distance from any point C (Ji, A:) to a point P (x, ?/) we have found the formula d = V(x ft) 2 + (y k)- ; all points (x, y} which satisfy this equation for a given value of d, and for (h, k) a fixed point, lie upon a circle of which (h, k) is the center and d is the radius ; no point not on the circle satisfies this equation. (x ft 2 ) + (y k} 2 = r 2 is then the equation of a circle of center (h, k) and radius r. Any equation which can be put into this form represents a circle, for it expresses the fact that the distance from any point (x, y) whose coordinates satisfy the given equation, to the fixed point (h, k) is constant and equal to r. In parametric form, the two equations representing the circle with center (ft, k) and radius r are written : x ft = r cos 0. y k = r sin 0. If is given values the corresponding values of x and y determine points upon the circle (x ft) 2 + (y fc) 2 = r 2 . Illustrative problem. Find the equation of the circle of radius 5 ; center (3, 7). By the distance formula, taking (x, y) as any point on the circle, (z- 3)2+(y + 7)2 = 25, or z 2 - 6 x + y 2 + 14 y - 33 = 0. In parametric form the equations of this circle are, x 3 = 5 cos 0. y + 7 = 5 sin 0. 2. Reduction to standard form. Any equation of the type, x 2 + / + 2 Ox + 2 Fy + C = 0, or Ax 2 + Af + 2 Ox + 2 Fy + C = 0, represents a circle. The center and radius are determined by completing the square, as in the illustrative problem below. If the expression for the radius is zero, the circle reduces to a point ; if it is negative the circle is imaginary. 224 UNIFIED MATHEMATICS Illustrative problem. Find the center and radius of the circle, This equation represents a circle since it can be put into the form of a circle, as indicated herewith : 2(x2 + 3 X + f ) + 2(7/2 - J y + f |) = 15 + + -V. 2(x + |)2 + 2(j/ - |) 2 = *jp. (x + I) 2 + (y ~ I) 2 = - 6 5 (or 12.81). This equation states that the point (x, y) is at the distance -- from 4 the point ( - f , -J) ; this equation represents a circle with the center (-,!) and radius - - or - - or 3.58. 4 4 PROBLEMS Find the equations of the following circles : 1. Center (3, 4), radius 5. 3. Center ( 4, 0), radius 4. 2. Center (0, 0), radius 10. 4. Center (- 6, 6), radius 6. 5. Center (6, 8), radius 10. 6. Draw the circle of radius 10, center (0, 0) and estimate carefully its area on the coordinate paper. Find the centers and radii of the following circles ; time yourself ; the eight problems should be completed numerically within 12 minutes. 9. x 2 + y 2 - 39 = 0. 12. 2 z 2 + 2 y 2 - 5 x + 7 y 15 == 0. 13. 3 a; 2 + 3 ?/ 2 - 15 x + IT y + 9 = 0. 14. x 2 + 6 x + y 2 -10 = 0. 15. Draw the graphs of the preceding 8 circles, using only one or two sheets of graph paper; time yourself, keeping a record of the time. THE CIRCLE 225 16. Given x = 5 cos 0, y = 5 sin 6, locate 16 points on the curve, using the values sin = 0, sin 30 = .5, sin 45 = .707, sin 60 = .866 for these and related angles. 17. Given x = 5 + 5 cos 0, y = 3 + 5 sin 0, locate 16 points on this circle. 18. When 8 = 37, 43, 62, 80, and 85 find x and y in the preceding problems. 19. Through what point on the circle x 2 + y z = 25 does the radius which makes an angle arc tan 2 with OX, pass ? Photo by H. J. Karplnskl The Rial to in Venice A famous circular arch, 95 feet wide by 25 feet high. 3. To find the intersection of a line with a circle. Tangents. The intersections of the circle, x z + y z = 100, with any line as y = x + 5, are represented by the solutions of the two equa- tions regarded as simultaneous. Six problems are given here. 1. a 2 + ?/ 2 =100, 2. a 2 + y 2 = 100, y = x. y = x + 5. 226 3. UNIFIED MATHEMATICS 00, 5. y = x.+ 10. 4. x 2 + y z = 100, y = a; + 16. 6. a 2 + 2/2 _ 100, y = a -f fr. Solving, by substitution in each of the six cases indicated above : Olr Graphical solution, determining intersections of the circle, x 2 + y 2 with various lines of slope 1 1. gives 2 x 2 = 100, x 2 = 50, x = \/50 = 7.07, y = 7.07 ; 2. gives2x 2 + 10x + 25 = 100, x 2 + 5x-37.5 = 0; x = - 2.5 V6.25 + 37.5 = - 2.5 6.61 = + 4.11 or -9.11, i/ = 9.11 or -4.11; 3. gives x 2 + 10 x = 0, x = or 10 (by factoring, simplest), y = 10 or ; THE CIRCLE 227 4. gives 2 x 2 + 32 x+ 156 = 0, x 2 + 16x + 78 = 0, - 78 = _ 8 + V-14 = imaginary values, not any scalar values of x; 5. gives 2 x 2 - 16x - 36 = 0, x 2 - 8 x - 18 = 0, x = 4 V34 = 4 5.83 = 9.83, or - 1.83, y = 1.83 or - 9.83 ; 6. gives 2 x 2 + 2 fc x + (k 2 100) = 0. _ - 2(fc 2 - 100) Very evidently the solutions of 1 to 5 are all included under the solution 6, as special cases. Geometrically the lines of slope 1 are divided by the circle into three classes, viz. (a) those which cut the circle in two distinct points ; (&) those which do not cut the circle ; and (c) those which are tangent to the circle, or cut the circle in two coincident points. Evidently lines in 1, 2, 3, and 5 belong to the first class ; the line in 4 to the second class. To determine the tangents one must find the value of k for which 200 k 2 = 0, as only when 200 k 2 = are two points whose abscissas are given by - k + * V200 - fc2 and - * - - V200 - fc 2 coincident. In this a a a u case, k = 14.14, the lines y = x 14.14 are tangent to the circle a; 2 + y 2 = 100. The abscissa of the point of tangency is A; T 7.07, since it equals --- ^ KULE. To find the tangent with given slope to a given circle write the equation of the family of lines of the given slope, y = mx + k, and solve for the points of intersection with the circle; get the condition that the two points of intersection should be coin- cident. This gives the value of k for which the line y = mx + A: is tangent to the given circle. NOTE. The method will apply to any curve of the second degree. 228 UNIFIED MATHEMATICS 4. Circles satisfying given conditions. To find the equation of a circle which satisfies given conditions it is necessary to use the analytic formulas which we have derived combined with the geometric properties of a circle. In general call (h, k) or (x, y) the center of the circle and r the radius ; sketch the lines and points which are given and indicate roughly the probable position of the desired circle ; solve the problem geometrically if possible, or indicate the solution, and express the geometrical facts in algebraical language by using the preceding formulas. Circle through three points Determination of center by perpendicular bi- sectors of chords. 5. Illustrative problems. Find the equation of the circle through A (1, 2), B (0, 8), and C (7, - 1), (1) using the distance formula, (2) using the per- pendicular bisector of the line joining two points, (3) using the general equation (X fc) 2 +(y &) 8 =r* which may represent any circle, and (4) using the general equation Ax 2 + Af + 2Gx (1) Call the center P(h, *), then PA=PB, PB - PC, and PA=PC. The distance from A to P equals the distance from B to P, whence by the distance formula, - 1)2 + (* - 2)2 = V(h - 0)2 + (k - 8) 2 ; THE CIRCLE 229 Similarly, v II. ^/(h- l)2+(*-2)2 = v'(h - 7) expresses analytically the fact that PA = PC ; and III. V(&-0) 2 + (Jk-8) 2 = \/(A-7) 2 + (Jfc + 1)2 that PS = PC. Since equation III is derivable from I and II, it adds nothing new ; any two of these equations are sufficient to determine (7t, &) the center. Squaring in each and combining terms we obtain from I, 12 k - 2 h - 59 = 0, a straight line which is the locus of all points equidistant from A and B ; and from II, 4 A - 2 fc - 15 = 0. Solving, we obtain the one point which is equidistant from A, B, and C, h = 6.77. k = 6.05. _ _ _ r = V(6.77 - I) 2 + (6.05 - 2)2 = V(33.29 + 16.40) = V49709 = 7.05. The circle is (x - 6.77) 2 + (y - 6.05) 2 = (7.05) 2 . (2) The center of the circle is the intersection of the perpen- dicular bisectors of the sides ; finding the slopes of the sides, the mid-points, the slope of the perpendicular to each side, the equations of the perpendicular bisectors of AB and AC are found (point-slope) to be 12 y - 2 x - 59 = 0. 4x-2y -15=0. Examination shows that these are precisely in x and y the equations obtained in our first solution in h and k and from this point the solution proceeds as in (1). The student should explain the reason for this. (3) I. (x - A) 2 + (y - A:) 2 = i* is the equation of any circle, center (7i, ft), radius r. Substituting in this equation (1, 2), (0, 8), and (7, 1), gives, II. (1 - A) 2 + (2 - ky = r 2 . III. (0 - A) 2 + (8 - Jfc) = r 2 . IV. (7 - hy + (- 1 - fc) 2 = r 2 . V. Ill - II - 2 h + 12 A; - 59 = 0. VI. II - IV - 48 + 12 h + 3 - 6 k = or 4 h - 2 k - 15 = 0. 230 UNIFIED MATHEMATICS V and VI are seen to be in // and k pivdsrly tin- equations solved in method (1). and in method (-2) for x and // ;us variables. (4) I. Ac* + Ay n - + 2 Gx + 2 Fy + C = 0. Substitute in this equation (1, 2), (0, 8), and (7, 1) and solve for the values of G, F, and C in terms of A. II. A + 4A + 2G + 4F+C = 0. III. 64 A + 16 F + C = 0. IV. 49^1 + A + UG-2F+ C = 0. V. II - III - 59 A + 2 G - 12 F = 0. -VI. II -IV -45A+12G +6F = C 1 F These are the same equations in -- and -- , regarded as the un- A A knowns, as appeared above in h and k. 6. Tangency conditions. If a circle is to be tangent to a given line the distance formula (normal form) from a point to a line may be used ; if a circle to be found is to be tangent to a given circle, then the radius sought, plus or minus the given radius, must be equal numerically to the distance from center to center, according as the circles are tangent externally or internally. 7. Circle through the intersection of two circles. I. x 2 -+- y z + 10 # = 0, a circle of radius 5, center (5, 0). II. x* + y z 40 = 0, a circle of radius 7, center (0, 0). III. (x 2 + y- + 10 JT) + A-(x 2 + 2/ ! - 49) = 0. The third equation is satisfied by the points of intersection of curves I and II, for all values of A: (see page 83). For all constant values of k, III may be written (1 + k)x* + (1 + k)y* + lOx - 49 k = 0, and the form shows that this represents a circle. To deter- mine the circle through the intersections of I and II, and any other given point substitute the coordinates in III, and solve THE CIRCLE 231 for A- ; since a circle is determined by the three points, it is easily seen that every circle through the two points of inter- section of the given circle is included in the family of circles, (1 + fr) a* + (1 + A-) y 2 -f 10 x - 49 k = 0. The method of deter- mining A: to have the circle pass through some other given Common chord of two circles or radical axis point is precisely the same as in the similar problem with straight lines (page 83). For A' = 1, this equation reduces to the linear equation representing the common chord of the family of circles ; whether two given circles intersect or not, this line, whose equation is obtained by eliminating x 2 + y 2 between the two given equations, is called the radical axis of the two circles. 8. Geometrical property of the radical axis. (x h) + (y A-) 2 is the square of the distance from (x, y) to the center of any circle ; (a; A) 2 +- (y k)- r* is the square of the length of the tangent to the circle from any point outside the circle of center (h, k), radius r. x* _j_ y* .j. 2 Gx + 2 Fy + C is the square of the length of the 232 UNIFIED MATHEMATICS tangent from any point (x, y) to the circle whose equation ia x 2 + y 2 + 2 Gx + 2 /fy + C= 0, since the left-hand member is identical with the left-hand member when written in this form: (x + ) 2 + (y + F)*-(G* + F*- C')= 0. Note that if any secant PAB is drawn through P(x, y) then PA PB = -PT 2 ; hence the expression C gives the product of the two distances along any straight line from the point P(x, y) on the line to the circle. There is a Distances from a point to a circle On any secant through P, PAB, PA x P.B is constant. PA x PB = P2V 2 = (x - hy +(y- fc) 2 - i*. correspondence to the normal form of a straight line, since the left-hand member there also represents a distance. is an equation which is satisfied by any point from which tan- gents drawn to the two circles a* + f- + 2 G& + 2 F$ + C, = 0, x* + y* + 2 G& + 2 F z y + C 2 = 0, THE CIRCLE 233 are equal in length. Hence, the radical axis is the locus of points from which the tangents drawn to the two circles are equal in length. 9. Radical center of three circles. Given three circles, each of the three pairs of circles which may be formed from the Radical axes and radical centers Radical center of the three circles, 1,2, and 3. Radical center of the tjiree circles, 1, 2, and 3'. three has a radical axis ; the three radical axes pass through a common point, as may be easily shown by Sec. 4, Chapter V. a. X 2 + 2 + b. c. d. (a-c) e. (b-c) radical axis of a and b. 2 -F 3 )y+C-C 3 = radical axis of b and c. 234 rXIFIKI) MATIIKMATICS /. (d + )or(a-c) 2(G 1 - 9 ice d + e = gives a straight line through the intersec- tion of d and e, and since d + e = gives the radical axis of a and c, the latter line passes through the intersection of the two former radical axes. 10. Limiting forms of the circle equation. (x h)- + (y k)- = r- represents a real circle when r 2 is positive. (x K) 2 -\-(y A-) 2 = represents a point circle; the only real point which satisfies this equation is the point (h, k). (x }()- + (y k)- = r 2 , r a real quantity, represents an imaginary circle : no real point satisfies this equation, since every real value of x and y makes (x h) 2 positive and (y k) 2 positive. PROBLEMS ON THE CIRCLE 1. Find the center and give radius to 1 decimal place of each of the following circles ; plot ; find the three radical axes and the radical center. a. a? + .v 2 + 6x 8y 16 = 0. b. HINT. 3(z 2 j x ) + 3(j/ 2 + 5 y ) = 7 ; complete squares inside parentheses and note that 3 times the quantity added within each of the parentheses must be added on the right. c. 2? + y 2 - 6 -8 = 0. 2. Plot the following two circles and determine their common chord : what is its length '.' a. * + y* - 10 x- 100 = 0. 6. ^ + ^ + 10^-100 = 0. 3. Write the equation of the family of circles a. With center on ;c-axis, passing through the origin. //. With center on y-axis, passing through the origin. c. Passing through the origin. THE CIRCLE 235 d. With center on 3 x -i y 5=0, radius 5. NOTE. 3 h 4 k - 5 = 0. 4. What limitation is imposed upon the coefficients A, G, JP, and 07 in ^ + ^ + 2 Gfc + 21fy+ 07=0, a. if the circle passes through (0, 0) ? (1, 1) ? b. if the circle has its center on the axis of x ? y-axis ? c. if the circle is tangent to the aj-axis ? y-axis ? tangent to z-3 = 0? d. if the circle is tangent too; y 5 = 0? 5. Find the equations of the circles through the following three points : a. (0, 0), (6, 0), (0, 8). b. (1,5), (-3,1), (7, -3). c. (0, 0), (8, 2), (15, 3) ; use two different methods. 6. Find circle tangent to 3 x + 4: y 25 = 0, and passing through (2, 3) and (5, 1). Note the two solutions. 7. Find the radical axis of each of the three pairs of circles a + .y*- 6s -8y- 10 = 0, a- 2 + y 2 - 20 x + 50 = 0, 2y? + 2 y z 4- Sx + 6y - 25 = 0. Find the radical center. Plot. 8. Find the tangents of slope 2 to the first circle in 7 ; find the normal and the point of tangency. 9. Find the circle of radius 5 tangent to the line whose equation is 4 x 3 y 9 = at (3, 1). 10. Find for what value of r the line 4# 3?/ 9 = is tangent to y? + y 2 r 2 = 0. Two methods. Find the point of tangency. 11. Find to one decimal place the points of intersection of the circle y? + y 2 20 x + 50 = with the line y = 2 x - 12. Plot. 236 UNIFIED MATHEMATICS 12. Use the trigonometric functions to find points of inter- section of Note that tan = 2, where is the slope-angle of the line. 13. Use trigonometric functions to find k, when y = 2 x + k is tangent to the circle x 2 + y 2 = 100. Draw figure ; note that tan = | where is the slope-angle of the normal. [ x = 3 4- 10 cos e, 14. Plot the circle < e ., . ., 1 1/ = 5 + 10 sin Q. [ x = 8 sin 0, 15. Plot the circle { [ ?/ = 8 cos 0. Note that is here the angle made with the j/-axis by any radius. 16. Find the equation of the complete circle of the circular arch of the Rialto, referred to the horizontal water line and the axis of symmetry of the arc as axes. The arch is 95 feet wide by 25 feet high. 17. Find the equation of the circle of which the arch of the Eocky River Bridge, 280 feet by 80 feet, is an arc, referred to a tangent at the highest point of the arc as avaxis and the perpendicular at the point of tangency as y-axis. Determine the lengths of vertical chords between the arc and the x-axis, spaced at intervals of forty feet. CHAPTER XV ADDITION FORMULAS 1. Functions of the sum and difference of two angles. The formulas for (a + 6) 2 and (a 6) 2 are illustrations of addition formulas frequently of fundamental importance in mathe- matical work. Thus 10* 10" = 10*+" is an addition formula leading to the whole theory of logarithms, which revolutionized computation processes. The question arises as to addition formulas in the case of the trigonometric functions after the functions have been defined. Just as the exponent formula 1Qz+v 10* . 10", which was first proved for positive integers, is extended to hold for all values of x and y, so the formulas which are established for sin (a + ft) and cos (a + ft) when a and ft are acute angles will be found to hold for all real values of a and . f .V 2. Geometrical der- ivation of sin (a + P) and cos (a + P) ; a and (J acute and a + p<90. Given a and ft, two acute angles whose sum is less than 90, to find sin (a + ft) and cos (a + /?) in terms of sin a, cos a, sin ft, and cos ft. 237 OB = r 238 UNIFIED MATHEMATICS On the figure let a and ft be two positive acute angles whose sum is less than 90, taken, for convenience, distinctly differ- ent from each other. Let OP make the angle <* with OX, and OB make the angle ft with OP, and thus + ft with OX. From B drop perpendiculars BA to OP and BN to OX ; from A on OP drop a perpendicular AM to OX; from A draw a parallel to OX cutting BN at R. On the figure, noting that OB is taken as r, we have the following evident relations : AB = r sin ft ; .R.B = J.J5 cos a = r cos a sin /3 ; OA = rcos/3-, AM= OA sin a = r sin a cos ? = r sin a sin ft ; OA cos a=r cos a cos ft. sm _ r sin a cos ft + r cos sin /? sin (a -f- ft} = sin cos ft + cos a sin ft. Similarly, cos (. + . T= whence _ r cos a cos ft r sin a sin ft t r cos (a-\- ft) = cos a cos ft sin a sin ft. Having established these formulas geometrically for a -f ft when < a < 90, < ft < 90, and + < 90, it now remains to establish that these formulas hold for all angles a and ft, including negative angles. This extension is made by employing the theorems of Section 12, Chapter VII. ADDITION FORMULAS 239 3. Generalization for any two acute angles. sin ( + ft) = sin a cos ft 4- cos a sin ft, cos (a + ft) = cos cos y8 sin a sin /!?. First we will show that when a and ft are cmy two acute angles the two formulas established above when a + ft < 90 continue to hold. The extension to any acute angles requires that we prove these formulas to be true further (a) when + ft = 90, and (b) when + ft < 90. Proof, (a) If + /8 = 90, /? = 90 a, whence sin /? = sin (90 a) = cos a ; cos ft = cos (90 a) = sin a. The two formulas then give, by substitution, sin ( + /?)= sin 90 = sin 2 a + cos 2 a = 1, cos (a + /?) = cos 90 = cos sin a sin a cos = 0. The sine of 90 is 1, and the cosine of 90 is ; hence our formulas continue to hold even when a -f- ft = 90. (6) a + ft > 90. Take the complements of a and /J to be respectively x and ?/, whence x = 90 and y = 90 ft. Evidently x + y will be less than 90, by the same amount that a + ft exceeds 90. Further, since x = 90 a and y = 90 ft, sin x = cos a, sin y = cos /8, cos x = sin , and cos y = sin ft. Now, sin ( + ft) = sin (90 - x + 90 - y) = sin (180 x + y) = sin (a; + 2/)> since sin (180 0)= shift Since x + y < 90, sin (a? + y ) = sin cc cos y + cos a; sin y, as established above ; making the substitutions for sin x, cos y, cos x, and sin y, we have sin (a + /?) = cos a sin /? -f- sin a cos /J. Q. E. D. Similarly, cos ( + ft) = cos (180 x + y) = cos (x + y), since cos (180 0) = cos 6, for any angle 6. But x and y are acute angles, whose sum is less than 90 ; 240 UNIFIED MATHEMATICS therefore cos (x + y) = -f- cos x cos ?/ sin x sin y = 4- sin sin /? cos a cos /?. Now cos (a + ft) = cos (x + y), or cos (a + ft) = cos a cos /3 sin sin /?. o. K. i>. 4. Extension of the formulas for sin (a + 0) and cos (a + P) to all angles without restriction. To show that these formulas hold for all angles it is necessary now to show that if either a or ft is increased by 90 the formulas continue to hold pro- vided that they hold for a and ft. Thus given sin (a + ft) =' sin a cos ft + cos a sin ft, cos (a + ft) = cos a cos ^8 sin sin ft, we wish to show that sin ( -f- ?/) and cos (a + y~) are given by similar formulas, when y = ft + 90. sin (a + y)= sin ( + ft + 90) = cos ('+ /?), = cos a cos ft sin a sin ft, but sin ?/ = cos ft and cos y = sin /?, whence substituting, sin (a -f- y)= cos a sin y + sin a cos y. Q. E. D. Similarly for cos ( + ?/), we find cos a cos y sin a sin y ; since a and /? enter symmetrically in the above formulas this proof establishes that a also could be increased or, by an entirely analogous procedure, decreased by 90, with the same formulas for the new values. This establishes the formulas for any two angles a and ft whatever. For since the formulas have been proved above to hold for any two acute angles and ft, the formulas hold for any obtuse angle and any acute angle since y, the obtuse angle, may be regarded as 90 + ft. This establishes the formulas for any angle in I and any angle in II ; now increase a by 90, thus establishing the formula for any two obtuse angles. Continuing in this way can be any angle in any quadrant, I to IV, and ft also an angle in any quadrant whatever, and the formulas continue to be true. ADDITION FORMULAS 241 After these formulas are established for all positive angles up to 360, another method of procedure to establish the formulas for all positive and negative angles is to note that any integral multiple of 360, k 360 with k a positive or negative integer, can be added to any angle without changing the functions of the angle involved in our formulas. Thus if ft is any negative angle, numerically less than 360, the functions of a + ( ft) are the same as the functions of a + (360 ft) which is the sum of two positive angles ; but the functions of 360 ft are the same as those of ft and after application of the formula the 360 can be dropped. In other words in these formulas any integral multiple of 360 can be added at pleasure and also dropped at pleasure, and in this way the formulas are established for all angles. Illustrative problem. Given sin a .45, cos ft = .68, find sin (a + /?) and cos (a + ft). sin a = .45 ; a can be in I or II since sin (180 ) = sin a. cos a = Vl - .45 2 = V/7975 = .893. cos ft = .68 ; ft can be in I or IV since cos ( ft") = cos ft. sin ft = VI - .68 2 = V(.32)(1.68) = .16 x V21 =. .16 x 4.58 = .733. sin a. - .45 determines either a, or a-.> cos (3 = .68 determines either Pi or p 2 There are strictly four problems, solved as follows : a in I, ft in I. a in I, ft in IV. sin = .45. cos a = + .893. cos ft + .68. 242 UNIFIED MATHEMATICS sin = + .733. sin = -.733. sin(a + ) sin( + ) = sin a cos + cos a sin . = .45 x .68 .893 x .733. sin (a + ) sin ( + ) = .45 x .68 + .893 x .733 = .306 - .655 = - .349. = .306 + .655 = .961. cos (a + ) cos (a + ) = cos a cos sin a sin . = .893 x .68 - .45 x .733 cos (a + ) = .607 -. .330 = .277. = .893 x .68 + .45 x .733 = .607 + .330 = .937. The two columns represent two solutions which have the three central values, sin a, cos a, and cos ft, in common. The student is expected to complete the solution, beginning as follows : a in II, in I. a in II, in IV. sin a = .45. cos ft = .68. cos = - .893. sin = + .733. sin = -.733. In general work only one case, indicating which solution is given. 5. Historical note. The formulas for sin(a + ) and cos(+) are closely allied to Ptolemy's theorem (c. 150 A.D.) that in any inscribed quadrilateral the product of the diagonals is equal to the sum of the products of the opposite sides. If a, 6, c, and d are the sides, in order around the quadrilateral, and e and /the diagonals, e/=oc + 6tZ; in the Greek trigonometry employing chords this theorem plays the same role that the formulas for sin (a + ) and cos (a + ) play in the trigonometry employing sines and cosines. A great French mathematician, Viete (1540-1603), the first to use generalized coefficients in algebraic equations, was the first to give these formulas, as sin (2 a + ) and cos (2 a + ) in terms of a + and a ; the ADDITION FORMULAS 243 modern form appeared in 1748 in the work of the Swiss mathe- matician Euler. PROBLEMS 1. Given a = 30, j8 = 45, find sin( + ) and cos(a-f /3). Check by tables. 2. Given a = 60, p = 45, find sin 105, and cos 105. Check by the preceding problem, and explain the check. 3. Given sin = f, and sin p = ^ } find sin ( + /?), when a and p are both acute ; find sin (a + /3) when a and p are both obtuse ; when a is obtuse, /? acute. 4. Given a and p acute angles, sin a = .351, cos p = .652, find sin ( -+- /3) by the formula and check with the tables. 5. Given sin 18 = .3090, cos 18 = .9511, find sin 36. 6. Given sin 18 = .3090, find sin 78. 7. Using the results of problem 1 for sin 75 and cos 75 with the data of problem 5, find sin 93 and cos 93 ; thus find sin 3 and cos 3. 8. What are sin (45 + a) and cos (45 + ) in terms of a ? 9. Find sin (60 + a) in terms of sin a and cos a. 6. The formulas for sin (a p) and cos (a P). If (3 is a negative angle, a p comes directly under the a + /3 formula as H- ( (3) ; if (3 is any positive angle greater than 360, p can be reduced to less than 360 by subtracting 360 (or some multiple of 360) without affecting the functions of ^8 or of p ; if (3 is positive and less than 360, the functions of ft will be the same as the functions of a +(360 (3), since this simply adds one complete revolution to a p. Hence sin ( p) = sin ( + 360 - P) = sin a cos (360 - 0) + cos a sin (360 - /8) = sin a cos ( /3)+ cos a sin ( /8). cos (a - p) = cos a cos (360 - /3) - sin a sin (360 - 0) = cos a cos ( /8) sin a sin ( /?). 244 UNIFIED MATHEMATICS Substituting in these formulas, cos ( ft) = cos ft, and siu ( /3) = sin ft we obtain the subtraction formulas : sin ( /?) = sin cos ft cos a sin ft, cos (a ft) = cos a cos /? -f- sin a sin /?. 7. Tangent formulas. tan ( + /?) = ^ __ 1 - tan a tan /3 1 + tan a tan ^ Since sin ( + /?) = sin cos ft -f cos a sin ft, and cos (a + ft) = cos a cos ft sin a sin ft, for all angles a and /3, without restriction, it follows that tan ( + ft) = sin fe + ) = sin cos ff + co3ce sin/3 cos (a + /?) cos a cos /8 sin a sin /8 ' for all angles a and /3. Dividing numerator and denominator of the right-hand ex- pression by cos a cos ft, we have 1 - tan tan ft o- M i / tana tan 8 Similarly, tan (a 8) = l + tanatan 8. Functions of double an angle. The formulas for sin (a+ft), cos (a + ft), and tan ( -f- ft} hold if ft = a, whence sin (2 a) = sin ( + ) = sin a cos a + cos a sin a = 2 sin a cos a, sin (2 a) = 2 sin a cos a. Similarly, cos (2 a) = cos 2 a sin 2 = 2 cos 2 a 1 = 1 2 sin 2 a. By division and simplification, or directly from tan ( + ft), 2 tan a tan (2 a) = 1 - tan 2 a Note that whether a be regarded as positive or negative, i.e. as obtained by positive or negative rotation, as + a, or + a 360, 2 a has the same terminal line as 2 720. ADDITION FORMULAS 245 NOTE. See the preceding list of problems, and use numerical values as there computed. 1. Given a = 45, ft = 30, find sin ( ft) and cos (a - ft), checking by the tables. 2. Given a = 60, ft = 45, find sin (a ft) and compare with problem 1. Find tan (a + ft), tan (a ft), and tan 2 a. 3. Given sin a = f and sin/2 =-fa, ^ n< ^ s ^ n ( a ~ P) when and ft are acute ; find sin (a ft) when a and /3 are both obtuse. Explain the result ; find sin ( ft) and cos (a ft) when a is obtuse and ft is acute. Interpret. Find tan (a ft), tan (a -(- /?), and tan 2 a for a and /3 in I. 4. Given a and ft acute, sin a .351 and cos ft = .652, find sin ( ft) and check by the tables. Find tan (a ft). 5. Given sin 18 = .3090, cos 18 = .9511, and sin 15 from problem 1, find sin 3 and cos 3. 6. Find sin 42 as sin (60 - 18). 7. Express sin (60 a) and cos (60 a) in terms of func- tions of a. 8. Find the value of sin (60 + ) sin (60 a). 9. Find the value of cos (45 -f )+ cos (45 a). 10. Show that sin (a + ft) sin (a ft) = sin 2 a sin 2 ft. 11. Find a value of cos (a + ft) cos (a ft), similar to the preceding. 12. Given tan a = 1.4, find tan 2 a. 13. Given cos 2 a = .63, find sin a and cos a ; are there two solutions ? 14. Given that one line cuts the x-axis at an angle a such that tan a = 3, and another line cuts the avaxis at an angle /3 such that tan ft = \, find the tangent of the angle between the two lines by assuming that they intersect on the avaxis. Check by using the tables to find the slope angles of these lines. 246 UNIFIED MATHEMATICS 9. The tangent of the angle between two lines. Given any two lines as y 3 a; 5, ?/= x 7, it is evident by plane geometry that the angle between them is the same as the angle between y = 3 x, y = x, lines parallel to these given lines through the origin. The word "be- tween" implies no dis- tinction as to priority of either line : thus the angle may be taken as either a positive or nega- tive acute angle, or the corresponding supple- mentary angle. Thus if the lines were inclined to each other at 30, the angle might be con- sidered as -(- 30, - 30, + 150, or -150; the tangent of the angle would then have the Angle between two lines Parallel lines through the origin make the same angle. V3 i 3 value given by the expression To distinguish between the two lines we may say that we wish the angle from the line of slope + 3 to the line of slope 1, or in the general case, from the line of slope m 2 to the line of slope mi ; by analogy with our use in defining the angle which a line makes with the cc-axis, when we say the angle which the line y = x makes with y = 3 x we mean the angle obtained by re- Angle between two lines volving the line whose slope is = e l 2 , 0i > &z- ADDITION FORMULAS 247 3 so as to make it coincide with the line whose slope is 1. Calling the angle whose tangent is 3 (written tan" 1 3 or arc tan 3, meaning the angle whose tangent is 3), 2 , and the angle whose tangent is 1, O l} we find that the angle from the 2 line to the 0j line is < = d l 2 . tan ^ tan (ft- ft) = tanft-tanft - = I 1 + tan Oi tan 2 1 If the two lines are parallel the angle is 0, hence tan = 0, and mi w 2 = 0, or mi = w 2 , as anticipated ; if the lines are perpendicular tan< becomes infinitely large, and for finite values of m t and m 9 _ (excluding lines parallel to the axes), the denominator 1 + m^m^ = 0, or m 2 = -- -, i.e. the slope of a perpendicular is the negative reciprocal of the slope of the given line. When one line is parallel to the ?/-axis, its slope ra 2 (or m { ~) is infinite, but the angle between the two lines can be obtained by dividing numerator and denominator of tan by m 2 (or ra^, giving tan or -- when ra 2 ap- proaches infinity, for the tangent of the angle made by a given line with the y-axis. tan 4 = m i- m 2 gives the an- 1 -f- /7Zi/7Z 2 gle from the ra 2 line to the m t line. If # 2 > Oi, is negative, but the formula < = O t 2 still holds. 10. Functions of half an angle, cos (2 a) = cos 2 a sin 2 a for all values of a. Substitute x for 2 a, and ^ for a. 2 248 UNIFIED MATHEMATICS cos x = cos 2 sin 2 - l.oos' whence Half-angle relations Similarly, cos ? = V|(l + cos a). M sin - = cos cc). tan ft = /T- cos a; ^ /(I - cos aQ(l - cos x) ? 2 *l + cosa; * 1 cos 2 x + if x is in I or II, and if a; is in III or IV ; the formula X 1 COS X 1-1 tan - = - is one 111 which 2 since the sin x takes care of the algebraic sign ; and so also sin a; x tan - = , both 2 1 + cos x by rationalization. Note that if x is regarded as a positive angle, less than 360, x sin - is + ; but the same position of the terminal line is obtained by x 360 ; -' and ~ + 180 have sine and (x \ x cosine opposite in sign, but tan f - 180 ]= tan - Since cos ( x) = cos (x~) it must be stated whether x is in I or IV, or in II or III ; i.e. cos x alone does not locate the angle x. If a; in I is regarded as a positive angle, - is + acute, and sin - and cos - are positive ; if x in I is regarded as a negative reflex angle, '- is negative obtuse, and sin - and cos are both _ a +i ADDITION FORMULAS 249 negative ; in either case tan - is positive. Similarly if x is & taken in II, III, or IV, the formula takes care of all positions, proper account being taken of the algebraic sign of the radicals. PROBLEMS Find the angle between the following lines : 1. y = 3 x 5, and y = x 7. 2. y = 3 x 5, and y = x 7. 3. 2y-3x-7 = 0, and 3 y + 4 x -5 = 0. 4. 3 y = 5 x 5, and y = 8 x 10. 5. 3 y = 5 as 5, x = 5. 6. 3y = 5x 7, y = 5. 7. In the preceding 6 problems, find the tangent of the angle made by the first line with the second line, i.e. the tangent of the angle obtained by rotating the second line until it coin- cides with the first. Why is it that the sense of this rotation is immaterial ? 8. In the above problems check by finding from the tables the trigonometric angles involved. 9. Find the pencil of parallel lines making an angle of 30 with each of the lines in problem 1 ; find the one of the family through (-3,5). 10. Find the pencil of lines making an angle of 45 with each of the lines in problem 3 ; find the particular one through the origin. 11. Find the pencil of lines making an angle of 90 with each of the lines in problem 4. 12. Given sin 30 = .5000, cos 30 = .8660, and tan 30 = .5774, find sin 15, cos 15, and tan 15. 13. Find sin7|, cos 7-|, tan 7^-, using half-angle formulas. 14. Given sin 45 = cos 45 = .7071, find sin 22, cos 22|, and tan 22. 250 UNIFIED MATHEMATICS 15. Use sin (a ft) formula to obtain sin 7| and cos 7f , from the functions of 30 and 22^-. Compare with problem 13. 16. Given sin 18 = .3090 and cos 18 = .9511, find sin 12 and cos 12. 17. From the functions of 12, compute the functions of 6, and then the functions of 3 and of 1|, using half-angle formulas. 18. Compute sin If and cos If by the difference formulas, taking If as 7f 6. 19. Compute the functions of f from the functions of 1. 20. Find by interpolation sin 1 and cos 1 from the com- puted values of the functions of f and If. Compare with the tabular values. 21. Make a table of values of the sine, from to 45 in- creasing by If intervals. CHAPTER XVI TRIGONOMETRIC FORMULAS FOR OBLIQUE TRIANGLES 1. General statement. Employing elementary theorems of plane geometry it is possible to construct any triangle when given the three sides, or two sides and an angle, or one of the three sides together with two of the angles ; in trigonometry the corresponding problem is the numerical solution, not sim- ply the graphical, of the types of triangles mentioned. The trigonometric solution which has been given of the different types of right triangles, with unknown parts, can be applied to effect the trigonometric solution of any oblique triangle; but in general, these methods do not give convenient formulas for computation. As the general triangle is fundamental in surveying (note the term " triangulation"), in astronomical work, and in many problems in physics, more convenient formulas than those given by right triangles are a practical necessity. In general the laws and formulas of plane trigonometry connect directly with proposi- tions of plane geometry ; the effort is to express the inter- dependence of the angles and sides in the form of equations involving the trigonometric func- tions of the angles. The vertices of any triangle A, B, C; a, p, -y; a, b, c being lettered A, B, C, it is con- venient to designate the corresponding angles at these ver- tices by a, ft, and y, respectively, or by A, B, and C, if no 251 252 UNIFIED MATHEMATICS confusion of meaning is possible ; the sides opposite A, B, and C are designated by a, &, and c, respectively. 2. Cosine law. If the two sides of a triangle are given, the third or variable side, opposite the angle a, between the two .HJ 6' \C' h I. A \ - - - ^s . J-^ \ V -/ r ~\ \ 5 \ ^N j 1 -T/'C "/ 3-L ( ( _i- _I_J-^. / _L_ - V-^- - \ _^ \ \ - ]|- t 1 ! /V --J . ^ L - \ :/:: \ \ \ \ S SV ^Sr - ka . ^--. ^a V -l--t V -/ :*:: A \ o \ \- \ \ s -'/ /; A i> j/: .1 as a 2 = b 2 + c 2 - 2 be cos a given sides, evidently changes as a changes. Let b and c remain fixed. Let M be the foot of the perpendicular from C upon AB ; then AM = b cos a, for any angle when the direction AB is taken as positive. Further in every position MB = AB- AM = c - b cos , for in every position AM + MB = AB. The altitude MC = h = b sin a. Hence, SC^ = JO 2 + MC 2 ' (c b cos a) 2 + (6 sin ) 2 = c 2 2 be cos a + & 2 (cos 2 a + sin 2 a). a 2 = 6 2 + c 2 - 2 6c cos a. All limitations upon a are removed by the different types of figures. Hence for any angle a, a 2 = b 2 + c 2 2 &c cos a gives the length of the side a, opposite a. in terms of the other two sides and a. M falling outside B ~ . Formula unchanged SmC6 a and " ma ^ re P resent an 7 Slde and the opposite angle of any given tri- angle, 6 and c being the other two sides, our formula may be stated as follows : FORMULAS FOR OBLIQUE TRIANGLES 253 TJie square of any side of a triangle is equal to the sum of the squares of the other two sides less twice their product into the cosine of the including angle. Or, Tfie cosine of any angle equals the difference between the sum of the squares of the two including sides and the square of the side opposite, divided by twice the product of the including sides. If a, b, and c are the sides of any triangle, with a, ft, and y the corresponding opposite angles, we have the following relationships : a 2 = &2 + c , _ 2 bc CO s a, ' b 2 = c 2 -f a 2 2 ac cos ft, c 2 = a 2 + b 2 2 ab cos y ; 7,2 _i_ C 2 _ a 2 or cos a = cos ft = cos y = '2bc C 2 + a 2 _ ft2 2ac 3. Cyclic interchange. Any formula which has been de- rived, without imposing any limitations upon a, b, c, a, ft, or y, connecting a, b, c, and trigono- metric functions of the angles a, ft, and y, will continue to hold if a and b and, at the same time a and ft, are interchanged ; or if I ( are changed to to , and to , such changes effect ? X simply a re-lettering of the figure. Cydic interchange The change of a into 6, b into c, and c into a is called a cyclic interchange of the letters a, 6, and c. Note that cyclic interchange gives the second formula from the first, and the third from the second. 254 INIFIED MATHEMATICS In the figures, a in the first is chosen as an acute angle, but this limitation is removed by deriving the same formulas for a a right angle and for a an obtuse angle ; c is taken as longer than b, but interchanging b and c in our derived for- mula leaves the formula unchanged ; assuming 6 and c equal would involve no change whatever in our proof; and if b is assumed greater than c, a fourth figure can be drawn in which M falls beyond B on AB produced ; but the formula a 2 = 6 2 -j- c 2 2 be cos remains the same, as the student may easily verify. PROBLEMS In the following problems use .866, .707, and .500 for the cosines of 30, 45, and 60, respectively. 1. Given b = 140, c = 230, a = 60, compute a. Kefer back to the section on extraction of square root, page 23. 2. Compute a when a = 30 and 45, 90, 120, 135, 180, when b = 140, c = 230. 3. Given a = 155, c = 234, ft = 35, compute b. What changes in b are effected by changes of 10' in ft ? 4. Given a = 155, c = 234, compute ft when b = 172. What is the maximum change in ft which an error of 1 unit in a, b, and c could introduce, ft being computed to minutes ? Take 1551 234| with 1711 ; take 1541, and 2341, with 1721 Note that (155i) 2 and (154|) 2 differ from (155) 2 by about 155; similarly with the other values ; if the squares are found by logarithms it is well to look up log 155.5 and log 154.5 at the same time as log 155, etc. 5. In problem 1, find cos ft, and then ft, taking for a the value obtained there. 6. In problem 1, find cos y and y, using the computed value of a. Check by summing ft and y with the given angle. 7. Given a = 200, 6 = 150, c = 300, find a. What change in a would a change of 1 in a effect ? Suppose that a, b, FORMULAS FOR OBLIQUE TRIANGLES 255 and c are given only to two significant figures, i.e. a is between 195 and 205, b is between 145 and 155, c is between 295 and 305, compute a and discuss limiting values. 8. Compute the third side in the following 5 problems, using logarithms for squaring ; time yourself in the exercise. Fifty minutes should be ample time .for the 5 problems ; devise a convenient form and use it in each example. a. Given a = 366, b = 677, y = 15 10'. b. Given a = 423, c = 288, ft = 35 15'. c. Given b = 627, c = 816, a = 100 41'. d. Given a = 635, c = 341, ft = 67 38'. e. Given c = 184, b = 295, a = 130 54'. 4. Sine law. A circle may be circumscribed about any triangle ; let the radius be designated by R. The figure shows Sine law : sin a sin sin y that if A is an acute angle, sin = -2-? = JL ; if is 90, this Jf 2 R formula is still true, as a equals 2 R, and the formula gives sin 90 = 1 ; if is obtuse, the figure gives sin (180 a) = -^-, a whence sin a = . 2R 256 UNIFIED MATHEMATICS Therefore without any limitation whatever upon a, sin a = 2R' interchange of letters gives sn ? = sin y = 7 _ 2R' c '2R Whence a Sine law : a. obtuse sin a sin B sin y This formula states that in any triangle the ratio of the side op- posite any angle to the sine of that angle is constant, and this ratio is numerically equal to the diameter of the 'circumscribed circle. -n, ,, sin a sin 8 sin y , . c ,, urther, - = - = L } O r the ratio of the sine of any a b c angle to the side opposite is constant. Note that if 2 R is regarded as the chord of 180 of the circle in which the triangle ABC is inscribed, the proposition states, in effect, that in any circle any chord is proportional to the sine of the inscribed angle which intercepts the arc of the chord. The formula may be stated : sin a _ sin B _ sin y _ sin 90 _ sin 30 _ sin k a b c 2R R ~ chord all of the chords being chords of the circle circumscribed about the triangle. The ratio of the sine of any central angle in a circle to the chord of double the angle can readily be shown to be constant, 5. The sine law historically. The sine law was discovered by an Arabic (Persian, by birth) mathematician, Nasir al-Din, FORMULAS FOR OBLIQUE TRIANGLES " 257 at-Tusi, who lived 1201-1274 A.D. To him we owe the first systematic treatise on plane trigonometry, an achievement made possible by the combination of the Greek trigonometry using chords with the Hindu trigonometry employing sines. To Europeans the sine law was communicated by the great Ger- man mathematician and astronomer, Regiomontanus, in his work on trigonometry, De Triangulis, the first published sys- tematic treatise ; it was published at Nuremberg in 1539, many years after the death of Regiomoutanus, who lived 1436-1476. PROBLEMS 1. Given a = 150, b = 200, a = 30, find sin (3 using natural functions. 2. Given a = 150, a = 30, ft = 45, find 6, using natural functions. 3. Given a = 150.4, b = 214.3, a = 31 10', find sin ft employ- ing logarithms. 4. Given o = 150.4, a = 31 10', ft = 44 16', find b by loga- rithmic computation. 5. In the formula, a 2 = b 2 + c 2 2 be cos a, substitute the values as given in problem 1 and solve for c. Note that there are two solutions. What is the explanation ? 6. Time yourself in solving the following set of 6 problems, applying the sine law ; make a type form of solution and use it in each problem. Thirty minutes should be sufficient for the 6 problems. a. Given a = 366, 6 = 677, a = 15 10'. Find sin ft and ft. 6. Given a = 423, c = 288, y = 35 15'. Find sin a and a. c. Given a = 627, a = 100 11', ft = 43 15'. Find 6. d. Given 6 = 816, ft = 67 18', y = 34 9'. Find c. e. Given c = 635, ft = 130 14', a = 20 12'. Find b. /. Given b = 284, a = 40 10', ft = 35 15'. Find c. 258 UNIFIED MATHEMATICS 6. Half-angle formulas. As the circumscribed circle has yielded a formula of great value trigonometrically the in- scribed circle may be examined trigonometrically with the hope of a similar result. The bisectors of the three angles of the triangle meet in a point which is the center of the in- scribed circle ; let this circle be drawn and let L, M, N, be the points of tan- gency, then AM=AN, BL = BN, OL = OM = ON = r, radius inscribed circle CL CM, being tangents from an exterior point. Evidently the six segments mentioned make the perimeter, 2 s, of our triangle ; 2s = a + 6-f-c; adding above we have that AM+ BL + CL = AN+ BN+ CM= s, but BL+CL = a, and BN+ AN= c,; whence AM = s a, CM= s c, and similarly BN=BL = s-b ON= OL = OM= r. tan _ = s a a. -1- /-> ' tan ~B = '. 2^ s-b 2 s c 7. Area. In the preceding section the area of the given triangle is easily determined in terms of r and s, for the area equals the sum of the three triangles on a, b, and c as bases, each having the altitude r. .. A = % r(a + b + c) = rs. FORMULAS FOR OBLIQUE TRIANGLES 259 However, if the three sides are given, this formula does not enable us to determine r without using a further formula to determine A, A = ^bc sin a = ^ ac sin ft = 1 ab sin y. This type of formula for the area is applicable when two sides and the included angle of a triangle are given or found. & 2 4- c 2 a 2 A= 4- 6c sin a can be combined with cos = -- - in such 2 be a way as to eliminate a, giving A in terms of the three sides. A* = $ & 2 c 2 sin 2 a = & 2 c 2 (1 - cos 2 a) . = & 2 c 2 (1 - cos a)(l -f cos a) c 2 - qg\ 6 2 + c 2 - a 4 V 2&c Wa - 6 2 + 2 6c - c 2 \/6 2 + 2 6c + c 2 - a 2 ' 4 V 26c 2bc c )2 _ a g] 16 6 2 c 2 _ (a 6 + c)( + ft c)(6 + c a) (6 + c + a). 2 2 2 2 But The above formula for A 2 may be written, A 2 = s(s a)(s &)(* c) A = Vs(s d)(s 6)(s c). -, ,, A Further A = rs, whence r = = s This value of r is employed with the half-angle tangent formulas of the preceding article to determine the angles of a triangle when the three sides are given. 260 UNIFIED MATHEMATICS 8. Newton's check formula. A formula which involves all of the sides and all of the angles of an oblique triangle is particularly desirable as a check formula to be used upon the results obtained by direct application of the sine law or in a solu- tion obtained by right triangles. Such a formula was devised by Sir Isaac Newton and appeared in his Arithmetica of 1707 ; our proof follows the lines of that by Newton. B 3 rf 2 ?j s .At P .1 F Let ABC be any triangle ; from C draw the bisector CE of the angle -4(7J5 or y ; extend 5(7 to F, making CF=CA=b; AF is parallel to CE, by plane geometry ; angle CFA = angle BCE = i y. Now angle .BvlF= + i y = 90 - 1( - ), since ! a + !/? + ^y= 90. Applying the sine law to the triangle BAF, we have the desired formula : a + b cos ^(g (3) c sin \ y FORMULAS FOR OBLIQUE TRIANGLES 261 By drawing the bisector of the exterior angle, and drawing a parallel from A, a second useful check formula is obtained : a b __ sin |(q P) c cos i -y Noting that 1 y = 90 ^(a + /?), division of the second equation by the Newtonian, member for member, gives a + b ten (a + p) this symmetrical formula is known as the tangent law. Cyclical interchange gives in each one of the above two corresponding formulas. 9. Historical note. The formula A = Vs(s d)(s b)(s c) was first given by Hero of Alexandria, first century A.D., a teacher of mathematics and mechanics in what was probably a kind of technical school at Alexandria in Egypt ; it is called Hero's formula, An extension of this formula is given by Bhaskara, a Hindu mathematician of about 1000 A.D. Bhaskara's formula gives the area of any quadrilateral which is inscribable in a circle, i.e. with the opposite angles supplementary, as L- A = V(s a)(s b)(s c)(s d). The triangle may be regarded as a special case with d = 0. 10. Reflection and refraction of light. Rays of light, like rays of heat and sound and elec- tric rays of various types, travel in straight lines from the source. Rays of light emanating from the sun travel in nearly parallel rays, since the point of con- vergence, the source at the sun, is at SO great a dis- Reflected ray travels shortest path tance from the earth. A ray of light which meets a polished plane surface, a mirror, is reflected at an angle which is such 262 UNIFIED MATHEMATICS as to make the total path from the source (L) to the reflect- ing surface and then to a second position (R) the shortest possible. LSR is the shortest distance from L to S to R if the angle of incidence i, made by the original ray with the normal, to the surface at S where the ray strikes, is equal to the angle of reflection r. Evidently LSR = LSR' ; the straight line joining L to R', a point symmetrically situated to R with respect to the polished surface, is shorter than any other line, for any other broken line LS'R = LS'R' is greater than the straight line LSR' and hence greater than LSR. If the ray of light meets, not a polished surface but some transparent medium, other than that in which the ray is traveling, the ray of light is not continued in the same straight line in which it starts but it is broken, or refracted, continuing on its path in a straight line which makes a differ- ent angle with the normal than does the original, incident ray. It is found by physical ex- periments that the angle of refraction, the angle of the refracted ray with the nor- mal, bears a simple relation to the angle of incidence, sin i sinr = Jc, wherein k depends Refracted ray of light upon the nature of the two media through which the light is passing. Thus for a ray of light passing from air, a rarer light medium, to the denser water the value of k is |, sin i _ 4 sinr 3 A student who thoughtfully examines this formula will be reminded of the sine law, which does indeed give a very simple construction for the refracted ray when the constant k is known. Let two concentric circles be drawn whose radii bear to each other the ratio, , of the index of refraction. In the FORMULAS FOR OBLIQUE TRIANGLES 263 figure the ratio is taken |, the index of refraction for light from air to water. Extend LO, the incident ray, to L', cutting the circle of smaller radius. From L' drop a line parallel to the normal N'O to cut the larger circle in R. Connecting R LO is the incident ray ; OR is the refracted ray with gives the refracted ray. In the triangle OL'R, the Z OL'R = 180 - 1, and the Z ORL' = / r, of refraction ; by the sine law sin (180 - Q _ sin i _ 4 sin r sin r 3 From water to air the index is f , it being found that if the refracted ray is replaced by an original ray, this new ray in the second medium will be refracted along OL, the path of the incident ray with which we started. The construction for the refracted ray in air for a ray of light emanating from the water, RO, is entirely similar to the preceding. RO is extended to R 1 on the larger circle. From 264 UNIFIED MATHEMATICS R' a parallel R'L is drawn to the normal to cut the smaller circle. OL is the refracted ray. Evidently if sin r = f sin i, sin r is always less than f. If a ray of light starts from any point within the arc A' T wherein T is the intersection of the vertical tangent to the smaller circle with the larger circle it cannot be refracted into the air at O, and the whole light is reflected at 0. This property of the light rays is utilized in certain spectroscopic work. Thus in the case of a glass prism, index of re- Glass reversing prism Index of refraction f Angle of incidence,45. fraction f , if light strikes the plane sur- For any angle i greater face at an angle of incidence greater than 41 48', since sin 41 48' = .6666, or |, all the light will be reflected ; this type of prism is used in projecting lanterns. than a, sin a = f , the beam is reflected. PROBLEMS 1. Given a 9, b = 14, c = 19, find the area of the triangle, using Hero's formula. 2. Given a = 9, b = 14, c = 19, find a, using the cosine law,. 3. Given a = 9, b = 14, c = 19, find the area by the formula A = %bc sin a. 4. Given a = 9.34, & = 14.31, c = 19.27, find the area by Hero's fcrmula, using logarithms. 5. Given a = 9.34, b 14.31, c = 19.27, find a by the cosine law, and then find the area using the formula involving sin a. 6. In the two triangles above find r, the radius of the in- scribed circle, using r s = A. 7. In the two angles above find tan i , tan ^ ft, and tan ^ y, using the half-angle formulas. Find the angle sum in each case. FORMULAS FOR OBLIQUE TRIANGLES 265 8. Draw circles with radii two inches and three inches and show how to construct the refracted rays of light passing from air into glass at angles of incidence of 30, 45, and 60. 9. For what angle will a ray of light passing from glass into water be reflected, and not refracted ? The index of re- fraction of light passing from glass into water is -|. Draw the figure. 10. Find the angle of refraction of rays of light passing from air into water, Tc = 1 .33, when the angles of incidence are 31 15', 37 18', 44 25', 67 10', 83 15'. For which of these angles is the course of the ray changed by the greatest amount ? 11. Suppose the rays in problem 10 to pass from air into glass, solve for the angles of refraction. 12. Construct two of the figures in both problems 10 and 11, and check graphically the results obtained above. CHAPTER XVII SOLUTION OF TRIANGLES 1. Solution of triangles given two angles and one side: cap type. With surveying instruments the simplest method of locating the distances from two fixed points to a third inacces- sible point is to determine the length AB and the angles a and ft, at A and B respectively, wherein A and B are two points from which (7 is visible. Using the sine law, sin a sin ft sin y we select the equation - = - , sm a sin y b c or = -- -, since in each of these only one unknown sin ft sin y quantity appears. The third equation - = - , not in- , sin a sin ft dependent of the other two, is used as a partial check upon the computed values. As a more complete check use Newton's formula, cos a - c sin ^ y This form of triangle appears in the classical problem, whose solution by plane geometry is ascribed to one of the seven wise men of Greece, Thales of Miletus, sixth century B.C. The prob- lem is familiar to the surveyors, being used in determining distances across a stream, or to an inaccessible point. The astronomer has the same problem in locating the distance of fixed stars using two observations, at different points in the earth's orbit, of the angle made by lines from the earth to the 266 SOLUTION OF TRIANGLES 267 star and to the sun ; for simplicity, the two points of observa- tion may be considered as taken at the extremities of the diameter of the earth's path. In locating batteries by the sound waves this type of triangle is employed ; two or three observers at different points can locate an enemy battery by this method within a radius of fifty feet or thereabouts. 2. Type form of solution : cap type. The form of the solu- tion is important ; follow the given form closely. Given a = 65 11', ft = 38 24', c = 175 feet. Find a and b. c sm (t / 1 1 r i_i / . i o/ c (written from the formula = sin y sm a sin y should not be set down). which sin y sin a , = -bc sma= 1 c 2 sin a sin logc = + log sin a = 2 sin y a = 65 11' B = 38 24' y = 76 25' = 2.2430 9.9580 - 10 12.2010 - 10 -log sin y= 9.9876 10 log a = 2.2134 a = 163.4 log a = 2.2134 + log sin ft = 9.7932 - 10 12.0066 - 10 -log sin a = 9.9580 - 10 log b = 2.0486 - 10 But log b = 2.0486 by above computation, which checks. Two angles logc + log sin B - log sin y log b b logc 2 + log sin a + log sin B -log sin y log 2 A 2A A and a side given = 2.2430 = 9.7932 - 10 12.0362 - 10 = 9.9876 - 10 = 2.0486 = 111.8 = 4.4860 = 9.9580 - 10 = 9.7932 - 10 14.2372 - 10. 9.9876 - 10 4.2496 17,760 8880 268 UNIFIED MATHEMATICS The check which we have used is only partial as an error in y or siiiy would be carried through the work without showing up in the check. The Newtonian formula gives a real check upon the computation. Check. q& c sin i y a + b = 275.2 log (a + b) = 2.4396 log c = 2.2430 .1966 a _ = 26 47' log cos |(a - ft) = 9.9880 y = 76 25' log sin y = 9.7914 .1966 which checks. NOTES. The whole form of solution is placed on paper before the logarithms are inserted. Place the given angles in vertical column and obtain the third angle by noting the angle which added to the given angles makes 180 ; thus, here note first that to complete 11' and 24' to 1 takes 25' . Add this 1 to the 8 and 5, the units of our given angles, making 14 ; complete by 6, which is written in its proper place, to 20. Carry the 2 tens, to the tens, making 11 tens, or 110, requiring 7 tens (written in the proper place) to complete to 180. Look up logc, i.e., log 175, writing this immediately in all places where it occurs ; for the area, it is simpler to calculate 2 A and divide by two than to divide by subtracting log 2 in the work, log c 2 = 2 log c, which is set down in its place. Finish, as far as possible, with the logs of numbers before taking up the logs of trigonometric functions. log sin 65 11', log sin 38 24', and log sin 76 25' should be found in the order in which they occur in the tables, to avoid useless thumbing back and forth ; any value found should be immediately inserted wherever it occurs in the form. PROBLEMS 1. Prove the sine law by using perpendiculars dropped from a vertex to the opposite side. 2. Given c = 350.4, a = 36 14', ft = 100 24', find b and a, by the sine law. 3. Given a = .03504, a = 36 14', ft = 100 24', find b and c, by the sine law. SOLUTION OF TRIANGLES 269 4. Solve completely the following 5 triangles ; take the time of your solutions ; write the complete form of solution for each problem, in turn, before inserting any logarithms. The five problems should be completed within one hour and 20 minutes using the rough check by the sine law. As a separate exercise check all by Newton's formula, timing yourself. a. a = 627 a = 100 11' /8 = 43 15' 6. 6 = 816 = 67 18' y = 3409' c. c = 635 =130 14' a = 20 12' d. & = 284 a = 40 10' = 35 15' e. a = 366 a = 15 10' = 95 14' 3. Given two sides and the angle opposite one: aba. type. Given b, a, and a to construct the triangle geometrically. AC is laid off of length b and the line AX is drawn so as to make Z. CAX = a. Since a must lie opposite to , a is taken as Given two sides and the angle opposite one The side opposite the given angle must always be greater than, or equal to, the corresponding altitude. radius and with C as center an arc is swung to cut the side AX. Since the shortest distance from (7 to AX is the length of the perpendicular CM, if a is given less than this perpen- dicular there is no solution. If a is given equal to the per- pendicular there is one solution; if a is greater than the perpendicular the arc cuts AX in two points, but unless a < 6 the one point of intersection to the left of A will not repre- sent a solution. The perpendicular is of length b sin a ; if a 270 UNIFIED MATHEMATICS is equal to or greater than 90, there will be one solution if a > b, and none if a j< b, for the greater angle lies opposite the greater side. By plane geometry then, we have the fol- lowing scheme, indicating whether one solution, two solutions, or no solutions are possible. a J> 90, a <_ 6, no solution. a _> 90, a > b, one solution. a < 90, a < 6 sin a, no solution. a < 90, a = b sin , one solution. a < 90, b sin a < a < b, two solutions. a < 90, a > b, one solution. Trigonometrically, by our formulas, we would arrive at these facts, but a student who is not able to observe the geometrical relationships is not likely to be able to interpret the trigonometric formulas. When the sine of an angle is given, the angle may be either in I or II, a or 180 a if a is either angle which satisfies the relationship. Then, if a < b sin a, sin ft will be greater than 1 and there is no angle satisfying the relationship ; if a > b, a > ft (greater angle, greater side opposite), and only the acute angle ft can be taken ; if a < b, both values of ft can be taken. PROBLEMS 1. Given a = 30, a = 150, b = 60, 70, 75, 100, 150, 180, and 200 respectively ; draw the figures and determine the number of solutions in each case. Solve for ft in each case where it is possible. 2. Given a = 90, a = 150, b = 75, 100, 150, 200. Discuss. 3. Given a = 150, b = 75 ; a = 20, 30, 45, 60, 80, 90, 120, 150. Discuss the solutions, geometrically and trigouo- metrically. SOLUTION OF TRIANGLES 271 4. Solve the following eight problems, having one or two solutions, and time yourself. Use the following form of solu- tion. The eight problems should be completed within one hour. a. Given a = 366, b = 677, = 15 10' ; solve for ft. a log 6 = + log sin a = log a = log sin /3 = or simply ft = , if there is only one solution. 6. Given a = 423, c = 288, y = 35 15' ; find a. c. Given b = 376, c = 804, y = 68 20' ; find (3. d. Given b = 650, a = 830, = 98 56' ; find ft. e. Given a = 67.2, c = 40.4, y = 24 49' ; find a. /. Given b = .0188, c = .0196, y = 100 14' ; find ft. g. Given a = 504.2, c = 1763, = 12 39' ; find y. h. Given b = 3,245,000, c = 2,488,000, ft = 80 28' ; find y. 4. Type form : aba type with two solutions. Form of solution when two solutions are found. Given a = 187, b = 235, a = 37 15'. a sin a Or Newton's check formula, b + a _ cos ^ (ft a) c sin 7 log 6= 2.3711 + log sin a = 9.7820 10 12.1531 -10 log a = 2.2718 log sin = 9.8813 10 272 UNIFIED MATHEMATICS ft = 49 83' a = 37 15' .71 = 93 12' loga= 2.2718 + log sin 7! = 9.9993 10 12.2711 - 10 log sin a = 9.7820 10 log ci = 2.4891 Ci = 308.4 log& = 2.3711 + log sin 7! = 9.9993 - 10 12.3704-10 log sin ft = 9.8813 10 logd = 2.4891 Compare with values for log c\ (and lo 2 AI ab sin 71 log a = 2.2718 + log b = 2.3711 4- log sin 71 = 9.9993 10 log 2^ = 4. 6422 2 AI = 43870 AI = 21935 02 = 130 27' a = 37 15' 72 = 12 18' loga= 2.2718 + log sin 72 = 9.3284 10 11.6002- 10 log sin a = 9.7820 - 10 logC2= 1.8182 c 2 = 65.8 log 6= 2.3711 + log sin 72 = 9.3284 10 11.6995-10 log sin 2 = 9.8813 10 logC2= 1.8182 c 2 ) found above. 2 A 2 = ab sin 72 log a = 2. 27 18 log b = 2.3711 log sin 72 = 9.3284 - 10 log 2 A 2 = 3.9713 2 A 2 = 9360 A 2 = 4680 5. Given two sides and the included angle : type airy. The method of solution here given is the solution by right triangles, since that involves no new formula and no greater amount of computation than the common solution employing a new tan- gent formula ; the tangent formula is given in Section 8 of the preceding chapter. Solution by right triangles. m Given a = 280, b = 240, 7 = 35, h = a sin 35, x = a cos 35, z = b a cos 35. asin35 ;c= asin > ; check, c = & sin ^ z b a cos 35 sin a sin /3 SOLUTION OF TRIANGLES 273 2 A log a + log sin 35 : log h -. - log Z : log tan a -. a . ?' log a : -f- log sin 7 : log sin a . logC : C : log b : + log a : + log sin 7 log 2 J. : 2 A A -. ab sin 7 2.4472 9.7586 - 10 2.2058 : 1.0253 1.1805 : 86 13' : 58 47' ; 2.4472 : 9.7586 -10 2.2058 - 10 ; 9.9990 - 10 ; 2.2068 161.0 2.3802 : 2.4472 : 9.7586 - 10 : 4.5860 : 38550 = 19270 log a = 2.4472 + log cos 35 = 9.9134 - 10 log x = 2.3606 x = 229.4 z = 10.6 Check. log 6 = 2.3802 + log sin 7 = 9.7586 12.1388 -logsin/3= 9.9321 logc= 2.2067 This problem also occurs frequently in surveying. It permits the determination of the direction and length of a tunnel through a mountain by means of the location of some point from which both ends of the proposed tunnel are visible. The dis- tance and direction from a given point to a second point, past some barrier, are determined by this method. Thus the distance from B to C through woods can be found, if some point can be located from which both C and B are visible. The distance and direc- tion to invisible points are constantly needed in artillery fire ; another method of finding distance and direction of the target is that of finding an observation point from which both the gun and the target, invisible at the gun, are visible to an observer. Distance across a barrier 274 UNIFIED MATHEMATICS PROBLEMS Using the form of solution above, solve for the side opposite the given angle. 1. Given a = 3846, & = 4977, y = 3810'. Find c. 2. Given b = 4.832, c = 8.973, a = 108 56'. Find a. 3. Given a = .0485, c = .0682, (3 = 58 38'. Find 6. 4. Using the form of solution given above, find the side opposite the given angle in the following five problems ; time yourself, and compare with the time for the same five problems solved by the cosine law (page 255). a. Given a = 366, 6 = 677, y = 1510'. b. Given a = 423, c = 288, ft = 35 15'. c. Given 6 = 627, c = 816, = 10041'. d. Given a = 635, c = 341, 0=67 38'. e. Given c = 184, 6 = 295, =13054'. 6. Discussion of checks and methods. The procedure by logarithms as with physical measurements involves numerous approximations. That two values of log c, in the check and in the solution, or by two different methods of solution, do not agree precisely is a frequent result of correct computation. However, the disagreement will be within certain well-defined limits, depending entirely upon the range (number of places) of the logarithm tables which are used both for numbers and for angles ; the tabular difference should be noted, mentally, and any discrepancy between check and computation should be examined as to its effect upon the value of the computed quantity. Thus no error in our computation (page 273) accounts for the difference between log c = 2.2068, and log c = 2.2067, nor does this here affect the value of c. However the angle of 86 13' is so near to 90 that the tangent grows very rapidly ; the tabular difference here is large, and might easily affect our result, through the inevitable inaccuracy of ordinary interpolation in this neighborhood. SOLUTION OF TRIANGLES 275 7. To determine the angles of a triangle when the three sides are given : abc type. tan a - r V( 8 _ a )( s _ 6)(s c) 2 s- a tan * 8 r s 4-& + c 2 s-b tan -v r 2 2 ' 8-C a = b = Check. a = 7 = 28 = 8 = s a = a + /3 4-7 = log (s - a) = + log (s - 6) = + log (s - c) = - log s = Check by noting sum of s a, s 6, s c which equals s. logr = - log (s - a) = log r 2 = logr = logr = -log (-&) = log tan J a = log tan \ /3 = = = logr= logr = log (s - c) = lo g* = log tan i 7 = 1& -A- 7 = Complete the solution of a problem of this type, form as above, taking a = 4320, 6 = 6840, and c = 8630. TIMING EXERCISES 1. Employing the form of solution as above, solve the follow- ing four problems for the angles a, ft, and y, timing yourself ; write down the complete form necessary for the solution of each problem before using the logarithm table. a. Given a = 320, 6 = 640, c = 580. 6. Given a = 44.8, 6 = 76.2, e = 70.4. c. Given a = 4.49, 6 = 8.87, c = 9.13. d. Given a = .0624, 6 = .0688, c = .0731. 276 UNIFIED MATHEMATICS 2. Solve the following four problems for , /3, and y, taking note of the time required. b = 640.6, c = 580.4. b = 7461, c = 5395. 6 = 2.346, c = 3.045. = 2.34xlO- 6 , c = 2.87xlO- 6 . a. Given a = 320.4, &. Given a = 3482, c. Given a = 1.835, d. Given a = 1.43 x 10" 6 , 3. How would the solution of problem 2 d be changed if a, 6, and c were given as 1.43, 2.34, and 2.87, respectively ? PROBLEMS Type problems. Solve for the other parts required for the solution of each problem. a b c a. 33 15' make a note of the time 1. 8294 6788 2. 8294 6788 3. 8294 6788 4. 8294 6788 5. 8206 6. 8206 7. 8206 6009 8. 356 9. .03267 33 15' 9645 33 15' 67 C 33 25' 15' 33 15' 67 25' 133 64 15' 10' 235 .05431 63 40' 10. 83 x 10 6 , 67 x 10 6 , 54 x 10 6 . 11. Given two forces of magnitude 384 and 276 acting at an angle of 38 with each other. Find the angle which the re- sultant makes with the larger force and the magnitude of the resultant. Note that the problem is simplified by re- garding the line of the larger force as an axis of reference ; the second force adds to this a component 276 cos 38 ; the 276 sin 38 Resultant of two forces vertical component is 276 sin 38 tan = 384 + 276 cos 38' the magnitude of the resultant, r, is by the sine law, SOLUTION OF TRIANGLES 277 r - = 276 sin (180 -38) sin' 276 sin 38 whence r = : sin < or r = V(276 sin 38) 2 +(384 + 276 cos 38) 2 , as hypotenuse of a right triangle ; or r = V(276) 2 + (384)' - 2 x 384 x 276 x cos 38, by the cosine law. 12. Given two forces of magnitude 684 and 450, acting at an angle of 64. Find the resultant in magnitude and direction, graphically and trigonometrically. 13. From an aeroplane moving horizontally at rate of 120 miles per hour (176 feet per second) a bullet is shot at right angles to the path of the aeroplane with a velocity of 2800 feet per second. What is the resultant velocity in magnitude and direction ? 14. A cylindrical trough of horizontal length 20 feet and with the ends semicircles of radius 2 feet each, contains how many gallons of water for 1 foot in depth, for 1^ feet, for 2 feet? 15. A typical oil tank is 30 feet long and has a diameter of 8 feet. Compute the volume in barrels (see page 94) for each foot of depth. Do not carry beyond tenths of a barrel. 16. What angle does ^ = 2# + 12 make with the #-axis? What angle does 3 y 4 # 20 = make with the a^axis ? Find the area and the other angle of this triangle, formed by the two lines and the avaxis, by trigonometrical methods. Find the angle between the two Hues by the formula, tan = ^ *- and check. Find the area by analytical 1 methods to check upon the trigonometrical solution. 278 UNIFIED MATHEMATICS Moving aero- plane 17. In the figure AB represents the move- ment in 30 seconds of an aeroplane moving parallel to OX at rate of 120 miles per hour ; Z AOX is measured as 83 30' and Z BOX is measured as 74 48' ; find the distance OB and the position of the aeroplane at the end of the next 30 seconds if it continues on its present course. 18. In problem 17 discuss the percentage effect of an error of 1 in angle A OX and in angle AOB, respectively. 19. Given that observers A and B at the ends of a battleship 340 feet long observe an object at angles of 106 30' and 74 48' respectively with the line AB. Find BO and AO. Solve this problem also graphically. 20. A and B are observation stations on the shore, 10 miles apart and may be assumed to be on an east and west line ; R, the battery, is three miles east of A and one mile south. Given that a battleship P is observed from A at an angle of 68 12' and from B at an angle of 59 02' with AB ; find the coor- dinates of P; find the range from R and the angle made by RP with the east and west line ; find the distance using the for- mula for the distance between two points. This problem is solved in actual practice graph- ically on large plotting boards. Using i inch to the mile, how closely could you approximate the distance ? 21. In the preceding problem, suppose that the observa- tions reported at the end of 1 minute are from A, -(38 08', and Shore battery observations SOLUTION OF TRIANGLES 270 from B, 59 12', locate the direction of movement of the ship. 22. If the battleship of the two preceding problems is 600 feet long and broadside to A, what angle does it subtend 23. When guns are tested at Sandy Hook or other proving grounds, the actual range for any angle of elevation of the gun is obtained by coincident observation of the splash of the shell by several observers located in towers along a line roughly parallel to the line of fire of the projectile. The li'ii e ^ Iffiire- -E- Observation towers at proving grounds shell is loaded with a slight charge of high explosive in order to give a splash of some magnitude. For convenience in our figure, we have assumed the towers on a north and south line spaced as indicated ; the height of the tower is regarded as negligible. Given that observers at A, B, C, D, and E observe the angle of the splash with the north and south line, the azimuth angle, as 2 20', 21 24', 27 16', 41 35', and 68 54', find the distance and direction of S, the splash, from G, the gun, which is 400 yards due east of A. Note that any two observers give the position of S ; the substantial agreement of three observers is taken as sufficient. Compare the solu- tions. Determine the coordinates of S with respect to a horizontal axis through BCDE and a vertical axis through AG. Assuming that the gun was pointed south the deflection, 280 UNIFIED MATHEMATICS to the east here, is termed the drift ; it is due to wind and other causes ; determine this angle. It is not essential that the towers be in a straight line ; the distance and direction of lines between adjacent towers is carefully measured. A large plotting board is lised to obtain a graphical solution. London Bridge Five pure elliptical arches ; central one 152 feet by 37 feet 10 inches, and the others 140 feet by 37 feet 2 inches. CHAPTER XVIII THE ELLIPSE 1. Parametric equations of an ellipse. The parametric equations of the circle, with center at the origin x = r cos 6, y = r sin 0, and of the circle with center at (h, Jc) x li = r cos 0, y Jc = r sin 0, if modified to read, and x = T! cos 6, y = r. 2 sin 0, h = r i cos 0, k = r sin 6 respectively, give a curve which is closely related to the circle. This curve is called the ellipse ; r t and r 2 are called the major and minor semi-axes of the ellipse, and the circles obtained 281 282 UNIFIED MATHEMATICS with radii i\ and r 2 are called the major and minor auxiliary circles of the ellipse. An ellipse with its major and minor auxiliary circles x TI cos 6 . x = r 2 cos . . gives the larger circle. gives the smaller circle. y t = TI sin & y - r-i sin 5 x = TI cos e . gives the ellipse. y r 2 sin P, Q, and R are called corresponding points. Eliminating between the two parametric equations of the ellipse gives + - = cos 2 + sin 2 = 1. fi z T r2 2 Writing, as is customary, a and 6 for r t and r. 2 this equation becomes THE ELLIPSE 283 If the ellipse had the center (fi, k) and the two values a and 6 corresponding to r t and r. 2 , the equation of the ellipse would be written (x h) = a cos 0, (y Jc) = b sin 6, in parametric form ; and a 2 6 2 in so-called standard form. If we assume, as is sometimes convenient, that a is the radius of the larger circle, we would have x h = b cos 6, y k = a sin 0, & 2 a 2 as the equation of an ellipse whose major axis is vertical. Note particularly that the terminal side of the angle B does not pass, in general, through the point on the ellipse which corresponds to that angle, but the angle is made by a line passing through the corresponding points on the major and minor auxiliary circles. PROBLEMS 1. Construct the ellipse x = 10 cos 0, y = 6 sin 6, by finding the points on the ellipse given by = 0, 30, 45, 60, 90, and the corresponding points in the other quadrants. 2. Construct the ellipse x = 10 cos 6, y = 6 sin 0, using corresponding points on the major and minor auxiliary circles to locate points on the ellipse. 284 UNIFIED MATHEMATICS 3. How does the ellipse x 3 = 10 cos 0, y + 2 = 6 sin 0, differ from the preceding ellipse ? 4. Locate 10 points 011 the ellipse x 3 = 6 cos 6, y + 2 = 10 sin 6, and draw the curve. 5. Prove that a is the largest value and b the smallest of a line from to any point on the ellipse. This gives the reason for the names, major and minor axes. 6. Show that every chord of the ellipse through is bi- sected ; O is the center of the curve. 7. The five arches of the London Bridge are of true elliptical shape; the central arch has the width 152 feet (=2 a) and the height 37 feet 10 inches (=6). Find the equation and plot 12 points on this arch ; b may be taken as 38 feet. The adjacent arches are of length 140 feet and height 37 feet 2 inches. Write the equations. In each case take the major axis as avaxis and minor axis as y-axis. 2. Properties of the ellipse. The equations x = a cos 0, y = b sin 9, wherein a > &, can represent any ellipse whatever, when the axes of the ellipse are taken as the axes of reference. The geometrical peculiarities of this ellipse will be characteristic of all ellipses, provided, of course, that no limitation is placed upon a and b (except that a may be taken as greater than b). Each ordinate of the ellipse x = a cos 0, y = b sin 0, THE ELLIPSE 285 is a constant proportional part of the corresponding ordinate in the major auxiliary circle, x = a cos 0, y = a sin 6. An ellipse with its major and minor auxiliary circles x = TI cos 9 . x = T-L cos 6 . . gives the larger circle. gives the smaller circle. y = TI sin 6 y = r z sm i . ,, ... gives the ellipse. y = r z sm 6 & P, Q, and E are called corresponding points. Take any point (a cos 0, b sin 6) on the ellipse, the corresponding ordinate in the circle is y e = a sin ; but y t =b sin = - (a sin 0) = - y c . Conversely, if the ordinate of a a any point on a given curve is always a constant proportional 286 UNIFIED MATHEMATICS part of the ordinate of a corresponding point on a given circle, for equal values of the abscissa, the curve is an ellipse ; the ratio need not be less than unity as the figure clearly indicates, a vertical ellipse being represented when the ratio is greater than unity. This property of this ellipse is equally true of any ellipse, replacing the terms ordinate and abscissa, where given or implied, by perpendiculars to the major and minor axes, re- spectively, of the curve. If the major auxiliary circle is rotated about OX as an axis through + an angle a, cos a = ~ , the a projection upon the plane of the original position will be the ellipse x = a cos 6, y = b sin.0; evidently the x of any point on the projected curve may be taken as the x of a point on the project- ing circle, or x = a cos 6. The ordinate on the pro- jected curve is in a con- stant ratio, cos = - , to a Elliptical section of a circular cylinder t ^ e or dinate on the circle. Further any plane section of a cylinder with circular base is an ellipse, for with the same abscissas, the ordinates of the curve of section bear a constant ratio - to the correspond- cos a ing ordinates on the circular base. THE ELLIPSE 287 Let the plane cut the axis of the cylinder in 0, and let BOX, on the diagram, be the intersection of the cutting plane with the plane of the circular section through (paral- lel to the base of the cylinder). Note that angle PMQ is constant. Similarly if a circular cone is cut by a plane cutting all the elements of the cone, the section formed is an ellipse; the geometrical proof is rather too complicated to give here but an analytical proof is indicated in Chapter XXXII, Section 6. The ancient Greeks studied the properties of the ellipse entirely from the point of view of the curve as a plane section of a cone. The Greek theoretical work concerning the properties of conic sections made it possible for Kepler to discover that the path of the earth about the sun is an ellipse, and for Newton to formulate the law of gravitation. The properties mentioned are intimately connected with the applications of the ellipse in engineering problems. From the property of the ordinates it follows that the area of an ellipse is - Ti-o 2 = nab. A = irab. a The proof is strictly by a " limit process," and may be made reasonably evident by dividing the semi-axis into 25, 50, 100, equal parts and drawing two series of rectangles on these equal subdivisions about the corresponding ordinates to fall entirely within (or entirely without) the ellipse and the circle, respectively. As the subdivisions are increased in number the one series of rectangles has the area of the quarter-ellipse as a limit ; it differs from this area never by an area as great as the rectangle of height b and base one subdivision ; so also the sum of the second series of rectangles has the quarter-circle as a limit, never differing from it by an area as great as the rectangle of height a and base one subdivision ; but the sum of the series of smaller rectangles always equals - times the a sum of the series of larger rectangles since the bases are equal 288 UNIFIED MATHEMATICS and the altitude of any one of the smaller is - times the alti- a tude of the corresponding one of the larger ; evidently, then, the limits of these sums are in the same ratio. In the dia- gram above the ruling of the paper divides the semi-major axis into 25 parts, and gives the rectangles. The Colosseum in Rome A famous elliptical structure, 615 feet by 510 feet, by 159 feet high ; the arena is an ellipse 281 by 177 feet. A rectangle of dimensions 2 a and 2 b may be circumscribed about the ellipse. The sides of this rectangle are tangents at the vertices of the ellipse ; the middle lines parallel to the sides are the axes of the ellipse ; the horizontal sides of this rectangle cut the major auxiliary circle in points which cor- respond to four symmetrically located points on the ellipse ; sin = - , cos = \/l = ^ , whence a * a 2 a x= Va 2 b 2 , y = , are the coordinates of these points on a this ellipse. The points on the major axes ( Va 2 ft 2 , 0) are the foci of the ellipse, and enjoy special properties with 289 respect to points on the ellipse, the path of the earth. The sun is at one focus of A O A< PF = e PZ defines the ellipse The constant e is called the eccentricity. 3. Standard definition of the ellipse. The locus of a point which moves so that its distance from a fixed point called the focus is in a constant ratio, e, less than 1, to its distance from a fixed line called the directrix, is an ellipse. Let the fixed point be Fand D'D the fixed line. Through F drop a per- pendicular to the direc- trix D'D meeting it in R and use this line as cc-axis. By definition, then, the ellipse will be the locus of a point which moves so that PF = e PZ, wherein PZ is the perpendicular from P to the directrix. Two points of our curve will be found to lie upon the avaxis ; the two points are the points A and A' which divide the segment FR internally and externally in the ratio e (taken as-| on our figure). Take the middle point of A' A as the origin, designating A' A, which is a fixed length dependent upon FR and e, by 2 a. Then OA = OA' = a. AF=e>AR. A'F=e-A'R. AF+ A'F = e(AR + A'R). 2 a = e(AR + OA' + = 2 e - OR, whence A'F- AF= e(A'R - AR). 20F=2ae. OF=ae. F is (ae, 0) and D'D is x - = ; e 290 UNIFIED MATHEMATICS PF= e PZ, taking P as (x, y), gives in analytical language, or formulas, V( - ae)2 + (y - O) 2 = e . (x - - } = (ex - a). x 2 -f 2 e 2 + i/ 2 = e 2 2 + a 2 . /}- ,; -H -- ^ - = 1. a 2 a 2 (l-e 2 ) Let ft 2 = a 2 (l - e 2 ), a 2 whence + ^ = 1. (1) 2 2 But this equation is satisfied by every point x a cos 0, y = b sin 0, determined as we have indicated above, and conversely ; our two definitions are equivalent to each other. The form of equation (1) shows that the curve is symmetrical with respect to the two lines which we have chosen as axes ; i.e. since x and x give precisely the same equation to deter- mine y, the curve is symmetrical with respect to the y-axis ; similarly since y and y give precisely the same values for x, the curve is symmetrical with respect to the cc-axis. The coordinate axes are axes of symmetry of the curve. By symmetry with respect to the y-axis we mean that if the right half of the curve is folded over on the y-axis as an axis, i.e. revolved about it as an axis (or axle) through an angle of 180, the two sides will coincide throughout. Hence corresponding to F l and D'D, there is another focus F 2 and a corresponding directrix D Z D' 2 , enjoying precisely the same properties with regard to the curve as F : and D'D. THE ELLIPSE 291 By symmetry, and since Similarly, and, since ^ = e P&, P 2 F 2 = e P 2 Z. = e tag Kir Symmetry of the ellipse with respect to its axes 4. Sum of the focal distances constant. Taking any point PI(XI, yi~) on the ellipse, the focal distances Pi-F\ and P\F 2 are equal to a ex v and a + ex 1} respectively. For PiFi = e P t Z = eix l J = ex a, which is a negative V e j distance since PI and are upon the same side of the line = - (ex v + a). As D'D. Similarly P& = e - P& = e positive values P^ and PjF 2 are a exi, and a + ex l ; these may also be derived by the formula for the distance between two points, noting that (x l} y^) is on the curve ; PF : -f PF 2 = a ex l + a -\- exi = 2 a. 292 UNIFIED MATHEMATICS 5. Right focal chord. When x = ae, ?/2 = 52(1 _ e 2 ) = -, since & 2 = a 2 (l - e 2 ), jji 2 5 2 y = , giving as right focal chord. a a 7 2 The value y = is the value of y obtained in the parametric ct T Tft form when sin 6 = - , ?/ = 6 sin = a ' a ft 2 / x = a cos 6 = a-%/1 s Va 2 6 2 = ae, showing that the focus as we have now defined it coincides with the focus as first defined. 6. Standard forms of the ellipse. The equation (- ^- = 1 may be interpreted geometrically, = 1, in which CM and MP represent the distances cut off from the center on the major and minor axes of the ellipse by the per- THE ELLIPSE 293 pendiculars to the axes from any point on the ellipse ; CA and CB are the lengths of the semi-major and semi-minor axes. Given a horizontal ellipse with center at (h, k) and axes a and b, the relation CM* "Ml* _ 1 CA 2 CB 2 ~ becomes a 2 b 2 = x h and MP y k. Similarly a vertical ellipse with center (h, k) and axes a and &, respectively, has the equation (y k) 2 , (x h} 2 - + i ' = 1. since a 2 b 2 CM is here y k and JP is x h. m f (x-hY , (yk) 2 (y k) 2 . (x-hy 1 Type forms : ^ - ^- + Vz -- _ 7.\2 _ + ^ - *- = l represents an ellipse. a 6 2 t- + ^^ ' = represents a point ellipse or two imaginary straight lines through (h, k) ; the only real point which satisfies this equation is the point (h, k). 294 UNIFIED MATHEMATICS (y. Jf\2 (ft Jf\2 i '*- + ^ 1- = 1 represents an imaginary ellipse ; only imaginary values of x and y can make the sum of the two squares on the left equal to 1. The equation of any ellipse with axes parallel to the coordi- nate axes may be written, = k, wherein a 2 6 ka 2 and kb* represent the squares of the semi-axes. As k approaches zero the ellipse diminishes in size, and when k = the equation represents the point (x 1} y^; when k becomes negative the ellipse becomes imaginary. 8. Illustrative problems. a. Plot the ellipse 4 x z + 16 x + 9 y* 18 y 75 = 0. First write in form to complete the squares, 4(x 2 + 4 x ) + 9(y 2 2 y ) = 75. Complete squares : 4(x 2 + 4x + 4)+ 9(j/2 2 y + 1) =75 + 16 + 9. 4(x + 2)2 + 9(y - 1)2 = 100. Write in standard form : (x + 2V 25 ft2 _ lOQ _ 20 a ~ 45 ~ 9 Plot the center, extremities of major and minor axes ; foci ; extremities of right focal chords ; at least one further point, obtained from the original equation and selected so as to give a point approximately midway between the extremity of a focal chord and the corresponding extremity of the minor axis ; by symmetry three other points are obtained. In this case x = gives desirable further points. 9 \/81 + 675 = 1 = 1 K27.5) = 1 3.06 = 4.06 or - 2.06. THE ELLIPSE 295 b. Plot the elliptical arch of a bridge, arch 125 feet wide and 37 feet high. Plot points for every 10 feet of the span ; and compute to T X 7 of 1 foot. |||g|jap : ^ ; ; 1 ; hi & -I _l -,,(-- 1 IE 3 ? :fe : r :I 1 -i:" 2 ..S s g; :|: 2 "~~. X.1 5> ^ gj ^ / "- ._ ...j - > - K S -- i -- -v ti^-^l - S i II i -, .-/ in jo :,,, _4o :,(.) 60~" Elliptical arch, 125 foot span by 37 feet high Scale, 1 inch to 40 feet. Horizontal measurements are from the middle of a quarter-inch square. * _L y \ (62.5) 2 372 = (62. 5) 2 ; & 2 =372; ae = -372; ^- = J^_. a 62.5 62.52 Here compute 37 2 and 2/2 = 372 _ /JLV . X 2, V62.5>/ /_37_y \62.5y , and multiply the latter by x 2 = 100, 400, 900, 1600, 2500, and 3600 ; extract the square root of the difference ; use four-place logarithms. log 37 = 1.5682 log 62.5 = 1-7959 log quot. = 9.7723 10 log quot. 2 = 9.5446 10 ( V- \Q2.b) .3504 = 1369 - 35.04 ; y w = -f- 36.51 2/20 2 = 1369 - 140. 16 = + 34.93 2/ao 2 = 1369 - 315.36 ; y 30 = + 32.46 2/40 2 = 1369 - 560.64 ; y^ = + 28.43 I/so 2 = 1369 - 876.0 ; y x - + 22.20 yeo 2 = 1369 - 1261.44 ; y = + 10.37 296 UNIFIED MATHEMATICS EXERCISE. Draw the major auxiliary circle on half-inch coordinate paper, and compute corresponding ordinates in the 37 preceding arch as -f - of the ordinates on the circle ; e.g. for 62.5 S7 x = 10, find y graphically on the circle and multiply by - 02. o PROBLEMS 1. Plot the ellipse 9 x 2 + 36 a + # 2 6 ?/ = ; what are the coordinates of the foci ? 2. Plot one quarter of the ellipse - + -^ = 1. 147 2 59 2 3. Plot an elliptical arch, width 233 feet, height 73 feet, plotting at least 10 points spaced at 20-foot intervals from the center. These are the dimensions of the arch of the Walnut Lane Bridge in Philadelphia; the arch is approximately an ellipse. 4. Plot the upper half of an ellipse, giving a vertical ellip- tical arch 13 feet 1| inches high, and 9 feet 2 inches wide. This represents the upper portion of an elliptical sewer used in the city of Philadelphia. 5. What limitation is there upon the values of A and B if the equation, A& + By 2 + 2 Gx + 2 Fy + C = 0, is to represent an ellipse ? 6. Put the following equations in standard form, complet- ing the square first and reducing to standard form by division. Time yourself. y* 6y4:3= 0. c. ox*-17x + 10y 2 - 100 = 0. d. 5*2 + 127/2-117 = 0. 7. Plot the preceding five ellipses, choosing an appropriate scale. Plot the extremities of major and minor axes, the THE ELLIPSE 297 extremities of the right focal chords, and one other point obtained by computation, together with the three points sym- metrical to the computed point. 8. Determine a 2 and ft 2 to one decimal place, in the follow- ing three ellipses : a. 17 x 2 + 43 y* = 397- b. 5x 2 -17x + 10y*-35y = 0. c. l(x - 2) 2 + 3(y - 3) 2 = 39. 9. In the three ellipses immediately preceding determine ae, , and - to one decimal place. a e 10. In each ellipse of problem 8 determine x when y = 2. 11. Using the data of problems 8, 9, and 10, plot the three ellipses of problem 8. 12. In the ellipse -^- + f- = 1, find the foci. What is the 100 36 distance of the point whose abscissa is + 3 from each focus ? of the point whose abscissa is 4, 5, 6, 7, x ? 13. Put the following equations in standard form and dis- cuss the curves : a. x*-6x + y 2 +8y-10=Q. b. z 2 - 6z + 4?/ 2 + 8 + 11 =0. c. B 2 - d. x 2 - e. x 2 14. The path of the earth about the sun is an ellipse with eccentricity .01677 ; this may be taken as ^ in the following computations. If the major axis of the earth's orbit is 185.8 million miles, determine the focal distance, i.e. from the sun to the center of the path ; determine also the minor axis. If a scale of one-half inch to fifteen million miles is taken, at what distance will the point representing the sun be from the center 298 UNIFIED MATHEMATICS of the path ? What will be the difference in length between major and minor axes? The path of the earth about the sun The distance of the sun from the center of the ellipse is represented by ^ 5 of an inch, on this diagram. 9. Tangent to ellipse of given slope. (See page 225.) To find the tangent of slope 2, y = 2 x + k is solved as simultaneous with the ellipse equation. An equation whose roots are the abscissas of the two points of intersection is found and the con- dition is used that the two points of intersection be coincident. a; 2 - = 1, a* b 2 y = 2 x 4- k. bW + a? (4 x* + 4 kx + fc 2 ) a ? x z (b* + 4 a 2 ) + 4 a z kx -f- a, 2 A; 2 - = 0. = 0. _ V4 a 4 fc 2 - (fr 2 + 4 a & 2 + 4 a 2 - 2 - 2 2 fc 4 a 2 - ft 2 ) 6 2 + 4 a 2 THE ELLIPSE 299 Put 52 + 4 a 2 _ fc2 _ o. ft 2 = 6 s + 4 a 2 , k = V6 2 -f 4 a 2 , y = 2 a; V& 2 + 4 a 2 are the two tangents of slope 2. Similarly the tangent of slope m is obtained by solving (2) y = mx as simultaneous, and writing the condition for equal roots. Clearing (1) of fractions, & 2 a 2 + a 2 ?/ 2 - a 2 ^ 2 = ; substituting from (2), 6 2 cc 2 + a 2 (m 2 a 2 -f- 2 Jcmx + A; 2 ) a 2 6 2 = (6 2 + a 2 2 ) x 2 + 2 fca 2 wwj + (a 2 A; 2 - a 2 6 2 ) = 0. _ - ka*m Vfc 2 a 4 r/i 2 (6 2 + a 2 m 2 )(a 2 fc 2 - a 2 6 2 ) 6 2 + a 2 m 2 Putting the discriminant equal to zero, A; 2 = 6 2 + a 2 w 2 . A; = V6 2 4- a 2 m 2 , whence y = mx are the two tangents of slope m to This method of obtaining the tangent applies to any curve given by an equation of the second degree. 10. Focal properties of the ellipse. The perpendicular from the focus (ae, 0) on any tangent of slope m meets it in the point, whose coordinates are found by solving the equations of these lines as simultaneous. 300 UNIFIED MATHEMATICS The equations, y = mx + Va 2 2 + 6 2 , of the tangent, and may be written y = (x ae), of the perpendicular, m y mx- = V 2 m 2 + 6 2 . my + x = ae. The perpendicular from the focus upon any tangent to an ellipse meets it on the major auxiliary circle The point of intersection satisfies both these equations ; further it satisfies the equation obtained by squaring and add- ing both members of each of these equations : my + + ma = ara = a 2 m 2 + 6 2 + a 2 6 2 = a 2 (l + m 2 ). a 2 + f = a 2 ; hence the point of intersection of the perpendicular from the focus on any tangent lies on the major auxiliary circle. Note that the above demonstration applies equally well to the perpendicular from the other focus (ae, 0) and equally well to the other tangent of slope m, y'= mx Va 2 m 2 -f- 6 2 . THE ELLIPSE 301 11. Tangent to an ellipse at a point PI(X { , y^) on the ellipse. The method outlined is general, being applicable to any alge- braic curve. The point PI(#I, yi) on the curve, considered as fixed during the discussion, is joined to a neighboring point PZ( X I 4- h, y t + ft) on the curve, which second point is then made to approach (x l} y^) along the cwve. The slope of the chord joining P x - is the slope of the chord h Jc to P 2 , -, is found not to change h indefinitely, but is found to ap- proach a definite limiting value as h and k approach zero, i.e. as P 2 approaches P x along the curve. This limiting position of the chord is the tangent at P l ; this line can be shown in the case of the ellipse (or any curve given by an equation of the second degree) to cut the curve in two coincident points at (x 1} y^) and in no other point. The method is applied in parallel columns to a general problem and to a particular problem. Tangent to the ellipse, - a 2 & 2 at PI(X I} . 25 4 " at Pi(3, f ) on the ellipse. Take the second point PZ(XI + h, y l -f- Jc) on curve P 2 (3 + h, f + k) on curve Substituting, + hy + a?( yi + A;)' - a 2 & 2 = 0. 4(3 + /i)2 + 25(f + fc)* - 100 = 0. + 2 62*0?! + 6 2 /i 2 + a 2 y 2 + 2 d^ky l + aW - a 2 6 2 = 0. But &V + aVi 2 - a 2 6 2 = 0. 36 + 24 h + 4 fc 2 + 64 + 80 k + 25 fc 2 - 100 = 0. 302 UNIFIED MATHEMATICS i + 52^2 Subtracting (and canceling) . , + a 2 & 2 = 0. 24 ft44 fc 2 +80 k 4- 25 fc 2 =0. &(804-25 k}= -ft(2444 h). k = 24 4 4 h h 2a?i/ i + a?k h 80 + 25fc' The chord PP Z is given by : y -i = (*-!). y-| = 7(*-3). ft Q II Since P 2 is on the curve, the chord equation may be written : 1111,= ! + Wh -! - + 24 + 4 h , ox (x 3). v Let P 2 approach P Y along the curve ; h and A; both approach 0, i.e. can be made just as small as you please. Thus if h is made .01 in our numerical prob- lem, k will be about .003, which can be obtained by solving the quadratic JEfM -i- -/r4, ^jffl^Fi^ ^ for fc. It is evident that the con- stant here may be regarded as 24 h + 7i 2 and that as ft = 0, this constant approaches zero, and one root of k approaches zero. Xote that the second value of A; approaches 2 y 1} and corre- sponds to the fact that the given value of x, x v + h, is the abscissa of two points, on the upper and lower parts of the curve, respectively. Since h and k both approach zero, - is the slope of the chord PiP 2 and 24 + 4ft =24 and - 80 4 25 fc = - 80 THE ELLIPSE 303 and the slope of the chord approaches more and more nearly, and as near as you may please to make it, by taking h (and thus A;) small enough, _24 ~80' Hence the chord approaches a limiting position, given by N 8 24, which may be written a* yi y = aV + Vx? y - 1.6 = - .3(x - 3) = ft262 ' or y-1.6 = - .3(x - 3). a?y iy - a*&2 = 10 >/ + 3* - 25 = 0. 40 ?/ + 12* -100 = 0. By precisely this method, step for step, the tangent to Ax* + By* + 2Gx + 2Fy + C = 0, is found to be Axjx + Bi h y + G(x + a;,) -f- ^(y + 2/0 + C = 0. PROBLEMS 1. Find the tangents of slope -f 2 and of slope 3 to each of the ellipses in problem 6 of the preceding set of problems ; the five problems, tangents of slope + 2, should take not to exceed 30 minutes ; note that after substituting 2 x + k for y it is better procedure, surer and quicker, to combine terms by in- spection rather than to expand each binomial before combining. This means to pick out the terms containing a; 2 , for example, and write the sum of these coefficients directly. 2. Draw at least three of the tangents of slope + 2 in the preceding exercise, and three of slope 3, each to its conic as previously drawn. Find the point of tangency algebraically and graphically. 304 UNIFIED MATHEMATICS 3. Without plotting the ellipse itself plot 12 tangents of slope 0, 1, 2, 3, 4, 5, 6, and 10, and of slope ^, 1, 3, and 6 to the ellipse 9x z + 25y 2 = 900 ; note that these give a fair outline of the ellipse. 4. In each of the ellipses of problem 8, page 297, find the tangents at the point whose ordinate is 2, employing the results of problem 10 of the same set, and using the formula, Ax& + By v y +Q(x + x l ) + F(y + y t ) + G'= 0. 5. Derive by the method outlined in section 11 of this chapter the tangent to the ellipse 9 x z + 25 y 1 = 900 at the point (8, 3.6) which is on the curve ; at ( 6, 4.8) on the curve. 6. In the ellipse 9 x 2 + 25 y 1 = 900, verify that the perpen- dicular from the focus upon any tangent to the ellipse meets it on the major auxiliary circle. Note that the converse is also true. This gives a method for drawing the tangent to an ellipse from a point outside the ellipse ; explain. 12. The tangent to an ellipse at a point on the ellipse con- structed from the tangent to the auxiliary circle. Let (x 1} y^ be any point on the ellipse ; the tangent is ^ -+- ^^ = 1 ; the a 2 a?-intercept, x i} of this tangent is , obtained by solving X L ?L- -|_ IM. = 1 as simultaneous with y = 0. Evidently x, = a 2 b- Xi depends only upon x\ and a, not involving b or y. Hence this value would be unchanged if b were taken equal to a. The tangent to the major auxiliary circle x 2 -f-?/ 2 = a 2 at (x l} y z ) on the circle is x^x + y?y = a 2 , and the intercept of this tangent a 2 on the ic-axis is also . This gives the following rule for y l drawing a tangent to an ellipse at any point on the ellipse : Construct the major auxiliary circle to the given ellipse; find the point P 2 on the circle having the abscissa of the given point; at the point P 2 construct the tangent to the circle, cutting the THE ELLIPSE 305 x-axis at T; the line joining T to P v on the ellipse is tlie tangent to the ellipse. Tangent to an ellipse constructed from the tangent to the auxiliary circle a the corresponding point Note that even if 6 is greater than a, this construction gives the tangent, but the circle ce 2 + y 2 =a 2 is then the minor auxiliary circle of the given ellipse. 13. The tangent to an ellipse bisects the angle between the focal radii to the point of tangency. Let ^ + M = 1 be the a 2 6 2 tangent at (x 1} y t ) which intersects the a;-axis at T (, 0\ 22 ^ ' Hence, a-ex,, 306 UX I PIED MATHEMATICS .-. PiT bisects the exterior angle of the triangle since it divides the opposite side into segments proportional to the adjacent sides. The normal bisects the interior angle between the two focal radii ; if the normal be drawn, each focal radius makes the same angle with it. Tangent to an ellipse constructed from the focal radii to the point of tangency Any ray of light or sound striking a reflecting surface is reflected in the plane of the normal to the surface and the original ray in such a way as to make the angle of incidence (i.e. between normal and original ray) equal to the angle of reflection. Hence rays starting from FI in our figure con- verge at F 2 . This is the principle of " whispering galleries," in which the rays of sound starting from a point FI converge at another point F 2 , making audible at F. 2 whispers at F t ; at intermediate points the conversation may not be audible as there is no reenforcement by convergence. PROBLEMS 1. In the ellipse 9 x 2 + 25 y 1 = 900, give the two graphical methods for drawing the tangent at the point (6, 4.8) on the ellipse. THE ELLIPSE 307 2. In the ellipse 25 x 2 + 9 y 2 = 900, give the graphical methods for drawing a tangent at the point (4.8, 6) on the ellipse. 3. Construct the three ellipses : 100 + 36 = ioo + ioo = L _*!. H L y!_ = i 100 144 Draw the tangent to each of these ellipses at the point whose abscissa is + 6 ; find the equation of each of these tangents and prove that they intersect on the avaxis. 4. Find the equations of the two focal radii to the point x~ v 2 (6, 4.8) on the ellipse - + = 1 ; find the bisectors of the 100 36 angles between these focal radii ; find the bisector of the angle which does not include the origin, and prove that it .coincides with the tangent at (6, 4.8) to the ellipse. 5. If an elliptical arch is to be in the form of the upper half of an ellipse, find the equation and plot ten points, given that the width of the arch is to be 100 feet and the height is to be 40 feet. 6. If the preceding arch is to have the dimensions as given, but is to be constructed as the upper quarter of a vertical ellipse, find the equation of the curve. Note that you have a point (50, 40) which is to satisfy the equation of the curve which can be written with only the denominator of x 2 as un- known. Compare this arch with the preceding one as to beauty of design. 7. Find the lengths of the ten vertical chords of the arch in problem 5, dropped from the tangent at the top of the arch and equally spaced horizontally. 308 UNIFIED MATHEMATICS 8. By the method of article 13, find the tangent to the curve given by the equation, xy=15, at the point (3, 5) on the curve. 9. By the method of article 9, find the tangent of slope 2 to the curve given by the equation, xy = 15. 10. Write the equations of the three ellipses of problem 8, page 297, in parametric form. OJ 2 y2 11. In the ellipse, - + ^r = 1, find where lines from 100 oO (50, 30) inclined to the horizontal axis at angles of 15, 30, 45, and 60 respectively, cut the ellipse. Find the lengths of these lines from (50, 30). See problem 5. 12. If in the ellipse of problem 6, supporting chords are drawn diagonally between parallel vertical chords, computed in problem 7, each from the upper point of the right-hand chord to the lower point of the left-hand chord (on the right side of the ellipse), compute the lengths of these chords. From Tyrrell's Artistic Bridge Design Alexander III Bridge in Paris A parabolic arch ; span, 107.6 m.; rise, 6.75 in.; width, 40 m. CHAPTER XIX THE PARABOLA 1. Definition. The ellipse has been defined (page 289) as the locus of a point which moves so that its distance from a fixed point, the focus, is in a constant ratio less than one to its dis- tance from a fixed line, the directrix. If this constant ratio is taken equal to one, the curve generated by the mov- ing point is called a parabola. PF = e PZ, e < 1, defines an ellipse. PF = PZ defines a parabola. PF = e PZ, e > 1, defines a hyperbola. 2. Equation of the parabola. Through the focus draw the perpendic- ular FQ to the directrix ; take this line as avaxis. Take FQ, which is constant, as 2 a. On the axis chosen only one 309 --Z- -F- F, the focus; ZZ', the di- rectrii A point equidistant from Fand ZZ' moves on a parabola. 310 UNIFIED MATHEMATICS point is found which is on the given curve ; the mid-point 0, dividing the segment QF in the ratio 1 to 1, is such that its distance from F, the focus, equals its distance from the directrix. Through O take a perpendicular to OX as the y-axis. Evidently F is the point (a, 0), and the directrix is the line x + a = 0. Take P(x, y) any point which is on the given curve, i.e. any point equally distant from F and from the line, giving = PZ. PF = V(x a) 2 + y 2 , distance between two points. PZ = x + a, distance from a point to a line. Note --ip gives the distance as negative ; but it is not necessary to take account of the sign as in the simplification this expression is squared. Equating, PF = PZ, gives V(x a) 2 -f y z = x -\- a. x* 2 ax + a 2 + y 2 = x 2 + 2 ax + a 2 . y 2 - = 4 ax. 3. Right focal chord. When x a, y = 2 a, giving the total length of the right focal chord as 4 a ; the coefficient of x in y 2 = 4 aa; represents the length of the right focal chord. 4. Geometrical properties. The curve is symmetrical with respect to the x-axis, for, assigning any value to x, you find for y two values V4 ax ; numerically equal but opposite in sign, or lying symmetrically placed with respect to the OX-line, which consequently is here the axis of the curve. The y-axis, x = 0, is tangent to this curve since, solving, ?/ 2 = 4 ax, x = 0, as simultaneous, gives y 2 = ; y equals zero twice, or the two points of inter- section of x = with y 2 = 4 ax are coincident. The point THE PARABOLA 311 (0, 0) is the point on the axis of symmetry, y = 0, which corresponds to itself ; this point is called the vertex. The line y = cuts the curve in only one finite point, given by y = 0, and x = ; the other point of intersection of y = with the curve is at an infinite distance. Given a as positive, negative values of x lead to imaginary values of y. Hence all points on the curve lie to the right of the tangent at the vertex. As x increases without limit, y increases also without limit. This curve extends, we may say, to infinity. In plotting a parabola, the vertex, the ex- tremities of the right focal chord, and at least two other points, should be plotted. The quantity y z 4 ax is evidently negative for points in- side the curve, zero for points on the curve, and positive for points outside the curve. 5. Finite points and the infinitely distant point on the parabola. - To plot carefully y z = 8 x, note that 4 a = 8, whence a = 2 ; indicate the vertex and, 2 units to the right, the focus ; 4 units (2 a) above and below the focus locate the extremities of the right focal chord which has the length 4 a ; take values of x at appropriate distances from the vertex and focus to give the portion of the parabola desired ; in y z = 8 x, x = 0, 1, 2, 3, 4, 6, and 8 give sufficient points to plot the curve for our purposes. The parabola ?/ 2 = 8 x is intersected by the line y = x at (0, 0) and at (8, 8) ; y = ^ x cuts this parabola at x = and at x = 32 ; y = .1 x cuts at x = 800, y = 80. These values are obtained by solving the equation y 2 = 8 x as simultaneous with each of the linear equations. y=.Qlx Solvm > y> = 8x gives .0001 x z = 8 x, x = or x = 80,000. If one centimeter is taken as 1 unit, y = .01 x cuts the parabola y 2 = 8 x at the vertex and at a distance of 80,000 cm., nearly ^ mile, from the vertex. As the line y = mx moves nearer and nearer to the 312 UNIFIED MATHEMATICS A line parallel to the axis of a parabola cuts it in a point infinitely distant axis y = 0, the second point of intersection moves off farther and farther, without limit. This is the meaning of the ex- pression that the axis of the parabola, and by similar reason- ing any line parallel to the axis, " cuts the curve at an infinite distance." The methods given above for plotting y 2 = 8 x apply to any parabola y 2 = 4 ax. PROBLEMS Plot the following parabolas carefully : 1. y 2 = 4 x, from x = to x = 8. 2. y z = x, from x = to x = 10. 3. y- 12 x, from x = to x = 12. 4. y 2 = T \ x, from x = to x = 100. THE PARABOLA 313 5. s = 16 t 2 , or t 2 = T T s, taking OS as the horizontal axis and taking J inch to represent 10 units of s (distance in feet) and | inch to represent 1 unit of t (time in seconds). This gives the distance fallen from rest in time t by a freely fall- ing body. 6. Find the intersections of y = x, y = ^x, and y = ^x with the curves of problems 1, 2, and 3. 7. Solve s = 800 1 with s = 16 1 2 ; s = 800 t is the space cov- ered by a body moving with uniform velocity 800 units per second. What is the physical meaning of the values obtained for the point of intersection ? 8. A mass rotated on a cord exerts a force of tension on the cord, F = Given ra = 1 pound, and r = 10 feet ; draw the graph for velocities of 1 to 100 feet per second, taking F on the horizontal axis. Compute the corresponding number of revolutions per minute for v = 10, 20, 50, and 100 feet per second. What is the relation between v and n, where n is the number of revolutions per minute ? Indicate on the vertical scale a second scale giv- ing n. 6. Geometrical interpretation of y 2 = 4 ax. The equation y 2 = 4 ax may be inter- preted geometrically as follows : MP- = 4 a VM, the square of the perpendicular from any point on a parabola to the axis is equal to the rectangle formed by the length cut off on the axis from the ver- The square on PM equals the rectangle with VM and RFC as sides tex by the perpendicular, with a constant line of length 4 a, the length of the right focal chord. 314 UNIFIED MATHEMATICS 7. Standard and limiting forms. Given that the axis of a parabola is parallel to one of the coordinate axes, the relation, ._]____ !*: S:::::::| i leads to 2 ard forn MWffllH ff g I : ::::::::::: M]?F i:||:: ^r? parabola. Thus ii figure a drawn, ha r -- i : Af : -- K\ Q q VPVT !j :: J::::|l:i::::::: :::fl::t:i^3"S-(t:^ ::::::::::;::::::::::: imfii axis para E^ctl rjq/rt - - a-axis ar W 1 ::::::: to the ris ![pp = |!E|! lation MP 2 ^ leads to z --/-- j? 2 M :::: j::::::::::M!^B :::::::::::: :|g[g :^::::::::::::: (y - 5) 2 = t |r Were F ffil:: (*, *), Vb \\W\~'." = * "" os h sin distance f = 4 a VM of the e upper abola is F( 2, 4 a = 4, to the opening A horizontal and a vertical parabola the point would be it is the m a point whose abscissa is h to a point whose ab- scissa is x ; similarly MP is y 7c ; now when the curve opens to the right VM is positive ; hence the equation is (y If} 2 4 a(x h}, with a positive. Were the axis parallel to the a;-axis, but the curve opening to the left, the equation would be : (y fc) 2 = 4 a(x K), with a negative. Similarly the curve on our figure which opens down is given by the equation (a? 3) 2 = S(y 2). In the general case, (x 7i) 2 = 4 a (y A;) has V(h, fc) as vertex ; the axis is x h = and is parallel to the y-axis ; THE PARABOLA 315 the curve opens up when a is positive and down when a is negative. (y - Kf = 4 a(x - h) - A)2 = 4 a (y - K) Standard forms of the parabola equation As a approaches in ?/ 2 = 4 ax, the parabola approaches more and more nearly to coincidence with the x-axis ; two coincident straight lines constitute a limiting form of the parabola. As a becomes larger, the parabola approaches the y-axis. 8. Tangent of slope m. y 2 = 4 ax. y = mx -\- Tc. + 2 Jcmx + fc 2 4 ax = 0. _ (2 a km) V(2 a fern) 2 ftv- whence, since A = 0, m y = mx + is the tangent of slope m to y* = 4 ax ; in ra 2 ' m is the point of tangency. 9. Tangent from an external point. For any given point ( x i> y\) outside the curve two values of m will be found for which y = mx -\ : will pass through (x 1} y^) ; hence there m 316 UNIFIED MATHEMATICS are, in general, two tangents which pass through a given point outside the curve. Xim t _ y im + a = 0, m = For points inside the curve, y? 4 ax^ is negative, and there are no tangents. 10. Tangent at a point (x l} j^) on the parabola. By the method of article 9 of the preceding chapter the tangent to the parabola, y z _ 4 aX} a t fa } y ^ on the curve is found to be y^y = 2 a (x -f j). Similarly the tangent to By* -f 2 Gx + 2 Fy + C = is found to be By\y + G(x + ccj) + F(y + y^) -\- C = ; and with the x 2 term present, Ax^x replaces in the above ex- pression the term 11. Illustrative example. Put in standard form and plot carefully y*-l2y + 6x-n = 0, completing the square inside the parenthesis. 4a=-f; a = -|. Plot F, ^, .KFC', and the further point where y = % ; 1 _ 8 . rp - - THE PARABOLA 317 Draw a smooth curve tangent to x = -, at F(\-, f ) through the points which are plotted ; here it would be well to find from the original equation the intercepts on the axes. PROBLEMS Put in standard form and plot : 1. 4# 2 12 y + 6x 11 =0. 5. (y 3) 2 = 8 x + 11. 2. 4 x 2 - 12 x -6y -11 = 0. 6. y z = 6x + ll. 8. Solve graphically, to 1 decimal place, x 2 + y 2 = 25 y z = Sx, by drawing both graphs to the same axes. Put in standard form and plot the equations obtained in the two following problems : 9. The formula for the height of a bullet shot vertically upward with a velocity of 800 feet per second, s = 800 1 16 1 2 . 10. The formula for the time of beat, in seconds, of a pendu- lum is t z = -I, taking g = 980 and / measured in centimeters ; 9 taking g = 32, / must be measured in feet. Compute corresponding values of t and I by logarithms, cor- rect to 3 significant figures. Would the diagram be changed if g is taken as 982 instead of 980 ? At sea level on the equator g = 978.1 cm./sec. 2 ; at Washington, 980.0 ; at New York, 980.2 ; at London, 981.2 ; or in feet/sec. 2 32.09, 32.15, 32.16, and 32.19, respectively. 11. The Hell-Gate steel arch bridge in New York is one of the largest arch bridges in the world. See the illustration, p. 353. The lower arc of the arch is a parabola, 977.5 feet as span and 220 feet as height of the arch. Write the equation of the arc, taking as ic-axis the tangent at the vertex of the parabola and as y-axis the axis of the parabola. Compute 4 a to 1 decimal place. The roadway is 130 feet above the 318 UNIFIED MATHEMATICS base of the arch ; compute the length of the roadway between the parabolic arcs. There are 23 panels or openings, spaced 42.5 feet apart at the centers ; compute the vertical lengths to the roadway from the arc of the parabola, also to one decimal place. Compute the approximate length of the parabolic arch it- self by computing the lengths of the twenty-three chords on the parabola ; note that only 12 computations are necessary ; do not carry beyond tenths of a foot, as hundredths would have little significance. Locate the focus and the directrix of this parabola. 12. Engineers use the following method for constructing a parabolic arch ; show that it is correct. Suppose that it is desired to construct a parabolic arch of width 100 feet and height 30 feet ; a rectangle 50 by 30 is drawn and the right- hand side is divided into 10 (or n) equal parts which are joined to the upper left-hand vertex of the rectangle (and parabola) by radiating lines ; the upper horizontal side is also divided into 10 (or n) equal parts and ordinates are drawn at these points ; corresponding lines intersect at points on the parabolic arch desired. Of the lines drawn the second (or rth) ordinate to the right of the vertex corresponds to the second (or rth) radiating line drawn from the vertex to the second (or rth) point of division from the top, on the right-hand side. 13. If an ordinary automobile headlight reflector is cut by a plane through its axis the section is a parabola having the light center as focus. If the dimensions of the headlight are 10 inches in diameter by 8 inches deep, locate the focus. 14. Locate the focus of a parabolic reflector, 6 inches in diameter and 4 inches deep ; 5 inches deep ; 6 inches deep. 15. The cable of a suspension bridge whose total weight is uniformly distributed over the length of the bridge takes the form of a parabola. Assuming that the cable of the Brooklyn bridge is a parabola, which it is approximately, find the equa- tion in simplest form ; the width between cable suspension points is about 1500 feet and the vertex of the curve is 140 feet, approximately, below the suspension points. THE PARABOLA 319 16. The Kornhaus Bridge over the Aar at Berne, Switzer- land, has for central arch a parabola ; the span is 384 feet and the height of the arch is 104. feet. If there are vertical columns spaced 24 feet apart, determine the length of these columns, assuming that the roadbed is 30 feet above the vertex of the parabola. If the floor of the roadbed is on a 2.7 per cent grade, determine the difference in elevation between the center of the bridge and the ends. 17. The parabolic reflector at the Detroit Observatory, University of Michigan, has a diameter, which corresponds to arch span, of 37.5 inches; the focal length of the mirror is 19.1 feet, from vertex to focus. Determine the height of the arch (or the depth of the reflector) ; determine the equation of the parabolic curve. The rays from a sun or star which strike this surface parallel to the axis of the parabola converge at the focus. What is the slope of this mirror at the upper point of the mirror ? At the point whose abscissa is ^ inch ? 18. To draw a tangent to a parabola from an external point you can proceed as follows : take the external point as center, the focal distance as radius, and describe an arc cutting the directrix ; from the point of intersection draw a line parallel to the axis ; the intersection point with the parabola is the point of tangency. Prove this method. 19. Find the tangents of slope + \ and .3 to each of the parabolas in exercises 1 to 7 ; time yourself. 20. Find the tangent to each of the parabolas in exercises 1 to 7 at the point on each parabola whose abscissa is + 2 ; the exercise should be completed within thirty minutes. Show the geometrical method of working one of these problems. 21. Find the tangents to the parabola in problem 1 from the point (2, 10) outside the parabola. Describe a geometrical method of working this problem after the graph of the parab- ola is drawn. 22. Plot to scale with the dimensions given the fundamental parabola of the Alexander III Bridge. CHAPTER XX THE HYPERBOLA 1. Definition and derivation of the equation. (See ellipse, page 289, and parabola, page 309.) PF=e-PZ, e>l. Take FX perpendicular to D'D as the avaxis, intersecting the given directrix at Q. Let A and A' divide the segment QF internally and ex- ternally in the ratio e (f in the figure). The mid-point of 320 THE HYPERBOLA 321 AA, is taken as the origin and the perpendicular through this point to X'X as the y-axis, OA = OA' = a. Precisely as in the ellipse, AF= e AQ, A'F=e-A'Q; whence AF + A'F = e (AA'). AF+OA'+OF=2ae, ' 2 OF=2ae, OF= ae, and, by subtraction, e F is (ae, 0) ; D'D is x - - = 0. o The relation PF e PZ gives the equation, V(aj - ae) 2 + y* x* + a 2 e 2 + y 2 = e 2 z 2 + a 2 . x 2 (l - e 2 ) + ?y 2 = a 2 (l - e 2 ). Up to this point the work is practically identical with the work in the case of the ellipse ; here, however, 1 e 2 is nega- tive, since e > 1. Hence we write this equation, Let 6 2 = a 2 (e 2 -l). /1-2 9/2 5 __ .v_ _. ^ a 2 6 2 2. Geometrical properties of the hyperbola, = 1. Since y and ?/ lead to the same values of x, the curve is symmetrical with respect to the -axis ; since x and x lead to the same values of y, the curve is symmetrical with respect 322 UNIFIED MATHEMATICS to the y-axis. Hence the intersection of the two axes is the center of the curve. 6 a Solving for y, this expression shows that the curve is symmetrical with respect to the ic-axis, for any value of x gives two values of y equal in value but opposite in sign. Since any value of x Symmetry of the hyperbola numerically less than a gives imaginary values of y, the curve lies wholly outside the region bounded by x + a = and x a = 0. When x = a, y = ; these are self-corresponding points on the horizontal axis of symmetry ; these points are called the vertices. When y = 0, x is imaginary ; the vertical axis of symmetry does not intersect the curve. Solving for x, x = - 6 + & 2 ; the curve is symmetrical with respect to the y-axis ; every real value of y gives two corresponding real values of x, symmetrically placed with THE HYPERBOLA 323 respect to the y-axis ; the vertical axis of symmetry does not cut the curve in real points. As y increases in value, without limit, so do the two corresponding values of x increase in value, numerically without limit. Since there is this vertical axis of symmetry it is evident, precisely as in the ellipse, that there is a second focus, F 2 ( ae, 0), and a corresponding directrix, x + - = 0. e The axis which cuts the curve is called the principal axis ; the other axis is called the conjugate axis. The lines x = a are tangent to the curve at the vertices. See page 310. 3. Right focal chords. The foci are the points (ae, 0) and ( ae, 0) ; when x = ae, y=- Va 2 e 2 a 2 = b Ve 2 1 ; a 6 2 but since 6 2 = a 2 (e 2 1), these values of y equal ; each a 2 6 2 right focal chord is of length - , and is constructed by erecting at the focus lines perpendicular to the principal, or transverse, axis of length on each side of the axis. a The foci are at a distance ae from the center ; now 6 2 = a 2 (e 2 1) gives ae = Va 2 + b 2 , which is the length of the diagonal of a rectangle of sides a and 6. 4. Finite and infinitely distant points on the hyperbola. yJ2 ii'l Given to plot the hyperbola - -- !j = 1, note that a 2 = 16, 16 49 & 2 = 49; 49 = 16(e 2 -l), whence e 2 -l=f, e 2 = f|, and It will be found convenient to draw the rectangle having as center and extending 4 units to the right and left of and 7 units above and below. The half-diagonal of this 324 UNIFIED MATHEMATICS rectangle has the length V65, and may be used to locate on the principal axis the two foci. X U Asymptotes and one branch of the hyperbola = 1 16 49 The line y = x cuts the curve at - = 1, 33a;2 =1; 16 49 ' 16 x 49 = 4.88. V33 33 The line y = mx cuts the curve in two points, whose ab- scissas are given by 49 16m 2 '' V49 - 16 m 2 16 49 THE HYPERBOLA 325 As 16 m 2 approaches nearer and nearer to 49 these two points of intersection move farther and farther off; when 49 16 m 2 = 0, m = |-, the two points of intersection of each of these lines with the hyperbola move off to an infinite dis- tance ; the lines y = |- x and y = ^ x are called asymptotes of this hyperbola, intersecting the curve in two coincident points both at an infinite distance. x^ w 2 b In the hyperbola - = 1. the two lines y = x and a 2 & 2 a y = x, or - 9- = and - + ^ = are asymptotes : note a a 6 a b that (- ^ l ( - + - 1 gives the left-hand member of the equa- \a bj \a bj tion of the hyperbola, when the right hand is unity. 5. Illustrative problems. Plot the hyperbola ^2__^ = 1 16 49 The rectangle of sides 8 and 14, parallel to x- and y-axes respectively, is plotted with its center at the origin. As noted above, the diagonals of this rectangle give the distance from the center on the transverse axis, here horizontal, of the foci ; the diagonals extended are the asymptotes, and to these lines the curve approaches more and more nearly as the curve recedes towards infinity. The right focal chords have the total 2 ft 2 49 length ; plot vertically above and below the foci. Take x = 6, a 4 this gives another point between vertex and right focal chord ; S = ^" 1 5 y = 7vT25 = 7(1.12) =7.84; 49 lo take x = 10, y = 7 v/OS = 7(2.29) = 16.03. Plot also the symmetrical points. PROBLEMS 1. Plot the hyperbola - - = 1. 49 16 2. Plot the hyperbola y = 1. 100 100 This type of hyperbola, a = 6, is called an equilateral or rec- tangular hyperbola. Why ? 326 UNIFIED MATHEMATICS 3. Plot the hyperbola ^--^- = 1, computing values re- o7 59 quired to one decimal place. 4. The equation of the hyperbola ^- = 1 may be put in parametric form, Noting that sec is x = a sec y = b tan 1 cos 6 , find by using logarithms five points 011 each of the above hyperbolas. Check on the graphs drawn. What is the geometrical significance of ? 5. Find the equations of the asymptotes of each of the pre- ceding hyperbolas and find to 1' the angle of inclination to the horizontal axis. 6. Compare the right-hand branch of the hyperbola x* v 2 81 ~ = 1, with the parabola ?/ 2 = (x 4) ; this parabola J-O y lo has the same vertex and passes through the extremities of the right focal chord. Prove this. Do these curves coincide in other points ? m i&r, Focal distances to Pi Ui, y\) on the hyperbola THE HYPERBOLA 327 6. Difference of the focal distances constant. Designate the right focus by FI and the corresponding directrix by DD', and the left focus by F Z) having D^D f as the corresponding directrix. Then the focal distances PF l and PF 2 are ex^ a and exi + a, respectively ; the difference, PF Z PF^ = 2 a, is constant. The hyperbola may be defined as a curve generated by a point which moves so that the difference of its distances from two fixed points is constant. 7. Standard forms of the equation of the hyperbola. ( - /Q* _ (y *0 2 _ i A TO a 2 & 2 (y _ fc)2 _ ( g A) 2 _ . ^^ 6 2 Precisely as in the ellipse, article 6, Chapter 18, the equation of the hyperbola |^ = 1, may be interpreted Of OM 2 MP 2 -i For any hyperbola whose axes of symmetry are parallel to the coordinate axes we obtain, from this relation, the equations a 2 6 2 and a 2 6 2 for a horizontal hyperbola, for a vertical hyperbola. Horizontal hyperbola 3s I 4- - m Vertical hyperbola 328 UNIFIED MATHEMATICS 8. Conjugate hyperbolas and limiting forms of the hyperbola equation. Given any hyperbola, X2_y2 =1 a 2 6 2 the lines a represent the asymptotes ; the equation a 2 & 2 represents a vertical hyperbola about the same rectangle and having the same asymptotes. Any two hyperbolas so related are called conjugate hyperbolas. Illustration. (z-3) 2 (y + 2) 2 , (z-3) 2 (y + 2) 2 '- = 1 and * 4 = 1 are con- 16 49 16 49 jugate hyperbolas ; the second is written in standard form **-~ i L^ '- = 1, wherein the focal distances ae from 49 16 the center, and the distances - of the directrices from the e center (3, 2) are obtained, regarding o 2 as 49 and i 2 as 16. The asymptotes of these conjugate hyperbolas are given by the equation i - isLJI L = 0, or by the equivalent in JLO ^rt/ separate factors, 7 fy> /A2 (y fc\2 The equation ^ * *t '- = 0, representing two real straight lines, is a limiting form of the hyperbola equation, (x - h) 2 (y + fc) 2 b -t- = m. Asm approaches zero, - remains con- a 2 b 2 a stant and the hyperbola approaches more and more nearly the L.T x h 11 k n j x h . y k n two straight lines = and 1- * : = 0. THE HYPERBOLA 329 9. The equilateral or rectangu- lar hyperbola. The hyperbola 22 y2 - = 1 is called an equilat- a 2 a 2 eral hyperbola since b = a ; it is also called a rectangular hyper- bola as the asymptotes are at right angles to each other. Since 6 2 = a 2 (e 2 1), the value of e in an equilateral hyperbola is V2 or 1.414; for e >V2, 5 2 > a 2 ; for e < V2, & 2 < a 2 . X;0- HfC 'A : . The equilateral or rectangular hyperbola 10. Illustrative problem. Put the equation 4 x 2 + 16 x - 9 y* - 18 y - 75 = in standard form and plot the curve. ) = 75 4x + 4)- 9(i/2 + 2y + 1)= 75 + 16 - 9 20.5 9.11 The center is at (2, 1); the hyperbola is of horizontal type; a 2 = 20.5 and a = 4.53 ; 6 2 = 9.11 and b = 3.02 ; ae = V20.5 + 9.11 = V29761 = 5.44 ; = ; a 4.53 Sari further convenient points are given by x = 5 ; substi- tuting in the original is easiest, giving 9 y 2 + 18 y - 105 = ; y=-l Vl Upper portion of right-hand branch of the given hyperbola = -l3.56. 330 UNIFIED MATHEMATICS PROBLEMS 1. Put the equation 4 x-2 + 16 x - 9 y* - 18 y + 107 = 0, in standard form and plot. 2. Plot one quarter of the hyperbola 147 2 59 2 3. Plot a hyperbolic arch, width 200 feet, height 60 feet, as part of a rectangular hyperbola. Assume the equation 2/2 _ x z a 2 } an( j no te that (100, a'+ 60) is on the curve. 4. What limitation is there upon the values of A and B, if the equation Ax 2 + By 2 + 2 Gx + 2 Fy + C = represents a hyperbola ? 5. Any equation of the form xy = k or (a-p + b$ + GI) (a& + b$ -f Cj) = k, has for its locus a hyperbola ; the lines obtained by equating the left-hand member to zero are the asymptotes. Plot the hyperbolas xy = 10 and (x 3 y)(x 4 y) = 50. 6. Put the following equations in standard form, completing the square first and reducing to standard form by division. a. b. c. d. 5 a 2 - 12 y 2 - 117 = 0. e. 3 x 2 - 24 z-4y 2 - 16^-52 = 0. 7. Plot the preceding five hyperbolas, choosing an appro- priate scale. Plot the extremities of the conjugate axes ; plot the rectangle and its diagonals ; plot the extremities of the right focal chords ; plot at least one further point, properly chosen to give the form of the curve, and its symmetrical points with respect to the axes. It is desirable to plot at least two of these curves completely ; the remaining curves need be sketched only in the first quadrant. THE HYPERBOLA 331 8. Determine a 2 and 6 2 to one decimal place in the follow- ing three hyperbolas : a. 17 x 2 - 43 y 2 = 397. 6. 5 a; 2 - 17 x - 10 ?/ 2 - 35 y = 0. c. 7( - 2)2 - B(y - 3) 2 = 39. 9. In the three hyperbolas immediately preceding deter- fi 2 CL mine ae, , and - to one decimal place. a e 10. In each hyperbola of problem 8 determine x when y = 2. 11. Using the data of the three preceding problems, plot the three hyperbolas of problem 8. /v.2 y2 12. In the hyperbola -- * = 1, find the coordinates of the 64 36 foci. What is the distance of the point whose abscissa is 12 from each of the foci ? of the points whose abscissas are 10, 11, 15 ? State the general form for this distance. 13. Put the following equations in standard form and discuss the curves represented by these equations : a. cc 2 6 x - y 2 6 y = 0. 6. x 2 -6a-4/ 2 - 36 CHAPTER XXI TANGENTS AND NORMALS TO SECOND DEGREE CURVES 1. The general quadratic in x and y. The general equation of the second degree in x and y is written, Ax 2 + 2 Hxy + By 2 + 2 Gx + 2 Fy + C = 0. The equations of the circle, parabola, ellipse, and hyperbola are special types of this general equation. Since none of these standard forms, representing curves of the second degree with axes of symmetry parallel to the coordinate axes, have an xy term, we find it convenient to discuss the general equation, with H=0, or Ax* -f By 2 + 2 Gx -f 2 Fy + C = 0. This represents one of the curves circle, ellipse, parabola, or hyperbola mentioned above, or some limiting form of the same, including pairs of lines and imaginary types. It can be shown that Ax 2 + 2 Hxy + By* + 2Gx + 2Fy + C=0 represents no new curve ; simply one of the above-mentioned types turned at an angle to the coordinate axes. 2. General equation of the second degree represents a conic section. Given a right circular cone, it can be shown by the geometrical methods of Euclidean geometry, that the section which is made with the surface of the cone by any plane is one of the curves above mentioned ; thus a plane parallel to the base cuts the cone in a circle, or in a point circle if through the vertex. 332 TANGENTS AND NORMALS 333 The cone is conceived as the whole surface determined by the straight line elements of the cone produced to infinity. A plane which runs parallel to only one of the elements cuts the cone in a parabola, or in two coincident lines if the plane passes through an element and a tangent to the circular base of the cone. A plane which cuts all the . elements in finite points cuts the cone in an ellipse ; this is a point ellipse when the plane passes through the vertex of the cone. A plane which cuts the cone parallel to the plane of two elements cuts it in a hyperbola ; if the plane passes through the vertex the hyperbola reduces to two straight lines. 3. Historical note on conic sections. The fundamental prop- erties of conic sections were discovered by Greek mathema- ticians nearly two thousand years before the invention of analytical geometry which was perfected by Descartes and Fermat, French mathematicians of the seventeenth century. A treatise on conies was written by Euclid (c. 320 B.C.), but it was entirely superseded a century later by a treatise by Apol- lonius (c. 250 B.C.) of Perga, whose treatise included most of the fundamental properties which we discuss. The proper- ties of the parabola connected directly with focus and directrix are not included in the eight books (chapters) on conic sec- tions by Apollonius, nor was the directrix of the central conies employed by him. Pappus of Alexandria (c. 300 A.D.), almost the last of the Greek mathematicians of any note, included these in his Mathematical Collections. The Greek mathematicians were interested in these curves for the pure geometrical reasoning involved. That the paths of the planets were conies they did not know ; nor did they know any practical applications of these conies. However, the fact that Greek mathematicians had studied these prop- erties made it possible for John Kepler and Isaac Newton to establish the laws of movement of the planets in the uni- verse in which we live. The men mentioned and Nicolas 334 UNIFIED MATHEMATICS * Copernicus, who reasserted the heliocentric theory of the uni- verse, were all thoroughly versed in the pure geometry of the Greeks ; their new theories were built directly upon this foundation of pure geometry. 4. Tangent of slope m to a second degree curve. Any line y = mx -f k cuts a curve given by an equation of the second degree in two real points, or in two imaginary points, or in two coincident points. The abscissas of the points of inter- section are given by the quadratic in x obtained by substitut- ing y = mx -+- k in the equation of the curve ; the two equations are solved as simultaneous equations. The condition for tan- gency is that the two points of intersection of the line with the curve shall be coincident ; this will be the case when the two roots of the quadratic in x, i.e. the two values of the abscissas of the points of intersection, are equal. When the intersections of a line with a quadratic curve are given by a linear, instead of a quadratic, equation, the mean- ing is that one point of intersection has moved off to an in- finite distance. As the coefficient of the square term of a quadratic approaches zero one root becomes larger and larger without limit ; see page 98. Parabola, y 2 = 4 ax, y = mx + k. Solving, a . == -(fcm-2a)V4a(q-7nfc) > m 2 For equal roots, or coincident points, k = . m , a . f a 2 a\ A. y = mx -\ is tangent to w 2 = 4 ax at [ , - - i m \m? m J yS. y2 Ellipse, (- = 1, and the line y = mx -f k. a 2 6 2 Solving, TANGENTS AND NORMALS 335 B. y = mx -\/a?m 2 -f- 6 2 is tangent to the ellipse ^ I y2 = 1 at x = T a ' m Va2m2 + v== a 2 6 2 a 2 m 2 + 6 2 Va 2 m 2 + 6 2 For every value of m there are two real tangents to an ellipse. Similarly C. y = mx Va 2 m 2 is tangent to the hyperbola, a 2 b 2 Va 2 m 2 - 6 s For values of | m | > - , there are two real tangents to a a hyperbola ; for | m | < - the tangents are imaginary ; for m = - , there is only one tangent and its point of tangency d is at an infinite distance, or, as noted before, the lines y = - x a are asymptotes of the curve. The method of this article is employed in deriving tangents ; the equations given under A, B, and C above are used mainly in proving geometrical properties of these curves. Note that if these equations are used as formulas, they apply only to curves of the type given; y = mx -\ gives the tangent only m to a parabola of the form y 2 = 4 ax (a may be positive or nega- tive). Similarly, y mx Va 2 m 2 + & 2 gives the tangent of slope m only to the ellipse, + y~ = 1, (a 2 may be less than ft 2 ). 336 UNIFIED MATHEMATICS PROBLEMS Find tangents of slope 2 and of slope 3 to the following three curves. Follow the method of article 4. 1. x 2 + y 2 10 x = ; find the points of tangency. 2. 3y 2 4 a; 6 y = ; find also the normal of slope ^. 3. x 2 + 3y 2 4ce 6y = ; find the diameter joining the two points of tangency. 4. xy 250. Find the tangents of slope 2, and the points of tangency. Find the tangent of slope m. For what values of m are the tangents imaginary ? Plot 10 points on this curve. Where do all points of this curve lie ? Find the intersections of y = .01 x -j- 5 with this curve. 5. Find the tangents at the extremities of the right focal chord of y z = 8 x ; where do they intersect ? Similarly in 7/ 2 = 4 ax. Is this true of any parabola ? Explain. 6. Find the tangents at the extremities of either right focal chord of + = 1 ; ' * y\ + &) a^ 80 ou the curve ; the chord PiP 2 k has the equation y y l = - (x a^). k Find the value of - conditioned by the fact that P 2 and P : fv both lie on the curve, by substituting fa, y^) and (o^ + h,y v + k) in the given equation and subtracting, member for mem- ber. k This value of - will be found, in general, to have a definite h limiting value as A; and h approach zero ; this limiting value is the slope of the tangent. The method outlined applies to any curve given by an algebraic equation. ?/2 = 4 ax P^X!, 2/j) on curve ; P 2 (ajj + 7i, yi + k) on curve ; y y\ = j(x *i)> chord joining P t P 2 . (yi + *) 2 = 4a(aj! + A), or y? + 2 ky l + k 2 = 4 ax 1 + 4 ah, since P 2 is on curve. y^ = 4 axi, since P x is on curve. -?/! + k 2 = 4 a/i, by subtraction. - = - - gives the slope of the chord joining P t to P 2 . h 2 y-L + k k Let h approach 0, k also approaches 0, but - always equals III and this value approaches more and more nearly to - or as a limit : this limit is the slope of the tangent. 2 yi yi y yi = (x x 1 ) is the tangent to y\ z = 4:ax Sit x on curve. TANGENTS AND NORMALS 339 This equation may be simplified, y$ = 2 ax + ft 2 2 aXi ; but ft 2 = 4 ax, whence fty = 2 a(x + x t ), which is the tangent equation. By precisely this method, the tangent to x 2 v 2 h = 1> at (xi, Vi) a 2 6 2 on curve has been found, in section 11, chapter 18, to be ~a? + ~P~ The tangent to = 1, at (x,, ft) on curve is ^ = 1. a 2 6 2 a 2 & 2 The tangent to Ax 2 + By 2 + 2 Gx -{-2 Fy + C=0, at (x l5 ft) on curve is The tangent to ^lx 2 + 2 Hxy + Bif + 2 Gx 4- 2 /fy + (7 = 0, at (x 1? ft) on curve is -G(x + x 1 ) + ^ + ft)+C'=0. All the preceding are embraced in the last formula, as special cases. The final form should be remembered and used as a formula. The above special forms for the equations of the tangents to conies given by equations in standard form may be used to derive tangential properties of these curves. Some of these properties are touched upon in the problems below and will recur in the next chapter. 340 UNIFIED MATHEMATICS 6. Tangential properties of the parabola. y* = 4 ax, any parabola P v VP 2 - y^ = 2 a(x -}- Xi), tangent at P\(x l} y^ on curve, cutting the axis at T, the directrix at Q, the vertex tangent at S. y y 1 = p. ( _ Xl ), normal P L N, cutting the axis at N. m -\t Tangential properties of the parabola T and ^V are obtained as the ^-intercept of tangent and normal, respectively; T is (- aj 1} 0) ; N is (^ + 2 a, 0). Hence FT = VM, the tangent cuts off from the vertex on the axis the same distance that the perpendicular from the point of tangency on the axis cuts off from the vertex. This gives a simple method of drawing a tangent to a parabola : drop the perpendicular P\M to the axis ; extend the axis from the vertex, making VT = VM, the length of the intercept; the tangent. TF=x 1 -\-a = TANGENTS AND NORMALS 341 Hence Z FTP 1 = Z FP l T, base angles of an isosceles A. Now Z FTPi = Z TP^Z, alternate interior angles of parallel lines, etc. Z FP l T/. TP L Z, i.e. the tangent bisects the angle between a focal chord and a line parallel to the axis ; the normal PiN bisects the supplementary angle FP^, making Z FP 1 N'= Z NP,R. Further S is the mid-point of TP l (since VS Y is parallel to PiJ/and bisects the side TM). .-. FS is perpendicular to TP l ; the perpendicular from the focus to any tangent meets it on the vertex tangent. QF being drawn, A QFP l = A QZP^ ', Hence QFP l is a right angle. Extend P^F to cut the parabola at P., ; draw P 2 Z 2 to the directrix, and P 2 Q. A P 2 Z 2 Q = A P Z FQ. Hence P 2 Q bisects the angle Z z PzF &nd is the tangent. Further Z P- 2 QPi is a right angle, since it is half of the straight angle about Q. Summary of tangential properties of the parabola 1. The tangent bisects the angle between the focal radius and a line parallel to the axis ; the normal bisects the in- scribed angle between the focal radius and a line parallel to the axis through the point of tangency. 2. The perpendicular from the focus of any parabola on any tangent meets it on the vertex tangent. 3. Tangents at the extremity of a focal chord meet on the directrix, and at right angles. 4. The focal chord is perpendicular to the line joining the focus to the intersection on the directrix of the tangents at the extremities of the focal chord. 342 UNIFIED MATHEMATICS 7. Tangential properties of the ellipse and hyperbola. /** '/" /*" iP" |- 2- = 1, any ellipse, ^- = 1, any hyperbola. (JL Ct a 2 & 2 ~ a* b 2 tangent at fa, y,) on curve, cutting the principal axis in th point T. ! ^^~ jr*-}*- ._ !s--S s __, < f -^^ -,~~- - ^sj^-v. + :::::::::::::;;; '7)\i __ s -t- "X, v-( L ' ^ i ' *,. v ii." J " LrTl j'/ii ' - '-" ^- j / \ - ''^N. ./ - T/^ /- / \ . ! -T T \' ;; -r-i/i T -- -.y-" - ' iv "r-j- 1 ^ SZ~ \ ~ i . / <_ _ S ^_ __ ^ ^T-rr-j/---l' k_ 1 at _ i ( ' : , ^*aei^ - V ' --^ ...... f4- l_j_ i . ; I .1.1 i , . . i ^ KJ ^ ' ! ' ^ =r: is ^v ' ' i lS^ ^^f T^Z ,-_ . ^v^v^ ^"r" -^ r ^^^ Tangential properties of the hyperbola The tangent bisects the angle between focal radii to the point of tangency. giving the normal at x t , y^ on curve. Tis -,0j In the ellipse, ,2 r A^ ' *^i? / In the hyperbola, = - ae; F^N= ae - &K* F 1 T=ae - ; F l N=Ae l TANGENTS AND NORMALS 343 The lengths i^JV and F 2 N are seen to be proportional to the lengths P^ and snce a + ex,. ae + ae Similarly ! * = * , since a ~ e = v T- T7t TvT nn . I ^ ^O ^ 2 ^ a + ex,_ a 2 . If a line from the vertex of a triangle divides the opposite side into segments proportional to the adjacent sides, the line bisects the angle of the triangle ; hence the tangent and normal at any point on the ellipse and hyperbola bisect inter- nally and externally the angle between the two focal radii to the point. Another method of constructing the tangent is to construct a 2 . In the ellipse the major auxiliary circle x 2 + y 2 = a 2 is drawn and the tangent at P 3 (x 1} y 2 ) on this circle x& -f ygf = a 2 has the intercept ; draw the tangent at (x lt y 2 ) to the circle, Xi cutting the X-axis at T ; connect T with P t on the ellipse. In the hyperbola x > a, so that this construction cannot be used ; from M(x l9 0) on the X-axis a tangent to the circle a: 2 + y z = a 2 intersects it at a point U whose abscissa is ^-, since j OT = a a x Summary of tangential properties of the ellipse, hyperbola, and parabola, regarding the parabola as having a second focus at an infinite distance on its axis of symmetry. 1. The tangent to an ellipse, hyperbola, or parabola bisects the angle between the focal radii to the point of tangency. 2. The perpendicular from the focus upon any tangent meets it on the circle having the center of the conic as center, 344 UNIFIED MATHEMATICS and passing through the principal vertices. In the parabola this circle has an infinite radius and so reduces to the tangent line at the vertex of the parabola. 3. Tangents at the extremities of a focal chord meet on the directrix. 4. The focal chord is perpendicular to the line joining the focus to the intersection on the directrix of tangents at the extremities of the given focal chord. PROBLEMS 1. Find the tangent to the curve xy = 25 at the point (5, 5) by the method of article 5. 2. Find the tangent to x 2 = 8 y at the point (5, - 2 g 5 -) by the method of article 5. 3. Find the tangents to the curves in problems 1-3 of the preceding set of problems at a point (x 1} y t ) on each curve by the general formula. 4. Find the tangent to the curve x 2 = 8 y at a point ( X D y\) on the curve ; note that (x 1} y^) satisfies the equation of the given curve ; find a second equation which the point (!, th) must satisfy if the tangent obtained is to pass through (5, 3) which is not on the curve ; solve the two equations as simultaneous and thus obtain the point of tangency of a tan- gent from (5, 3) to the given curve. 5. Find tangent and normal to the curve x 2 Wx 8y 5 = 0, at the point whose abscissa is 2 ; find the tangent of slope 2 to this curve. 6. What tangential property of parabolic curves makes them useful in reflectors ? Explain. Prove the property. 7. Write the equation of a hyperbola having the foci and x 2 y z vertices of the ellipse 4- ^- = 1 as vertices and foci, respec- ' * . ' tively ; find where these curves intersect ; write the equation of a tangent to each of the curves at one of the points of inter- section ; discuss these lines. TANGENTS AND NORMALS 345 8. Write the equation of the tangent at a point (x l} y^) to each of the following curves ; use the general formula ; time yourself. a. 4z 2 -6a; + 9?/ 2 + 5?/ = 0. b. 4z 2 -6a;-9y 2 -5?/ = 0. c. 4 x 2 6 x 5 y = 0. d. cc y e. -- h = 1 5 it is not necessary to clear of fractions as Y^g- and 2*3- can be thought* of as co-efficients of x 2 and y 2 . /. X 2 _ Q X -4y 2 - 8^ + 7 = 0. 9. Find the tangents to the first three curves in the pre- ceding exercise at the points where these curves cut the 10. Find one point on each of the curves of the eighth problem and write the equation of the tangent at that point. From Tyrrell, History of Bridge Engineering Elliptical arch bridge at Hyde Park on the Hudson The span is 75 feet and the rise is 14.7 feet. Note that the reflection completes the ellipse. CHAPTER XXII APPLICATIONS OF CONIC SECTIONS 1. General. Numerous applications of the conic sections, viz., circle, ellipse, parabola, and hyperbola, have been indi- cated in the problems given under the discussion of each curve. In general it is the tangential properties of the curves and the further geometrical peculiarities of these curves that make them so widely and so variously useful. The fact that simple geometrical properties are connected with curves given by algebraic equations of the first and second degrees in two variables seems to imply a certain harmony in the universe of algebra and geometry. 2. Laws of the universe. In 1529 the Polish astronomer- mathematician, Copernicus (1473-1543), rediscovered and restated the fact, known to ancient Greeks, that the sun is the center of the universe in which we live ; he conceived the 346 APPLICATIONS OF CONIC SECTIONS 347 planets to move about the sun in circular orbits. About a century later the great German astronomer, Kepler (1571- 1630), was able to establish the following laws of the universe : 1. The orbits of the planets are ellipses with the sun at one focus. 2. Equal areas are swept out in equal times, by radii from the sun to the moving planet. 3. The square of the time of revolution of any planet is proportional to the cube of its mean distance from the sun ; ip 2 ,-73 i.e. = , if T\ and T 2 are the periodic times of two planets, 7? df and dt and d 2 the diameters of their respective orbits. Kepler's work was made possible by that of all his pred- ecessors, particularly the Greek mathematicians who had so thoroughly discussed the properties of the conic sections, and further by the work of the Dane, Tycho Brahe (1546-1601), whose refined observations gave the necessary data. Newton (1642-1727) completed the work of systematizing the laws of motion in the universe in which we live, showing that the attraction of any two bodies for each other is in- versely proportional to the square of their distance apart and directly proportional to their masses. Newton showed further that this assumption leads to the elliptical motion in the case of the sun and any planet. The paths of comets which pass but once are known to be parabolas, or possibly hyperbolas with eccentricity close to 1. 3. Projectiles. The first approximation to the path of a projectile is a parabola. Indeed for low velocities, below 1000 feet per second, the path is almost parabolic even with air resistance. The parametric equations of the path of a projec- tile shot horizontally with a velocity of 1000 feet per second, neglecting air resistance, are, in terms of t, the number of seconds of flight, as follows : x = 1000 1, y = - 16 1\ 348 r N I KIED MATHEMATICS When a projectile is shot at an angle a with the horizonal, we have shown that there is a horizontal component of veloc- ity, v cos a, and a vertical compo- nent of velocity, tions of the path of this projectile shot from the ground as avaxis are as follows : /* 2 = 64.4 h, where h is measured in feet ; this also is a parabolic relation. The bending-moment at any given section of a beam sup- ported at both ends and uniformly loaded varies at different points on the beam, being greatest at the middle. These mo- ments are computed graphically in the case of a bridge, being given by a so-called parabola of moments. This parabola for a bridge of length /, uniformly loaded with a weight of w per foot, is given by the equation, M =i wP iw x 2 , wherein x is the distance from the center of the bridge. The parabola is plotted across the length I of the bridge, with the vertical ordinate at the mid-point representing the maximum moment. Thus if a bridge is 100 feet wide and uniformly loaded 2 tons per foot, the moment at any point x distance from the center of the bridge is given by the formula 3f = 2xl 7 - 1000s 2 Draw the corresponding parabola, choosing appropriate units. When a rotating wheel is stopped by the application of some force which reduces the velocity uniformly per second, the equations giving the number of revolutions before the wheel comes to rest correspond closely to the equations of motion of a body moving under the acceleration of gravity. $ = w t - 1 Jet 2 , 6 represents numerically the angle covered in time t seconds, the body having an initial rotational speed of w revolutions per second and the velocity being retarded every second by k revolutions per second. Here again we have an equation be- 360 UNIFIED MATHEMATICS tween and t represented by a parabola. The time in which this body comes to rest is obtained by dividing the initial velocity by the uniform decrease in velocity per second, i.e. by the acceleration (or retardation). The relation between pressure and volume of a perfect gas, temperature being constant, is given by the equation : p v = fc ; in words the volume is inversely proportional to the pressure. Plotting points gives points on a hyperbola of which the j>axis and the v-axis are the asymptotes. Such illustrations could be multiplied, but many relations of this character, e.g. the ellipsoid of inertia, require consider- able technical explanation which would go beyond the limits of this work. 9. Quadratic function. The graph of the quadratic function, cue 2 + bx -j- c, is the locus of the equation, y = ax 1 + bx -j- c. 11 \ 2a) 2 - 4 ac 4a a(x+ 6 Y. y ^ (a 4a b V_ l^ f 1 _&2_4 ac ^ 2a) ~ i y T" j )' a\ 4a y The graph of y = ax 2 -f- bx -f- c is a parabola ; x H -- = is 2a the axis. If a is positive the parabola opens up ; the vertex is VI -- , -- - i ; if & 2 4 ac is negative, the vertex is \ 2a 4 J above the o;-axis and no real value of x makes y = 0, since the graph does not cut the axis. If a is negative, the parabola opens APPLICATIONS OF CONIC SECTIONS 361 down ; if 6 2 4 ac is negative, the vertex is below the or-axis and again the graph does not cut the o>-axis. If 6 2 4 ac = the graph is tangent to the #-axis. Evidently 6 2 4 ac < is the condition that ax 2 + bx + c = should have imaginary roots ; 6 2 4 ac = is the condition for equal roots ; and & 2 4 ac > is the condition for real roots. PROBLEMS Plot to the same axes the graphs of the functions in prob- lems 1, 2, and 3. Discuss. 1. y = 5x>+2x- 7. 2. y = 5 x* + 2 x + 7. 3. y = 5o? 2 +2aj + . 4. If a wheel is rotating at the rate of 800 revolutions per second, and a force acting continuously reduces the speed each second by 40 revolutions per second, find the time in which it will stop, and the number of revolutions made during the retarded motion. Use the formula given in articled. 5. Given that 10 cubic centimeters of air are subjected to pressure, at a pressure of 1 atmosphere the volume is 10, hence pv = 10 is the equation connecting volume and pressure. Plot the graph of this for values of p from ^ atmosphere to 5 atmospheres' pressure. 6. Plot the parabola of moments for the Hell Gate Bridge, assuming a uniform loading of 2 tons per foot. Do not reduce tons to pounds, but use ton-feet as units of moment. The equation is M = 490 2 # 2 , taking 980 as the length of the bridge. 7. Plot the parabola of moments of the Panther-Hollow Bridge in problem 3 of the preceding exercise, assuming 2 tons per foot as loading. The equation is M = 180 2 y?. 8. Find the equation of a parabola whose focal length is 19 feet. Draw the graph to appropriate scale. This is the parabola which, revolved about its axis, gives the parabolic 362 UNIFIED MATHEMATICS reflector, previously mentioned, which is in use at the Detroit Observatory. (See page 352.) Find the equation of a hyperbola which has the same focus as this parabola, the axis of the parabola as transverse axis, and the second focus on the axis at a distance of 2 feet on the other side of the vertex. This is the hyperbola which, revolved about its axis, gives the hyperbolic mirror mentioned. The parabolic mirror has a diameter of 37 inches. Find the 1 Q K abscissa for the ordinate - , thus finding the depth of the 1_A mirror. 9. Plot the parabola y 1 = 70.02 x. This is the parabola which is fundamental in the construction of the Hill Audito- rium. (See page 352.) The plane of the floor cuts the side walls in this curve ; so also the intersection of the ceiling and a plane passed vertically through the main aisle of the hall. In the plans the computations of ordinates for given abscissas are made to the thirty-second of an inch. Compute the focal ordinate. This is the radius of the circular arch over the stage. Compute the ordinates for x = 21, 26, 31, 51, and 71, and express in feet and inches. 10. The Italian amphitheaters are, in general, elliptical. The Colosseum in Rome (see illustration, page 288) is an ellipse with axes of 615 and 510 feet. Draw the graph to scale. On the same diagram and with the same center and axes of reference draw the arena, of which the dimensions are 281 feet by 177 feet, to the same scale. Note that the minor axis of the arena is almost the " golden section" of the major axis, i.e. 177 is approximately a mean proportional between 281 and 281 less 177. Find the mean and compare. 11. The bridge at Hyde Park (see illustration, page 346) is elliptical, with a span of 75 feet and an arch height of 14.7 feet. Draw this elliptical arch to scale. CHAPTER XXIII POLES, POLARS, AND DIAMETERS 1. Definition. The straight line AXJ.X + By iy + G(x + a?,) + F(y + yi ) + C = is called the polar of P (x lf y^ with respect to the conic Ax* + By 2 + 2 Gx + 2 Fy + C = 0. The point P v (x l} y^ is called the pole of the line. 2. Fundamental property of polar lines. If the polar of P V (XI, 2h) with respect to the given conic passes through P 2 (#2, 2/2)? then reciprocally the polar of P 2 (x 2 , Vz) will pass through Pfa, 3h). This fundamental property of polar lines enables one to prove complicated geometrical theorems for conies with a minimum of machinery. The proof of the theorem is itself simple, for substituting in the polar of -P^a?!, y^, the co- ordinates ( 2 > 2/2)) we have that AXM + Bya a + G(x3. L. = 1 ? but the geometrical properties of the same curve are discussed with reference to the center (h, k) as origin, giving the equation The axes may be subjected to a translation, giving new axes O'X' and 0' T' parallel to the old axes ; or the axes may be turned through an angle a, giving new axes OX' and OY 1 about the old origin ; the two motions can be combined, execut- ing first the translation, usually, and then the rotation ; it is possible also to shift to new axes inclined at an oblique angle to each other, but the formulas involved are too complicated for an elementary work. 2. Translation of axes. Suppose the a>-axis fixed and the y-axis moved parallel to itself to a new origin 0' at distance 00' = h, from 0. Take P(x, y) as the coordinates of any point with reference to the original axes. Evidently, as the a^axis is unchanged, the y of this and every other point remains 371 372 UNIFIED MATHEMATICS the same. Let M be the foot of the perpendicular from P to the avaxis ; then by our fundamental property of the distances between three points on a directed line OM= oa + ax. But OM=x, OO' = h, the distance either posi- tive or negative 00'; while 0'3f = x f by defini- tion. Hence, whatever the position of P(x, y), Translation of axes we h&W, X = x' + h. Similarly, if the avaxis is shifted parallel to itself by an amount k. The two equations x = x' + h, transform any equation given with respect to any axes, to a set of parallel axes having the point (h, k) as origin. 3. Algebraic substitution in functions of one variable. THEOREM. Substitution of x' -f- h for x in any algebraic equa- tion of type a x r - + fl^sc"" 1 -j- a n _^x + a n = 0, n an integer, gives a new equation whose roots are h less than the roots of the old. The proof of this theorem depends directly upon the pre- ceding article. The substitution x = x' + h moves the y-axis h units, reducing the abscissas of all points by h if h is positive and increasing them by h if h is negative. ILLUSTRATION. If the graph of y = x 3 2 x 2 18 x + 24 is plotted, the substitution y = y and x = x' + 4 simply shifts the y-axis 4 units to the right, thus decreasing the numerical value of each root by 4. The new equation is y = (x 1 + 4)3 _ 2(x + 4)2 - 18(x' + 4) + 24 = x" + 10 x' 2 + 14 x' - 16. TRANSFORMATIONS AND SUBSTITUTIONS 373 Now whatever number substituted for x makes x 3 2 ar 2 18 x + 24 = 0, it is evident that 4 less substituted for x' will make (x' + 4)3-2(x' +4) 2 - 18(x' + 4)+24 = 0. Graphically, of course, as we have indicated, the y-axis has been pushed 4 units towards the right, and the abscissa of each point of intersection of the curve with the x-axis has been reduced by 4. Graph of y = x 3 - 2 x 2 - 18x + 24 Similarly, in the general equation above, when x' -f- h is sub- stituted for x, whatever number a satisfies the original equa- tion in x, a h will satisfy the new equation in x'. Substitution of x' + h for x in any algebraic equation forms a new equation in x' whose roots are h less than the roots of the given equation. This type of substitution is used to facilitate the computa- tion of roots of numerical algebraic equations. A simple method of constructing the new equation in nu- merical equations will be explained below, in section 11 of the next chapter. 374 UNIFIED MATHEMATICS PROBLEMS 1. Show that the formulas of transformation given trans- form the equations of ellipse and hyperbola having (h, k) as center to the simpler form without first-degree terms. 2. Transform the equation (y 3) 2 = S(x + 2) to the point (3, 2) as new origin, new axes parallel to the old. 3. By translation of axes transform the equation x*-4xy -6x + 8y 10 = into a new equation in which the first-degree terms are lacking. 4. Find the equation of the line 3y 4 a; + 6 =0, referred to parallel axes through the point (3, 2). 5. Compare the slope of a straight line referred to new axes by translation, with the slope referred to the old axes. Compare intercepts. Compare the slope of a tangent at a fixed point on any curve with respect to new and with respect to old axes. 6. Given the expression for the volume of 1000 cu. cm. of mercury at C. when heated to t C., v = 1000 + .0018 1 (see page 63), transform to parallel axes with the point (t = 0, v = 1000) as new origin ; find the new equation in v' and t. Does v' represent volume ? 13 1 7. Given v = 1054 H -- , the velocity in feet per second of \J& sound in air at t centigrade, transform to parallel axes with (32, 1054) as new origin ; discuss the equation. 8. The equation y? 2 x* - 18 x + 24 = (page 373) was found to have a root between x = 4 and x = 5 ; transform to parallel axes through (0, 4) and the new equation in x' will have a root between and 1. Compute this root to tenths by substitution. 9. The equation or* - 2 x z 18 x + 24 = has ' a further root between 1 and 2. Compute this root to one decimal place by the process explained in the preceding problem. TRANSFORMATIONS AND SUBSTITUTIONS 375 10. Find the roots of 2 y? + 6 x 2 10 x 8 = 0, as in prob- lems 8 and 9 by considering the graph of the equation y = 2x 3 + 6x 2 10 # 8, when referred to new axes. (See problem 4, page 99.) 11. Transform the following equations to parallel axes, having (h, k) as the new origin ; determine (h, Jc) so that the terms of the first degree shall disappear. a. 5 x 2 + 4 xy f- 8 x - 5 y - 10 = 0. b. 5x* + 4:xy + y 2 -8x-5y-W = Q. c. xy 7 x 10 y 5 = 0. a. 4 x z 6 * - y z - 8 y 10 = 0. 12. Transform the following equations to parallel axes having (h, Jc) as origin. Can you determine (h, fc) so that the terms of the first degree shall disappear ? Why not ? (See problem 13.) a . 4 #2 _ e x _ g y- 10 = 0. 13. Show that if an equation of the second degree contains no first-degree terms, the origin is the center of the curve by showing that if (aj b y x ) is any point on the curve ( x i} y$ is also on the curve. 4. Rotation of axes. The formulas for sin (a -f /?) and cos (a + ft) give very neatly the relations which exist between the coordinates (x, y) of a point referred to the old axes and the coordinates (x f , y') referred to the new axes. Take P(x, y) any point referred to the original axes ; let a be the angle of rotation through which the axes are turned ; let ft be the angle which the line OP makes with the x' or new a; -axis. By section 4, chapter 15, for all values of a and /?, cos (a + ft) = cos a cos ft sin a sin ft, OP cos (a + /8) = OP cos a cos ft OP sin a sin /3. 376 UNIFIED MATHEMATICS But OP cos (a + /?) = z; OP cos = a/; OP sin /? = ?/'; hence, x = x' cos a y sin a. Further, sin (a + ft) = sin a cos /? + cos a sin /?. Multiplying by OP, and substituting, y = x' sin a + y' cos a. These same relations might also have been obtained by pro- jection ; they hold for every position of the point P. x x' cos a y' sin , y = x' sin a + y' cos a, effects the rotation through the angle a, and refers any equation in two variables to new axes inclined at an angle to the old axes. 5. Every second-degree equation in two variables represents a conic section. Proof To prove this theorem we need only to show that the equation (1) Ax* + 2 Hxy + By* + 2Gx + 2Fy + C=0 can, by rotation of axes, be transformed to an equation of the (2) Ax* + W + 2 Gx + 2 Py + C = 0. Every equation of this latter type represents, as we have shown, either circle, ellipse, parabola, or hyperbola or some limiting form of one of these curves. Substituting, x = x cos a y sm , y = x' sin a + y' cos a, the equation Ax 2 + 2 Hxy + Ef + 2 Gx + 2 Fy + C = becomes A(x' cos a y' sin ) 2 -j- 2 #(#' cos a y' sin )(x' sin + y' cos ) + B(x' sin a + y' cos a) 2 + 2 (#' cos a y' sin a) -f 2 P(x' sin a + ?/' cos a) + (7 = 0. TRANSFORMATIONS AND SUBSTITUTIONS 377 Collecting terms, we have, (A cos 2 a -f- B sin 2 a + 2 H cos a sin a)x' 2 -\-(A sin 2 + B cos 2 a 2 # cos a sin a)y' 2 -f [ 2 A cos a sin a -f- 2 H(cos 2 a sin 2 ) -j- 2 5 cos a sin a]x'y f + (2 G cos a + 2 Fsin a.)^ +(2 ^cos a - 2 G sin a)/ + C = 0. Let A'x' 2 + 2 tf'zy + By 2 + 2 6? V + 2 F'tf + C = represent this equation. We wish to show that it is always possible to find an angle a for which H' becomes 0. 2 If Setting //' = 0, leads to the equation tan2= , noting that cos 2 a sin 2 a = cos 2 a and 2 sin a cos a = sin 2 a. Since H, A, and 5 are real numbers and since the tangent of an angle can have any value from negative to positive in- finity, it follows that there is always some angle 2 a, for which 2 ff ' tan 2 a = . There are in fact always two positive angles, A -D less than 360, 2 a and 2 a + 180, which satisfy the given re- lationship. By turning through a, or a -f 90, one half of either of these angles, the equation Ax 2 +2 Hxy-\-By z -{- = 0, reduces to an equation of the type A'x' 2 + B'y'* + = 0, with the coefficient of the x'y' term equal to 0. The angle a of turning can always be selected as a positive acute angle, since if tan 2 a is positive, 2 a may be taken as an acute angle, and if tan 2 a is negative, 2 a may be taken as an obtuse angle of which the half-angle a will be acute. Illustrative problem. What angle of rotation will remove the xy term from 3 x z -f- 6 xy 5 y- = 100 ? tan2 = -^ = , A- B 8 cos 2 a = | ; cos a = V\ (1 + cos 2 a) = VlO sina = Vj(l cos 2 a) = ^ L - v'10 378 UNIFIED MATHEMATICS We select a acute, as noted, hence the positive values of the radical are taken. The formulas of transformation become, x = -?-*' - -L^y 1 =J-(3x'- y'), VlO VlO VlO y = Vio y' = -r (' + 3 j/')- VIo Vio Substituting, we have, = 100. The hyperbola = 100, or 4 x' 2 - 6 y ' 2 = 100 In combining terms, do not write the expansion but preferably combine like terms by inspection. Here the coefficient of x' 2 is f J -f |g _ ^ ; o f x'y' the coefficient is ~~ H + i ~~ l or 0, which checks ; for y' 2 we have ^ f$ f. Our equation becomes, 25 16.67 This curve is plotted with reference to the new axes, inclined at an angle a, tan a = }, to the x-axis. The coordinates of a point (x', y') on this curve, considered with respect to the new axes, satisfy the new equation TRANSFORMATIONS AND SUBSTITUTIONS 379 _^ = 1 ; when considered with reference to the old axes as (x, y) , 25 16.67 the coordinates satisfy the original equation. Thus the coordinates of the intersection with the original x-axis (5.8, 0) satisfy the original equation ; this point with reference to the new axes has the coordinates 5.8 x3 x' = x cos a + y sin = VlO -5.8 y 1 = x sin a + y cos a = VlO or (5.5, 1.8). The values for (x', y') in terms of (x, y) can be con- ceived as obtained by rotating through the angle a. PROBLEMS 1. Find the equation of the curve xy 7x + 3y 15=0 when referred to axes making an angle of 45 with the given axes. Note that a is 45 ; sin a = cos a = = ; rationalize V2 denominators after substituting. Plot the new axes at the angle indicated and plot the graph of the new equation, obtained by substitution, with reference to the new axes. 2. Find the equation of the curve 9x z + 24 xy + 16 y 2 - 6x - 15 y = with reference to axes making an angle arctan ^ with the old axes. Note that sin = i and cos = f ; in substituting take the fraction i as a factor in the value of both x and y and, after substituting, combine terms by inspection without writ- ing each expansion separately. 3. Find the equation of the curve 59x 2 -2ixy + 66 f- +72a;-396?/ + 444 =0, when referred to new axes such that the new avaxis makes an angle whose tangent is f with the old axis of abscissas. 4. In the equation Axy 8x + 10y + 7 = Q make the gen- eral substitutions which effect the turning of the axes through an angle a, and determine a so that the coefficient of the x'y' term shall disappear. 380 UNIFIED MATHEMATICS 6. Nature of the conic Ax- +2 Hxy + By*- + 2 Gx + 2 Fy + C= 0. A central conic is one which has a point which is such that every chord passing through this point is bisected. If this point be taken as origin of coordinates, it follows that if (#', y') is on the curve ( x', y') is also on the curve. A substitu- tion, x = x' + h and y = y' + k, which causes the terms of the first degree in our equation of the second degree to disappear gives the equation Ad"- + 2 Hx'y' + By' 2 + C" = 0. Now what- ever point (x', y') satisfies this equation ( x', y') will also satisfy the equation, and hence the new origin is the center of this conic. The substitution x = x' + h and y = y'-\-k gives two linear expressions in h and k as coefficients of the new x' term and y' term, and these are set equal to zero and solved for h and k to determine the center. / 2Ah+2Hk + 2G=Q and 2Hh + 2Bk + 2F=0 are the two equations which determine the center. If the two equations which serve to locate the center repre- sent two parallel lines in h and k, the conic has no center and A TT is a parabola. This condition is that -- = -- , or that H z - AB = 0. When H*-AB=0, the terms Ax 2 + 2 Hxy + By* form the square of a linear expression in x and y. Further it is shown below that if H 2 AB < 0, the conic is an ellipse, and if H 2 AB > 0, the conic is a hyperbola. The conditions determining the nature of the general conic are as follows: . -T. A 1V H- AB < 0, ellipse, H 2 AB = 0, parabola, H 2 - AB > 0, hyperbola. These are the conditions that there should be no points on the curve at infinity, one point at infinity, and two directions giving infinite points. They may be derived by substituting TRANSFORMATIONS AND SUBSTITUTIONS 381 y = mx + Jc and determining values of ra for which the quad- ratic has infinite roots ; it follows that for these values of m the line y = mx -f k will meet the curve in points infinitely distant. For the ellipse the values of m will be imaginary, and H 2 AB < ; for the parabola the two values of m will coincide, and H 2 AB = ; for the hyperbola the two values of m will be real and different, representing the slopes of the two asymptotes, and H 2 AB > 0. A second and independent proof is given in the next article. There it is shown that the product A B' is positive when H 2 AB is negative ; but when A' and B' are of the same sign the product is positive and the curve in x' 2 and y' 2 , not involving x'y', is an ellipse. Similarly the product A'B' is negative when H 2 AB is positive, and the curve represented by the transformed equation is a hyperbola. 7. Central conies ; abbreviated process of transformation. Substitution method. Determine the center ; transform to parallel axes with the center as new origin ; determine a and substitute ; plot with reference to the final axes. Abbreviated method. Determine the center (h, K) ; trans- form to (h, K) as new origin ; determine A' and B' by solving as simultaneous the equations, A 1 + B' = A + B, - A'B' = H 2 -AB; select the pair of values of A' and B' such that A' B' will have the same sign as JET; plot the new equation with reference to new axes having the origin at the center determined and the axes inclined at an angle a with the old axes, a being such that tan 2 - ~ ll ~A^' Derivation of A' + B' = A + B ; - A'B' = H 2 - AB. * A' = A cos 2 a + B sin 2 a + 2 H cos a sin a. B' = A sin 2 a + B cos 2 a 2 H cos a sin a. By addition, A' + B' = A + B. 382 UNIFIED MATHEMATICS The proof that A'B' = H* AB is somewhat long but not difficult. To the product A'B' add H' 2 = [2 H (cos 2 a - sin 2 a) - 2 (A - B) sin a cos a] 2 , which does not alter the value since H' is taken to equal 0. The expressions will combine to H 2 AB. The student would do well to verify at least one of the coefficients. Since a is chosen as a positive acute angle, A' B' has the same sign as H, for A' B' = (A -B) cos 2 a + 2 H sin 2 a = 2 H( A ~ B cos 2 a + sin 2 A V 2H J Now sin 2 a is positive, and cos 2 a has the same sign as and hence the product of cos 2 a by '-- is positive ; hence A' B' is the product of 2 H by the sum of two positive quantities and so is positive if H is positive and negative if H is negative. The equations A' + B 1 = A + B -A'B' = H 2 - AB enable us to determine A' and B' by solving these as simulta- neous equations. Two solutions are found, and the solution is selected which makes A' B' have the same sign as H. Only the new constant term, when transforming to (h, k) as new origin, offers any extended computation. This constant term AW + 2 Hhk + Bk* + 2 Gh + 2 Fk + C may be written h(Ah +Hlc+G)+ Jc(Hh + Bk + F) + Gh + Fk + (7, which reduces to Gh + Fk -f (7, since the other two expressions within parentheses were set equal to zero to determine the center. Illustrative problem. Find center, axes, and plot the conic, 3 tf + 6 xy + 5 y 2 - 12 x 18 y - 24 = 0. TRANSFORMATIONS AND SUBSTITUTIONS 383 Substituting (x' + A, y' -f k) and selecting the coefficients of x' and j/', to set equal to zero, 6 h + 10 k - 18 = 0. Solving, k = f , h = + \. >x The ellipse 3 x 2 + 6 xy + 5y 2 - 12 x - 18 y - 24 = or 3 x' 2 + 6 x'y' + 5 y' 2 - = 0, or 7.16 x" 2 + .84 y" 2 = 40.5 2 C', the new constant, new (x', y 1 ) equation is = -6.|-9.|-24= ^. The + 5j/' 2 - = 0. Note that tan 2 a = =-3. A t A'+ B' = 8, - A'B' = 9 - 15 = - 6. Solving by substitution, -A' (8 -A') =-6, A'2 - 8 A' + 6 = ; A' = 4 VlO. B' = 4 T VlO. J.' B' has the same sign as H ; hence the upper algebraic signs are taken, A 1 = 7.16, B' = .84. 384 UNIFIED MATHEMATICS Our final equation is 7.16x"2+.842/"2 x"* y"* _ 1 5.66 48.3 y"* =1 (2.38)2 (6.95)2 Some computation is unavoidable, and, in general, in practical applica- tions the results are rarely expressible in small and convenient integers. PROBLEMS 1. Find the center, axes, and plot the conic, 5 a? - 6 xy + 3 if + 12 x - 6 y 30 = 0. 2. Plot the following conies by turning the axes through 2 ff an angle a, tan 2 a= , so as to eliminate the xy term, and thus obtain an equation to plot which can be put in stand- ard form. a. 4 x 2 + 4 xy + y z 6 x + 8 y 12 = 0. b. x 2 4:xy + y z + 2x Wy ll = 0. c. 41 x 2 + 2xy + 3y 2 -26x-32y-171 = 0. d. 4 xy - 3 y* - 7 x 10 y - 15 = 0. 3. Apply the abbreviated method explained in section 6 to the central conies in the preceding problem; compare the numerical work involved by the two methods. 4. Find five points on the first and second conies in prob- lem 2 by giving values to x and computing the corresponding values of y. 5. Find the intercepts with the coordinate axes of each of the conies in problem 2 and verify your graphical construction by these points. 6. In each of the conies of problem 2 find the points of intersection with the line y = mx + b ; determine the values of ra for which one of the points of intersection should be at an infinite distance. In the case of the hyperbolas real TRANSFORMATIONS AND SUBSTITUTIONS 385 values of m will be found ; substitute in turn each of these \values for m and determine for what value of b the second point of intersection will move off to an infinite distance. This determines the two asymptotes. Explain. 7. Apply the abbreviated method to the discussion of the following central conies, having the origin as center : a. x* + 2 xy + 4 y z = 16. b. 4 a 2 -6 ay -3 ?/ 2 = 10. c. 2 a 2 -4 a?/-?/ 2 = -9. d. 5 z 2 - 3 xy + ?/ 2 = 24. 8. In the hyperbolas of problem 2, use the results of prob- lem 6 to show that the directions of the asymptotes are given by the factors of the terms of the. second degree. 8. The hyperbola as related to its asymptotes. The equation of the hyperbola in simplest form, ^_^_1 a 2 & 2 ' may also be written, (bx ay)(bx -f ay) = a 2 & 2 , whence. bx ay bx -f ay a 2 6 2 ~~' Since bx ay = and bx -f ay = represent the asymptotes of this hyperbola, the final form states that the product of the perpendicular distances of any point on the hyperbola from the two asymptotes is constant. The converse proposition is also true, viz., if a point moves so that the product of its distances from two intersecting lines is a constant, the point moves on a hyperbola of which the two lines are the asymp- totes. The proof of the converse is simply that the bisectors of the angles between the two given lines could be selected as axes of coordinates and, in consequence, the two lines would have as equations, expressions of the form y mx = and y + mx.= 0. Any point which moves so that the product of its 386 UNIFIED MATHEMATICS distances from these two lines is a constant would satisfy the Cation y-mx . y + mx = k . Vl + m 2 Vl + m 2 but this equation represents a hyperbola, and consequently the given locus is a hyperbola. It follows from the above argument that the equation of any hyperbola differs by a constant from the product of the first-degree expressions which, put equal to zero, represent its asymptotes. The terms of the second degree in the hyperbola can be factored always into real linear factors in x and y (not necessarily rational so far as the coefficients are concerned) which as lines have the slopes of the asymptotes. (See problem 8 of the preceding list, and compare article 6.) A particularly simple type of hyperbola equation occurs quite frequently in practical problems and this type will be taken to illustrate the method which is, however, general. Illustrative example. Plot the curve x y = i- x This equation may be written The only term of the second degree is xy. Placing the factors equal to zero, we have x = and y = 0. The asymptotes are parallel to our co- ordinate axes. The equation can be written in the form (x -h}(y -k} - c. By inspection we note that the equation may be written (y+l)(x- !) = -!. The asymptotes are given by y + 1 = and x 1 = 0. The intersection point is the center of the given curve ; further points should be plotted by substitution of values in the original equation. This equation in i and d, i = - , represents the relation between a 1 d given rate of discount for any interval and the corresponding rate of in- terest. If a bank in lending money takes out interest in advance, giving TRANSFORMATIONS AND SUBSTITUTIONS 387 to the individual not the face of the loan but that amount less the interest upon that amount for the given interval for which the note is to run, the bank is said to discount, the note. The rate of interest which the indi- vidual pays is obviously greater than the rate d which is used as the dis- count rate ; the relation is .JL. 1 -d In plotting the graph of this curve you would be interested only in values of i and d between .01 and .10, and you would confine your attention to the first quadrant, taking i inch to represent .01 on both axes. PROBLEMS 1. Plot the curve j p- / y = 1000; show that it represents a hyperbola having the axes as asymptotes. This equation rep- resents the relation between the pressure and volume of a quantity of gas which at a pressure of 1 atmosphere has a volume of 1000 cubic units, the temperature being kept constant. 2. Discuss the nature of the following curves, without making any transformation of axes ; in the hyperbolas give the slopes of the asymptotes, and in the parabolas the slope of the axis. a. 4 x 2 - y* 8 y = 0. 6. 4 a 2 -Sy- 10 = 0. c. 4 x 2 - 4 xy - y 2 - 100 = 0. d. 4:X 2 4:Xy + y z = 100 y. e . 4 a; 2 4 a;?/ + ?/ 2 = 100. f. 4 a 2 - 4 a*/ + 4 y 2 = 100. g. 4 x 2 4 xy 10 x = 25. h. 4ajy 7aj + 10y 5 = 0. i. xy = 15. j, 4 a: 2 + 4 y 2 = 81. k. 3x 2 -12x-2y*-Wy-15 = 0. /. 3 a; 2 -12 a; + 2 ?/ 2 - 10 y- 15 = 0. 388 UNIFIED MATHEMATICS 3. Transform to new axes so as to simplify the following equations to plot ; select the appropriate method of substi- tution adapted to each equation. a. a: 2 + 12 xy + 4 y* 4 x 24 y 10 = 0. b. xi + 3xy-3y 2 -10x-15y + 2 = Q. c. 4 x z 4 xy + 7 y 2 10 y + 4 x 25 = 0. d. 4 z 2 - 12 a# + 9 y 2 6 x - 10 = 0. e. 2x z + xy + 2y 2 -6x + 6y 15 = 0. /. a; 2 + 4 *y - 2 y 2 - 8 x + 20 y 30 = 0. gr. a 2 3 ay 7z = 0. ft. 2 x 2 - 6 a? + 5 y 2 - 20 y - 10 = 0. 4. Given that an aeroplane covers a distance of one hun- 100 dred miles in t hours, its velocity "in miles per hour is , t 100 i.e. v = ; given that on different occasions the aeroplane L covers 100 miles in 48 minutes (.8 hours), 1 hour, 1 hour 6 minutes, l hours, 1 hour and 24 minutes, 100 minutes, and 2 hours, respectively, find the velocities and plot a curve giving the relation between v and t. Choose units so that you can read from the curve between the extreme values the velocity within 2 miles per hour when the time of flight for 100 miles is given. Note that the curve is a hyperbola. 5. Given that an aeroplane covers on one trial 100 miles in 48 minutes, on another trial 125 miles in 61 minutes, and 156 miles in 71 minutes on a third trial, how could you compare graphically the corresponding velocities ? 6. The air in an organ pipe vibrates in a manner some- what similar to the motion of a pendulum ; the number of such vibrations of the air in one second depends upon the length of the pipe and upon the velocity of sound in air ; the formula 7i = , v in feet per sec. and I in feet, gives quits - 1 TRANSFORMATIONS AND SUBSTITUTIONS 389 > for closely the number of vibrations. Plot the curve n = L I values of I from 1 to 20 feet, choosing appropriate units. The curve gives the corresponding number of vibrations for pipes of different lengths. (See section 3, chapter 26.) 7. Discuss fully the equation x* - 2 xy + y* 10 y = 0. 8. The curve of transition on a railroad track in passing from one straight track to another is sometimes taken as para- bolic, because of the fact that the slope changes uniformly Parabolic transition curve on a railroad track The parabolic arc is used for vertical as well as .for horizontal transition curves. with uniform increases of the horizontal length taken parallel to the tangent at the vertex of the parabola. Assuming that the track AV changes its direction by 60 to VB and that the transition points A and B from the straight line to the parabolas are taken on each track 500 feet from the point of intersection of the two directions, find the equation of the parabola. Note that the axis FFis inclined at an angle of 120 to the extension of AV; note that E, the vertex of the 390 UNIFIED MATHEMATICS parabola, is midway between Vand the point where the chord AB cuts the axis, since the tangent to a parabola cuts off from the vertex on the axis a distance equal to the distance cut off from the vertex on the axis by the perpendicular to the axis from the point of taiigency. Find the equation of the curve with respect to the axis of the parabola as ?/-axis and the line through V at an angle of 30 with AV as a?-axis ; then transform to A V as cc'-axis and a perpendicular to AV at V as y'-axis by turning through an angle of 30, using the fundamental formulas for rotation of axes. 9. Assuming that a railroad track changes its direction by 40, 30, 20, and 10 respectively, find the equations of the parabolic transition curves with transition points (A and B, as in figure) 500 feet from the intersection point of the two straight tracks. 10. In going over a hill the form of curve to which the track bed is rounded is often made parabolic. When the grade is the same on both sides of the highest point, the problem is precisely that of finding a parabolic arch. Assuming that in a horizontal distance of 5000 feet the hill rises 100 feet, find the equation of the parabola having the vertex at the highest point and passing through the point 100 feet lower at a hori- zontal distance of 5000 feet ; find the four intermediate ordi- nates at distances 1000 feet apart. 11. An iron wire of diameter .2 cm. and length I cm., subjected to a tension T caused by a weight W grams, when caused to vibrate through its whole length has the number of vibrations determined by the equation _1 / 2i\ '.0777T When the weight is fixed and the length is variable, this gives a hyperbolic relation between n and Z. For W = 500 grams the equation is approximately n = . Plot and discuss. TRANSFORMATIONS AND SUBSTITUTIONS 391 12. In the preceding problem suppose that I is fixed at 100 centimeters and that W varies between 100 grams and 2000 grams. What is the type of relationship ? What would be the curve obtained by plotting to iv- and w-axes ? 13. The deck of any large vessel slopes from both bow and stern downwards towards amidships. The vertical section of the deck from bow to stern consists. of two parabolas, having a common vertex at the middle of the ship. Plot the parabolas which are used for a vessel 400 feet long, having the highest point at the bow 8 feet above the vertex, and at the stern the deck 4 feet above the vertex. Use a different scale for y than for x, at least twice as large. 14. Name the following curves, giving such facts as you can by inspection : a. 3 x + 2 y -5 = 0. ft. (3a:+2y 5)(a? 3) = 10. b. 3x* + 2 y- 5 = 0. I. 3x* + 3y 2 = Q. c. 3 x 2 + 2 y 1 - - 5 = 0. m. xy - 7 x + 6y - 18 = 0> d. 3x*-5x = 0. e. x /. 3^4-3^ = 25. p . + = 5. g. 3x*-6xy+3y*-5x=0. y r-= li. t 3* + 2y + 5,0. r . F= 331.7 l- j. (3x+2y-5)(x-3)=0. A/ 273 15. The highway over the Michigan Central R.R.. tracks and over the Huron River, on the Whitmore Lake road near Ann Arbor, is rounded off (in profile) to a parabolic arc, rising 2.40 feet in a span of 240 feet. Show that the grade leading up to the arc should be a 4 % grade. Draw the arc to scale. CHAPTER XXV SOLUTION OF NUMERICAL ALGEBRAIC EQUATIONS 1. Solutions of algebraic equations. By a solution of an equation of the type a^x" -J- a 1 # n ~ 1 + a%x n ~t 4. ... a n _^c, + a n = 0, wherein n is a positive integer and 0$, a l} a 2 are constants, we understand a value which, substituted for x, reduces the left- hand member to zero. That such a solution always exists is proved by methods of higher mathematics. The theorem that every such rational integral algebraic equation has a root is called the fundamental theorem of algebra; it was first proved about a century ago by Gauss. The solution may be a real number or a complex number, and any constant coefficient may be real or complex ; the latter involves the square root of a negative quantity and so is not representable as the abscissa of any point on our axis of positive and negative real numbers. Certain types of algebraic equations are solvable in terms of the general constants which enter as coefficients. Thus ax + b = is solvable in terms of a and &, and ax* -+- bx + c = is solvable in terms of a, b, and c. It has been shown that the general cubic in one variable and the general biquadratic, or fourth degree equation, are solvable in this way, but the general equations of higher degree than the fourth are not solvable in this sense. The approximate numerical solution of the real roots of rational integral equations with numerical coefficients is readily obtained and we have indicated in Chapter II and again in the preceding chapter, section 3, problems 8-10, the general method by which such solutions are obtained by substitution. 392 NUMERICAL ALGEBRAIC EQUATIONS 393 Simplifications for purposes of computation will be explained in this chapter. 2. Continuity. The height of an individual is a continuous function of the age of the individual ; by this we mean that in passing from one height to another the individual passes through every intermediate height. A graph representing age as abscissas and heights as ordinates will be a continuously Four continuous graphs. One discontinuous Continuity in passing from a positive to a negative value. connected curve. Upon this curve corresponding to any selected age, a l} a period of time, there will be one and only one corresponding height, A 1} and corresponding to any second age, 02, a second ordinate, 7t 2 , representing height. The curve joining the two points (a } , Tij) to (a. 2 , A 2 ) will be continuous and every intermediate height between the two given will be found to be represented by the ordinate corresponding to some age intermediate between the two given ages. The rational integral function of x, in which n is any positive integer, is continuous between any two values of x, and will be represented by a continuous curve. This has been assumed in drawing the graph of y = x 3 2 x 2 18 x + 24, and in other graphs. The proof in- volves discussion somewhat too detailed and mathematically refined for an elementary course. The symbol f(x) will be used throughout the remainder of this chapter to represent a rational integral function of X of the type mentioned above. 394 UNIFIED MATHEMATICS 3. Graph of y =f(x) by location of points. Give to x the appropriate values, find the corresponding values of y, and plot the points, connecting by a smooth curve. (See pages 70-71.) Apply the remainder theorem, and employ synthetic division to determine values of the function corresponding to given values of x. 4. Remainder theorem and synthetic division. (See page 25.) When j\x) is divided by x a, the remainder obtained by con- tinuing the division until the remainder does not contain x is equal to the original expression with a put for x. To divide f(x) by x a, employing synthetic division, a. Arrange f(x) in descending powers of x and write the coefficients horizontally, including zero coefficients for missing powers below the highest power which occurs. 6. Write + a under x a, the divisor, placed at the left. Under the coefficients of f(x) as written leave space for a second horizontal row and draw a horizontal line. c. Under the coefficient of the highest power of x, below the horizontal line drawn, place this coefficient again. Mul- tiply by + a and add to the following coefficient to the right. Place the sum below the line, vertically under the second coefficient; use this number below the line as multiplier of + a, and add the product to the third coefficient and continue this process until you have placed numbers under every coeffi- cient (and the constant term) of the upper row. The final number which appears is the remainder and should be cut off by a vertical separator ; the numbers under the horizontal line are coefficients in order from left to right of the quotient when f(x) is divided by x a. Throughout this discussion a may be either positive or negative. NUMERICAL ALGEBRAIC EQUATIONS 395 ILLUSTRATIVE PROBLEM. Divide x 3 2 x 2 18 X + 24 by x 3, and use the remainder theorem to determine the value of this function of x when x = 3. x 3 - 2 x 2 - 18 x + 24 x _ 3) 1 - 2 - 18 +24 + 3) +3 +3-45 1 + 1 - 15 (- 21 x 2 + x 15 is the quotient and 21 is remainder; 21 is the value of x 3 2 x 2 18 x + 24 when 3 is substituted for x. Since x s _ 2 x 2 - 18 x + 24 = (x 2 + x - 15) (x - 3) - 21, we have, substituting 3, 33 _ 2. 3 2 - 18- 3 + 24=(3 2 + 3- 15)(3-3)-21 = 0-21. PROBLEMS 1. Locate ten points upon the graph of y =2 a^-j-3 x 2 9 x 7. Take the ten points between x 4 and x = + 4, including \ and \ ; use the synthetic division method of finding the value of y except for x = 0, x 1, and x 1. Plot the points and draw a smooth curve connecting them ; choose the y scale so as to keep the points on the paper. Locate the zeros of the function on the graph. 2. Plot the graph of the function 2 a? + 3 x 2 7 between 3 and +-3. 3. Plot the graph of the function 2 y? 9 x 7. Note where the graph crosses the axis of x, thus locating the roots of the equation 2^ 9^ 7 = 0. Factor 2 y? 9 x 7, dividing by the factor corresponding to the rational root which you have found; solve the resulting quadratic, and compare with the values found by the graph. 4. Plot the graph of the function x 4 -2 a? +-3 z 2 -18 x+21 ; select the appropriate interval to give the points of intersection with the ic-axis. 5. Plot the graph of y = x 4 3 # 2 21 ; locate the zeros of the function on the graph. Solve as an equation in quadratic form x* 3 x 2 21 = and compare the solutions obtained with the roots located graphically. 396 UNIFIED MATHEMATICS 6. Plot the graph of the function 4a^ 3ar+ .5 in tin- in- terval from 1 to + 1 ; substitute for x the values 1, .8, -.5, -.3, .1, 0, .1, .2, .3, .4, .5, .6, .7, .8, .9, and 1, finding the values in general, by the division method applying the re- mainder theorem. The roots of this equation represent the values of sin 10, sin 50, and sin 70. (See section 10, below.) 7. Locate one root between and .1 of the equation 4 s -3 a; 4- .05234=0, by substituting for x the values 0, .01, .02, .03, up to .1. The value .05234 is the sine of three degrees which we obtained in problem 5, page 245. One root of this equation gives the sine of 1. 5. Number of roots. A value of a b for which /(c^) = 0, is a root of f(x) = 0. The remainder theorem applies and conse- quently (x aj) is a factor of f(x) since the remainder when f(x) is divided by x a t will be zero. Nothing in our argu- ment requires that a l be a real number. Hence, dividing f(x) by (x a v ), a new equation of degree one less will be obtained. This equation, by the fundamental theorem of algebra, also has a root, 02, giving a quotient of degree n 1. The number of such factors corresponds to the degree of the equation, n. Every rational integral equation of the nth degree has n roots, and no more. For no further value of x could make the product, a(x a,i)(x a*i)(x a 3 ) (x a n ), equal zero without making one of the factors zero and thus coinciding with one of the roots given. 6. Graphical location of real roots. Any real root of a rational integral function of x equated to zero is a value of x which makes the ordinate in y =/(#) equal to zero. The points in which the graph of the function of x crosses, or touches, the a;-axis correspond to real roots of the equation, f(x) = 0, or zeros of the function. Our assumption of continuity enables us to formulate the following theorem : NUMERICAL ALGEBRAIC EQUATIONS 397 Betiveen any tico values x= a and x = b, for which the two corresponding values of f(x) are opposite in sign, there lies at least one real root of the equation f(x) 0. A: Fpur graphs passing continuously from y = - toi/= ; one graph with a 2 3 discontinuity Thus to change continuously from + \ to ^, or from any positive value to any negative value, the function must pass through all values in- termediate, including 0. At this point where the function of x is 0, the graph of y =/(x) crosses the axis. Illustrative problem. Locate the roots of Plot the graph of y = x 3 - 2 x 2 18 x -f 24 by location of points. Give to x values from 5 to +5, find the corresponding values of y, and plot the points, connecting by a smooth curve. (See page 71.) Between x=l and x = 2, /(x) changes from -f 5 to 12 ; there is a root between x = 1 and x = 2 ; between x = 4 and x = 5 there is a root, as/ (4) is 16 and /(5) is + 9 ; at x = 4 there is a root, as/( 4) is 0. 7. Slope of y=f(x). The (h, Jc) method of finding the tangent at a point (a^, y^) on a curve applies, as we have stated in Chapter 18, section 11, to the graph of a rational in- tegral function of x. Thus in y = x* 2x z 18 a; + 24, let (x 1} y^) be any point on the curve and (a^ + /*, y t + fc) a second point. It is desired to find the slope of the graph at (a;,, y^ and y, = a^ 8 - 2a;! 2 - 18a^ + 24, + fc =(a?i + h) 3 - 2(xj_ + h} 2 - l&fa + K) +24, since (x 1} y,) 398 UNIFIED MATHEMATICS and fa + h, y l + K) are on the curve. Subtracting the upper from the lower equation, member for member, we have, fc = 7t(3 xf -4^-18) + 7i 2 (3 a, - 2) + 7i 3 , = 3 x? - 4 xj. - 18 + h(3 x l - 2) + & h Let h approach zero ; the terms on the right containing h and h? will also approach zero, as the coefficients are constants. Limit - = 3 x? 4 Xj. - 18, as h = 0. h "When x l = 1, the slope of the curve is 19 ; when x t = 3, the slope is 3 ; when x = 4, the slope is + 14. A double root of any equation corresponds to a point at which the function is zero and the slope of the curve, ob- tained by the (h, k) method, is zero. 8. Slope and maximum and minimum points. When the slope is zero, the curve is for the instant parallel to the cc-axis. This is a necessary condition for a maximum or minimum point, i.e. a point at which the value of the function attains a greatest or a least value in some interval which includes the point. This may be accepted by the student as graphically evident. A formal proof depends on the methods of the calculus, and rests essentially on the method used in finding the slope. PROBLEMS See the preceding list of problems. 1. Find the slope at any point (x 1} y^) of each of the follow- ing curves and locate the maximum and minimum points on the curve by setting the slope equal to and solving for # x : a. y = b. y = 2x* + 3x* 7. e. y = 4a 3* + .5. c. y = x*-3x z -21. /. y = 4 x? - 3 x + .05234. NUMERICAL ALGEBRAIC EQUATIONS 399 2. Find the slope at any point (x i} y } ) on y = x* - 2 x 3 + 3 x 2 - 18 x + 21. This gives the slope m as m = 4 a?! 3 6 x-p -f 6 a^ 18. Plot the graph of y = 4 x 3 6 x 2 + 6 x 18, and note that the zeros of this function are the values of a^ for which the slope of the curve y = x* 2 y? + 3 x z 18 x + 21 is 0. These are values of # for which the original function has maximum and minimum values. 9. Historical note. The solution early in the sixteenth cen- tury of the cubic and biquadratic was the undisputed achieve- ment of a group of Italian mathematicians. Fiori, Tartaglia, and Cardan were involved in the solution of the cubic, while Ferrari, pupil of Cardan, solved the quartic. Not until the beginning of the nineteenth century was it shown that the general equations of higher degree are not solvable, this being the work of a brilliant young Norwegian named Abel. 10. The cubic applied to angle trisection. By higher mathe- matics it has been demonstrated that geometrical problems which can be solved by ruler and compass correspond alge- braically to problems whose solution can be effected by linear and quadratic equations and equations reducible to quadratics, i.e. by equations of which the roots will involve only quad- ratic irrationalities (square roots, and square roots of expres- sions involving only rational quantities and square roots). The trisection of an angle is a type of geometrical problem whose solution cannot be effected with ruler and compass ; it is possible to reduce the trisection of an angle to an algebraical problem, the solution of the cubic equation. Let the given angle which is to be trisected be denoted, for convenience, by 3 a. Since this angle is given, the value of its sine is known. If the angle is given by a geometrical drawing, the ratio of the perpendicular h dropped from a point at a distance r from the vertex on one side to the second side to r, 400 UNIFIED MATHEMATICS i.e. - gives the sine of the angle. Let the value of the sine of r the given angle be k. Given sin 3 a = A;, find sin . sin 3 a = sin (2 a -f a) = sin 2 a cos a + cos 2 a sin a = 2 sin a cos 2 a + (cos 2 a sin 2 a) sin a sin 3 a = 3 sin a 4 sin 3 a. ft = 3 sin a 4 sin s a. This equation is a cubic in the unknown sin a ; for convenience it may be written k = 3 x 4 x 3 , substituting x for sin a. There are, in fact, three solutions of the cubic and these three solutions correspond to the fact that k is the sine not only of 3 a, but also of 180 - 3 a, and n 360 + 3 a, and (2w + l)180-3. Thus the cubic which would give the sine of 10, trisecting the angle of 30, is .5 = 3x - 4a*, or Ix 3 - 3 x + .5 = 0. The same cubic would be obtained if it were desired to trisect the angle of 150, or of 390, or 750, . There are an infinite number of angles which have this same sine, .5, but there will be only three different values involved when the sine of the third part of each of these angles is found. In the equation 4 x 3 3 x + .5 = 0, the roots represent sin 10, sin 50, and sin 250. (See problem 6, page 396.) 11. Closer approximation to located roots. The method will be shown by a numerical illustration. The equation (1) a?-2z 2 -18a; + 24 = 0, of which the graph is given on page 71, evidently has a root between 4 and 5. To form the new equation whose roots are 4 less than the given equation, substitute x' + 4 for x, giving (2) ( x ' + 4) 3 - 2(x' + 4) 2 - 18( / + 4) + 24 = 0. Assume that this gives (3) x' s + Bx f * + Cx f + D= 0, in which B, C, and D can be obtained by expanding and combining terms in (2). The left- hand members of equation (3) and equation (2) are then identi- cal. Evidently, if x 4' is substituted for x' in (2), it will give the original equation, and consequently, if x 4 is substituted in (3), it will give the original equation. Substituting, we have (4) (x - 4) 3 + B(x 4) 2 + C(x 4) + D = 0, which is identi- cal with the original equation. If the left-hand member of this equation, i.e. the original, is divided by x 4, the remainder is D and the quotient is (x 4) 2 + B(x 4) + C ; if this quotient is divided by (x 4), the remainder is C ; if the quotient of the preceding division is divided by x 4, the remainder is B. x - 4) x 3 - 2 x 2 - 18 x + 24 + * + ** ~^ The continued division by D a? -4 is effected by the + 4+24 \ . synthetic process, explained 1 + 6 (+14 C in section 4, above. + 4 B (5) x' 3 + 10 x' 2 + 14 x r 16 = is then the equation whose roots are 4 less than the roots of the original equation. This should be verified by substitution and expansion. The original equation has a root between 4 and 5. Hence this equation has a root between and 1. By trial of tenths, .1, .2, .9, this equation is found to have a root between .7 and .8. Hence the original equation has a root between 4.7 and 4.8. Form the new equation whose roots are .7 less than the roots of (5). aj' _ .7)1 + 10 +14 - 16 (.7 + .7) + .7 + 7.49 + 15.043 1 + 10.7 + 21.49(- .957 + .7 + 7.98 1 +11.4(+ 29.47 .7 (6) z 3 + 12.1 z 2 + 29.47 z - .957 = 0. 402 UNIFIED MATHEMATICS Equation (5) has a root between .7 and .8 ; hence equation (6) has a root between and .1. By trial of hundredths, trying .02, .03, .04 it is found that this equation has a root .03 + , between .03 and .04 and evidently nearer .03. Hence our original equation has a root 4.73 + . In this way we can compute any real numerical root of a rational integral algebraical equation to any desired number of significant figures. ILLUSTRATIVE PROBLEMS. 1. Find the cube root of 1,624,276 to four significant figures. x 3 1,624,276 = 0. By trial, substituting 100, 200, found to have a root between 100 and 200. x-100) 100) for x, this is x' - 10) 10) x"-7) 7) 1 + + - 1,624,276 4-100 +10000 -1,000,000 1 + 100 + 10000 ( + 100 + 20000 624,276 1 + 200 ( + 30000 + 100 1 + 300 -+ 30000 - + 10 + 3100 - 624,276 331,000 1 + 310 + 33100 (- + 10 + 3200 293,276 1 + 320 (+36300 + 10 1 + 330 + 36300 - + 7 + 2359 + 293,276 270,613 1 + 337 + 38659 (- + 7 + 2408 22,663 1 + 344 (+ 41067 + 7 1 + 351 + 41067 - 22,663 By derivation, the roots are 100 less than the roots of (1 ) ; hence a root between and 100. By trial, substitut- ing, root between 10 and 20. By derivation, has a root between and 10. By trial, a root between 7 and 8. By derivation, root between and 1. By trial, between .5 and .6, and nearer to .5. Hence the root of the original is 117.5, which may be partially checked by four-place logarithms. NUMERICAL ALGEBRAIC EQUATIONS 403 2. Compute one negative root of 2 a; 4 + lOo 3 - 8 a 2 - 11 x + 19 = 0. Negative root between 1 and 2. x + 2) 2 + 10 -2) - 4 - 8 -12 - 11 +19 + 40 - 58 2 + 6 4 -20 - 4 + 29 (- 39 + 48 2 + 2 4 -24 + 4 (+77 + 77 - 39 - 13.728 + 37.9632 2- 2 4 (-20 -20 J - 2.88 X - .6) 2 - + .6) + 6 l.i Boot between and 1. By trial, be- 2 + 4.8 - 22.88 + 63.272 - (1.0368 tween .6 and .7. + 1.2 - 2.16 -15.024 2- 3.6 -25. 04 (+48. 248 + 1.2 - 1.44 2_ 2.4 (-26. 48 + 1.2 2 - 1.2 - 26.48 + 48.248 1.0368 By derivation has a root betwen and .1. By trial, be- tween .02 and .03. The original equation has a root 2 + .62+, or 1.38~, i.e. between - 1.38 and - 1.37. PROBLEMS See the two preceding sets of problems and use the results obtained. 1. Compute to three significant figures the largest positive root of the following equations, d. x* - 2 x 3 + 3 x 2 - 18 x + 21 = 0. 2. Compute by the process indicated the positive root of #2 _ 3 x 21 = to three decimal places ; compute the same by solving as a quadratic, and compare as to efficiency the two methods. - 404 UNIFIED MATHEMATICS 3. Solve the equation 4^ 3x-f- .5= 0, computing the smallest positive root to four decimal places. This is the value of sin 10 ; check by your table of sines. 4. In problem 5, page 245, you have computed the sine of 3 to four decimal places. Write the cubic which will give the sine of 1 ; compute the smallest positive root, and discuss to what decimal place it could be carried with propriety when the sine of three degrees is given to four decimal places. 5. Plot the graphs of the two equations { ~ ' 2 Note that the points of intersection give the solutions of the two equations regarded as simultaneous ; but solving the two equations as simultaneous equations, we are led by substitution to the cubic x $ (z 2 - 16) = 1, or x 3 - 16 x - 4 = 0. Solve the cubic and compare with the solutions obtained graphically. 6. Historical problem. The great Archimedes proposed the problem to cut a sphere by a plane in such a way that the two segments of the sphere should have to each other a given ratio. Archimedes showed that the solution could be obtained as the intersection of a hyperbola and a parabola. If the diameter of the sphere is taken as 10 and k as the ratio of the larger to the smaller segment, this problem leads to the cubic (k 11 x 3 300 x + 2000 > = 0, k+I in which x represents the distance of the plane from the center of the sphere. Solve to two decimal places when k = 2. The plane at a distance x from the center then trisects the sphere. 7. In the preceding problem show that the solution may be obtained as the intersection of a hyperbola and a parabola. 8. A famous problem of antiquity is the problem to dupli- cate a given cube, i.e. to solve geometrically x 3 = 2 a 3 , a being the side of the given cube. Long before analytical geometry was invented it was known that the solution could be given as the intersection of the parabola x 2 = ay with the parabola NUMERICAL ALGEBRAIC EQUATIONS 405 y* = 2 ax. Construct the graphical solution when a is taken as 10. The problem may also be solved by the intersection of either of the two given parabolas with the hyperbola xy=2a 2 . Verify. If two means, x and y, are inserted between a, and 2 a, i.e. - = - = , then x is the solution of the equation x 3 =2 a 3 . x y 2 a This method reduced the problem of the duplication of the cube to the problem of inserting between two given numbers, or lines, two geometric means. 9. The volume of a spherical segment, greater than a hemi- sphere, of height x + r, is given by the expression o the volume of a sphere is f Trr 3 . Find the segment of a sphere of water of radius 10 which will be equal in weight to a sphere of wood, radius 10, which wood is only .6 as heavy as water. This leads to the cubic equation or x 3 3 r>x + .4 r 5 = 0, and r + x is the depth to which the sphere of wood will sink when it is placed in water. Compute this depth when r = 10. 10. Ice is only .92 as heavy as water. Use the equations of the preceding problem, substituting .92 for .6, to find the depth to which a spherical iceberg of radius 100 feet, if one were possible, would sink in water. 406 UNIFIED MATHEMATICS ; Mi *\J CS -III 3 z 3 f v i^s o S o o : - CHAPTER XXVI WAVE MOTION 1. General. In nature there are two types of recurrent motion, somewhat closely connected mathematically, in which repetition of motion occurs at regular intervals. One type of this motion, in cycles as we may say, repeats the motion in one place, and is in a sense stationary. The tuning fork in motion moves through the same space again and again ; a similar movement is the motion of a vibrating string. Of this stationary type may be mentioned the heart- beats, the pulse, the respiration, the tides, and the rotation of a wheel about its axis. The second type of recurrent motion transmits or carries the vibratory impulse over an extent of space as well as time. The waves of the sea are of this character. Sound waves, electrical vibrations or waves, and radiant energy vibrations are transmitted by a process similar to that by which the waves of the sea are carried. Both of these types of motion are representable mathe- matically by equations involving a sequence of trigonometric functions. To the fundamental and basic function involved, y = siii x, we will direct our attention in the next section and to simple applications in other sections of this chapter. 2. The sine curve. As a radius vector of unit length ro- tates in a plane with uniform velocity about a center, the sine of the angle 6 fluctuates between 1 and 1. The variation of sine 6 may be represented by the movement on the y-axis of the projection of the vector, and this movement of the 407 408 UNIFIED MATHEMATICS Graph of y = sin 9 ; a pure sinusoid The length AA' equals the circumference of the circle ; the amplitude, vertical distance between highest and lowest points, equals the diameter of the circle. Graph of y = sin 2 6; a sinusoidal curve The frequency is double that represented in the preceding graph. y = sin 8 + sin 2 6; obtained by addition of corresponding ordinates in the two preceding curves This type of curve is obtained from a tuning fork having an octave overtone. WAVE MOTION 409 projection is termed simple harmonic motion, frequently abbreviated S. H. M. Precisely the same type of movement is given by the projection of the moving vector on the ovaxis, x = cos 6, or on any line in the plane of the motion, z = cos (0 t), wherein e is the slope angle of the line and z is the projection of the radius. If the vector completes one revolution, 2ir r , or 360, in 1 second, the period of the motion is called 1 second, and the frequency, or the number of repetitions of the complete move- ment or cycle in a second, is 1 per second. If the complete revolution is effected in 1 second, the period is second and the frequency 2 per second. The graphs of y = sin and of y = sin 2 represent under these conditions the progress of the ordinate for uniform changes in 0, i.e. for uniform changes in the time, since the rotation is with constant angular ve- locity. For convenience the angle is conceived as measured in radians and the radius is taken on the cc-axis as the unit to represent one radian ; the abscissa then corresponds either to the angle measured in radians or to the length of arc traversed by the end of the moving vector. In plotting y = sin 0, values of from to 360 or from to 2 ir r are plotted on the horizon- tal or 0-axis. Note particularly the points for which 6=0, 30, 45, 60, 90, - 180, . 360 ; or 6 3 2 Note that AA' on our diagram represents one complete cycle or period. For many purposes it is desirable to take t, the time (in seconds, usually), as the variable. The same graph then represents y = sin 2 irt, wherein AA' is taken equal to 1 and the horizontal axis is the Z-axis. The same curve represents y = sin 20 irt, if AA' is taken as T ^ of 1 unit of time. The upper curve in our diagram is a pure sinusoid, the distance AA' representing the circumference of the circle of which the maximum ordinate is the radius. 410 UNIFIED MATHEMATICS The two curves plotted should be carefully studied ; the lower "curve has double the frequency of the upper and one half the period. The swing, amplitude as it is termed, is the same; the amplitude is the algebraic difference between the maximum and minimum values of the function. Any curve representing y a sin 8 or a sin *8 or a sin (kQ -f e) is called a sinusoid. We shall find that the graphs of y = a cos 6, y = a cos k$, and y = a cos (kd + e) differ from the preceding only in position. For most purposes it is conven- ient to plot time in complete units on the ordinary coordinate paper, the unit depending on the period of time in question. For a com- plete rotation in one minute ten seconds might be taken as one unit on the horizontal axis with the radius as vertical unit, and the curve would differ very slightly from our curve. The highest and lowest points would fall then at 15 and 45 respectively ; 0, 5, 7.5, 10, 15, 20, and 30 seconds corre- spond then to 0, 30, 45, 60, 90, 120, and 180 respectively. Physicists and engineers com- monly draw the sinusoidal curves which are of frequent occurrence entirely from graphical considerations. The circle with the desired amplitude is WAVE MOTION 411 drawn ; the angle between the axes is bisected and re- bisected (as often as desired) ; an appropriate length for a complete cycle is taken on. the horizontal axis, and this is divided into just as many parts, usually 16, as the circum- ference of the circle is divided by the axes and the bisecting lines v which were drawn. At each point on the horizontal axis an ordinate is drawn and from each corresponding point on the circle a horizontal line is drawn to intersect the correspond- ing ordinate. Corresponding points have the same numbers if on the circle intersection points are numbered from the right-hand intersection with the horizontal axis counter-clock- wise and numbered on the line from the left-hand end of the horizontal length taken to represent the time of one cycle, as indicated on our diagram. The two upper figures, page 408, were drawn by this method. The student is urged to make both the graphical construction and the construction by using the numerical values of the sines from the tables. Compare also the work under Section 11, Chapter VII. PROBLEMS 1. Plot the curves y = sin& and y sin 30 on the same sheet of coordinate paper ; take 1 inch as radius and on the horizontal axis take 1 inch to represent 60. For a pure sinusoid, y = sin x, one unit on x should be the length of the radius ; then 3.14+ radians represents 180, the second point in which the curve y = sin x cuts the axis of abscissas. 2. Plot y = sin 2 wt ; note that t = y 1 ^, y 2 ^, corresponds to 36 and multiples ; take one unit for t as 6 times the radius chosen. 3. How could you interpret the curve of the preceding exer- cise as y = sin 4 wt ? 4. Plot 10 points of y = sin (& 30). This curve is similar to the preceding; it is 30 behind, we may say, the regular sine curve ; the " lag " is 30 ; the two curves y = sin 6 and 412 UNIFIED MATHEMATICS y = sin (0 30) are said to be out of phase, the phase angle of the second being 30. The " phase angle " is of particu- lar importance in the theory and practice of alternating cur- rents. 5. Plot the curve E = 110 sin 6. Note that if the horizon- tal scale be taken so that 1 inch represents 60 and the vertical scale such that 1 inch represents 110, the curve is precisely the first curve of problem 1. This curve represents the vari- able electromotive force (e. m. f .) developed by a generator which generates a maximum e. m. f. of 110 volts. To plot the curve no knowledge of electricity is necessary, but com- plete interpretation requires technical knowledge. 3. Sound waves. If a tuning fork for note lower C is set to vibrating, the free bar makes 129 complete, back-and-forth, vibrations in one second. By attaching a fine point to the end of the bar and moving under this bar at a uniform rate, as it vibrates, a smoke-blackened paper, a sinusoidal curve is traced on the paper. Our curve is traced by a bar vibrating 50 times in 1 second. The curve y = sin (50 X 2 irt) Tuning fork vibrations recorded on smoked paper. In 1 second 50 complete vibrations are made ; the vertical distance between the top and the bottom of the arcs repre- sents the distance moved by any point on the moving bar ; the motion is simple harmonic (S. H. M.). The period is Jg- second ; the frequency is 50 ; the amplitude is about -% inch. If the smoked paper were moved with uniform velocity under the vibrating bar in such a way as to cover 50 times the cir- cumference of a circle with radius J T of an inch, or 50 x 2 ?r X -fa inch per second, the curve traced would be almost a perfect sinusoid of the type y = sin 0. The points move of course on WAVE MOTION 413 arcs of curves, but the variation from a straight line is ex- tremely slight. Corresponding to each movement of the vibrating rod there is a movement of the air. As the bar moves to the right it compresses the layer of air to its right and that compression is immediately communicated to the layer of air to the right ; as the bar moves back and to the left, the pressure on the ad- Vibration records produced by the voice " a" as in " ate " ; "ou " as in "about " ; " r " in "relay "; "6" in "be"; and "a" in "father." The tuning fork record, frequency 60 per second, gives the vibration frequencies. jacent air is released and a rarefaction takes place. In ^ of 1 second you have the air adjacent to the rod compressed, back to normal, and rarefied; during this time the neighboring air is affected and the compression is communicated a distance which is the wave length of this given sound wave. In 1 second this disturbance is transmitted 1100 feet at 44 Fahrenheit. The wave length for this sound wave then is it|p- = 22 feet. The wave length is commonly designated by X. If v is the velocity, and t the time of one vibration, X = vt. 414 UNIFIED MATHEMATICS The notes of the key of C on the natural scale have the fol- lowing vibration frequencies : cdefgabc' 256 288 320 341.3 384 426.7 480 512 The intensity or loudness corresponds in the rod to the length of swing of the vibrating rod ; as this amplitude de- creases, the intensity of the sound decreases. For small am- plitudes the vibratory motion gives a convenient way of measuring small intervals of time. Thus on the above diagram if the tone of the note lower C, y = sin 256 irt, were represented, each complete wave would represent T ^ of 1 second ; each half or each arch would rep- resent YST f 1 second. Tuning bars, with periods J^ and -j-^ of 1 second are run electrically for timing purposes. The curve y = sin 6 4- sin 2 6 represents the combination in sound of two tones which differ by an octave. Precisely the type of curve which is represented by our diagram can be produced mechanically by the record of a vibrating tuning fork 1 which sounds not only the principal note but also the octave overtone, due to the fact that the bar vibrates about the middle point at the same time that it vibrates about the end. Vibrating strings also have multiple vibration, overtones and other tones. Harmony is the result, in general, when the vibrating instrument gives vibrations which are connected with the fundamental vibration by simple numerical relations, like that of the overtone. Thus the notes of the major chord, key of C, c, e, g, c, on the piano, have the vibration. frequencies in the ratios 4 to 5 to 6 to 8. 4. Helical spring. Similar to the vibrations of the air are those of a spiral wire spring which oscillates back and forth when a weight is suspended by the spring; the successive compressions and elongations of the wire correspond quite 1 See Miller, The Science of Musical Sounds, p. 188, for photograph. WAVE MOTION 415 closely to the condensations and rarefactions of the air. The position of the weight at any instant can be given by an equation entirely similar to the equation above of note C. Thus if the time of one complete vibration is ^ second, and the maximum displacement is 4 inches, the equation is y = 4 sin 4 irt r , This gives the elevation above and below the point at which the weight comes to rest. Perfect elasticity of the spring is assumed. 5. Light waves. Light waves have a much higher velocity than sound waves, 300 x lO 6 meters per second. The different wave lengths correspond to different colors, just as different wave lengths in sound waves correspond to different tones. The wave length of the light from burning sodium (Z), of the spectrum) is 0.5890 x 10~ 6 meters per second, and for other colors varies for the visible spectrum between .39 and .75 x 10" 6 meters. The vibration frequency of the sodium light is the number of these waves which occur in one second of time, hence since these waves cover 300 X 10 6 meters in one second the frequency n is such that n - A. = v, or n 0.589 x 10~ 6 = 300 x 10 6 , 300 x 10 6 whence n = - = 509 x 10 12 vibrations per second. Radiant energy is of the saine general nature with longer waves. Light waves differ from the sound waves in having transverse vibrations, not longitudinal. 6. Electricity. In electricity, particularly in the discussion of alternating currents, the sine curve plays a prominent role. The equations e = 156 sin 0, i = 4 sin 0, and p = ei = 624 sin 2 = 624( - 1 cos 2 6) = 312 - 312 cos 2 0, 416 UNIFIED MATHEMATICS represent respectively the electromotive force, e, measured in volts, and the current, i, measured in amperes, and the power, p, measured in watts of an ordinary electric current. In general, current and electromotive force are "out of phase " ; the equa- tions when the cur- rent lags 30 behind the electromotive force are, e = 156 sin 0, i = 4 sin (0-30). Sinusoids traced by electrical means Oscillogram of an alternating current in which current and e. m.f. are " in phase." On the two diagrams we have represented by a photographic process the magnitude of the current and electromotive force of an alternating cur- rent. The current is represented by the curve with the smaller amplitude. In the first illus- tration current and e. m. f. are " in phase," and under these conditions a maximum of power is developed ; in the second illustration current and e. m. f. are " out of phase," the current lagging behind the e. m. f . The power at any instant delivered by an alternating current is given by the product of the current and the e .m. f . at that instant. Employing the formulas, Oscillogram showing current curve (lower) lag- ging 90 behind e. m. f. curve WAVE MOTION 417 cos (a ft) cos a cos ft -f sin a sin /?, cos (a + ft) = cos cos /3 sin a sin ft, whence cos (a ft) cos (a -f ft) = 2 sin a sin ft, show that p= ie sin sin (0 30) may be reduced to [cos 30 - cos 2(6 - 15)]. tt Plot the curve showing the power at any instant, when e = 156 cos and i = 4 cos (0 30). Note that this power curve is also a sinusoidal curve but placed with reference to a horizontal line which runs 270 units above the x-axis. PROBLEMS 1. Plot the curves y = sin 256 IT f and y = sin 512 w f t using ^ inch for 1 on the vertical axis and 6 half-inches for yi-g- of 1 second on the t or horizontal axis. Treat the equations as y = sin 2 irt r , and y = sin 4 Trf, respectively, substituting for t, 0, .1, .2, .3, , .9, and 1 instead of -fa of T ^-g-, T 2 ^ of T ^-g-, . Note that the unit T | ? taken as 1 on the horizontal axis, disposes of the difficulty of the awkward fractions. 2. What is the frequency of the vibrations in the curves of the preceding example ? What are the corresponding wave lengths ? 3. How would y = cos 2 irt r differ from the curve for y = sin 2 -rrV ? f Write 10 values of y = cos 2 irf for t = 0, -fa, i> i> ' ?' TZ> i' '" ** Note that these angles correspond to 0, 30, 45, 60, respectively. 4. Use the equation cos 9 = sin (90 + 6) to show that y = sin lags 90 behind y = cos 6. 5. Draw the graphs of y = cos and y = cos 2 0; divide the arc of the circle into 24 equal parts and take the distance rep- resenting 2 TT" as divisible by 24. 418 UNIFIED MATIIKMATICS 6. Draw the graph of y = sin ( * + ) and compare with V 4 / y = sin 6. Discuss the corresponding motion of the moving point on the vertical axis. 7. The limits of hearing are for vibrations of 16 per second and 40,000 per second. What are the corresponding wave lengths ? 8. Plot on the same diagram the two curves, e = 156 sin 0, i = 4 sin (0 - 30). 9. In problem 8 find the value of e for each 30 to 360. This completes a " cycle " of values. The time of this move- ment in a 60-cycle system is ^ of 1 second. What is the value of t for the angles given, and also for 6 = 45, 135, 225, and 315 ? 10. On the curve, on the same axes as the preceding, i = 4 sin (6 30), read the values of Q for the angles 30, 45, 60, 90, to 360. These may represent current in the cir- cuit of problems 5 and 6 ; the current lags 30 behind the e. m. f. What interval of time is represented by the 30 lag ? 11. Plot to the same axes the curves, i = 4 sin (6 + 40), e = 156 sin 0. The curve of i here leads the curve of e by 40. In the case of i, what are convenient values of 6 to plot without using tables ? 12. Assuming that it takes -^ of 1 second for one com- plete cycle of i or e in problem 8, find the time difference represented by the 40 angular difference. Find angles ap- proximately corresponding to ^ ^, T ^, T ^, and ^ of 1 second. 7. Sine curve; circle; ellipse; cylinder. If a circular cyl- inder, such as the one in our diagram, is cut by any plane, the WAVE MOTION 419 intersection is an ellipse. Thus, the plane through AOB in our diagram, in- clined at an angle of 45^ cuts the cylinder in an ellipse whose equation is # JP_ t a? 2 a? 1. The circular base and any parallel section has the equation O 2 " f "o 2 = Taking the portion of the cylindrical surface be- tween the elliptical curve and the circular curve through the same center and unrolling it gives a pure sinusoid, y = a sin 6. In the figure PQ = QM, since the cylinder is cut at an angle of 45. But QM, the ordinate on the circle, equals a sin 9 and r Sinusoid developed by means of cylinder A plane intersecting all the elements of a circular cylinder cuts the surface in an ellipse the arc AQ, which will be the ab- scissa, is a6. Hence the curve, when rolled out, is y = a sin 6 and will give a complete arch of the sinusoid if a circular u ?P er and lower Portions of the surface are given. 420 UNIFIED MATHEMATICS Experimentally the student can develop this curve by roll- ing a sheet of paper about a cylinder and cutting out with a sharp knife the required portions. Piston-rod diagram AB, the stroke ; HC, connecting rod ; 0(7, crank arm. 8. Piston-rod motion. The common piston-rod motion of engines furnishes abundant trigonometric material, much of which is of sufficiently elementary character so that by the application of simple formulas problems of interest to the engineer can be solved. The essential features for our purposes are the piston head H, the connecting rod HO of length I, the crank arm OC of length r, and the stroke AB, which is the distance through which the piston head H moves. Were the connecting rod infinite in extent, the motion of H would be simple harmonic motion when C is rotating with uniform velocity about 0. In modern engines the ratio of I to r varies from 3 to 1, low, to 4.8 to 1, which is approximately that of a Ford engine. It is desired to find for each position of the piston head the angle a of the connecting rod, the angle of the crank shaft, and also the effective pressure, called the tangential compo- nent, of the connecting rod to turn the crank shaft. In the first place, when I : r = 4.8 : 1, the angle a never exceeds arc sin . Determine this angle in degrees. As the 4.8 pressure P at H is horizontal, only a portion, Pcos a, of this WAVE MOTION 421 pressure is communicated to the connecting rod. Discuss the variation in pressure due to the inclination of the connecting rod and note that it is relatively small. Of course the pressure of the gas in the cylinder chamber is not uniform and this Connecting rods and cranft arms in six-cylinder automobile engine : ratio I : r = 3 : 1 variation is much more serious than the variation due to the angle of a connecting rod. Find also the maximum vertical pressure P sin a on the cross-head support. Show that the length of the stroke AB is equal to the diameter of the crank circle. The position of the pfston head is indicated in decimal parts of the total stroke 2 r, as measured from A. In our diagram the piston head is at .75 of the stroke. To determine a and when the position of the piston head is given we solve a triangle in which the three sides are known. Thus on our diagram O0'= 4.8 ; HO = 4.3 ; HC= 4.8 ; and 0(7= 1. Solve this triangle and determine the angles a and 6. Commonly a diagram is drawn in which the angles 6 of the crank arm are plotted as abscissas and the piston displacements are plotted as ordinates. 422 UNIFIED MATHEMATICS PROBLEMS 1. Given = 0, 10, 20, 30, 60, 90, 120, 150, and 180, find the corresponding values of a and the positions of H as decimal parts of the stroke. Plot angles as abscissas and piston displacements as ordinates. How could you complete this to 360 ? Could you go on beyond 360 ? 2. Assuming that the piston rod. transmits a uniform pres- sure of 200 lb., find the effective turning pressure when = 30, 60, and 90. Resolve the force at C into two com- ponents, one normal to the crank shaft and the other along the crank shaft. The component normal to the crank shaft, the tangential component, is effective, i.e. does the work. The radial pressure is also computed and is used to determine friction loss. Find the values of the radial pressure corre- sponding to above tangential pressure. Compute in this case for. the given angles the pressure 'on the cross-head support by the connecting rod due to the com- ponent, P sin a. Connecting rod, crank arm, and cylinder on A. T. & S. F. locomotive 1444 The stroke is 30 inches and connecting rod 60 inches. 3. Draw the graph illustrating relative positions of the piston head and the connecting rod when the crank-pin is at the lowest point, in the locomotive illustrated above. 4. Find the number of strokes of the piston per minute when the train moves 60 miles per hour, given that the driving wheels are 57 inches in diameter. CHAPTER XXVII LAWS OF GROWTH 1. Compound interest function. The function S = P(l+ i) n is of fundamental importance in other fields than in finance. Thus the growth of timber of a large forest tract may be ex- pressed as a function of this kind, the assumption being that in a large tract the rate of growth may be taken as uniform from year to year. In the case of bacteria growing under ideal conditions in a culture, i.e. with unlimited food supplied, the increase in the number of bacteria per second is propor- tional to the number of bacteria present at the beginning of that second. Any function in which the rate of change or rate of growth at any instant t is directly proportional to the value of the function at the instant t obeys wha^t has been termed the " law of organic growth," and may be expressed by the equation, y = ce kt , wherein c and k are constants determined by the physical facts involved, and e is a constant of nature analogous to TT. The constant k is the proportionality constant and is negative when the quantity in question decreases ; c is commonly positive ; e = 2.178 .. The values of the function of x, ce kx , increase according to the terms of a geometrical progression as the variable x increases in arithmetical progression. 2. IT and e. A function can be found by methods of the calculus which is such that the rate of growth of the function 423 424 UNIFIED MATHEMATICS at any instant t, or x, exactly equals the value of the function at that instant. This function is given by the equation, y = e f or y = e?. The constant e is represented by the series wherein [3, ( [4 = 4x3x2 teger, represei down to 1. 2.71828; to 1( The curve equation is such that a curve the sic merical value that point. The graph that the slop< the negative that point. The graph that the slope ailed factorial 3, represents 3x2x1, and X 1, and, in general, \n, n being a positive in- its the total product of all the integers from n The sum of this series to 5 decimal places is ) places, = 2.7182818285. represented by the . , , , , , , , , , , , , , , , , t---- y = e x b any point on this >pe equals in nu- s the ordinate at ^ " ;: ~JtlT~ P I ' c- c-ttn'h - 1 i 4- - f- - - - " at anv point is - - 4* of the ordinate at / f 1 ,1 ^_ n 1 ' \-~ I 1 / oi y e** is such / ; L 1 i ' at any point ISA; : : : j f rr :::::::::::!? 1 l!|!l IIJJU II |||lll4^M 1 "^ ^M > U-I-I-L ::|::::::::|!! rrrt : 0: _ j: ^ 3:-- r= ! = ;4 ::: ^^S5 Graphs of y = 6? and y = 1(H times the corresponding value of the function at that point. Values of the .function y = e* may be determined by loga- rithms. LAWS OF GROWTH 425 Thus to find the points on y = e*, for which x = 2, 1, 0, .1, .2, .5, .8, 1, 2, and 3 respectively, we take first log y x log e ; since log e = .4343 logy = -.8686, y= .135 = 9.1314-10 log y = - .4343, y = .368 = 9.5657-10 log ?/ = 0, y = 1 logy= .0424, y= 1.103 log#= .0869, y = 1.222 logy= .2171, y= 1.649 logy= .3474, y= 2.225 logy = .4343, 2/= 2.718 logy= .8686, y= 7.390 a?= 3, logy= 1.3029, y = 20.09 Similarly, if y ce kx , log y = log c + kx log e, and these values are obtained by logarithms. The limit of the expression as n approaches infinity gives the value, e. When n is taken as a large positive integer, it can readily be shown that this expression has a value differing but slightly from e. e and IT may be called fundamental constants of nature ; in mathematical work as applied to statistics and to physical problems of varied kinds these constants often appear. 3. Natural logarithms. The first logarithms as computed by Napier were not calculated to any base, but were founded 426 UNIFIED MATHEMATICS upon the comparison between an arithmetical and a correspond- ing geometrical progression. However, a base of the Napierian logarithms can be established, and it is approximately -. e Graph of y = logx The ordinates represent the "natural logarithms" of the numbers represented by the abscissas. From the mathematical point of view, and indeed for many applications of mathe- matics to physical problems, the base e is preferable to 10 as the base of a system of logarithms. Logarithms to the base e are called natural logarithms. 4. Application. The most immediate application of a function in which the growth is proportional to the function itself is to the air. The decrease in the pressure of the air at the distance h above the earth's surface is proportional to h. h_ The expression P = 760 e 7990 gives the numerical value of the pressure in millimeters of mercury for 7t measured in meters. The negative exponent indicates that the pressure LAWS OF GROWTH 427 decreases as h increases. In inches as units of length of the mercury column, h in feet, P = 29.92 e~^. This is known as Halley's Law. The growth of bean plants within limited intervals and the growth of children, again between quite restricted limits, follow approximately the law of organic growth. Radium in decomposing follows the same law, the rate of decrease at any instant being proportional to the quantity. In the case of vibrating bodies, like a pendulum, the rate of decrease of the amplitude follows this law ; similarly in the case of a noise dying down and in certain electrical phenomena, the rate of decrease is proportional at any instant to the value of the function at the instant. PROBLEMS 1. By experiment it has been found that 1000 of the so- called "hay bacteria" double their number, under favorable conditions, in 20 minutes. Find the rate of growth per minute. Take n = 1000 e u , and determine Jc by substituting n = 2000, t = 20, and solving for Tc. Determine the number that would grow from 1000 bacteria in 1 hour ; in 1 day. 2. The cholera bacteria have been found, under favorable conditions, to double their number in 30 minutes. Determine the rate of growth per minute, and the number that would grow in one day from 1000. NOTE. The favorable conditions cannot be continued for such a period. 3. Assuming that P=760e~i^ find the value of h which will reduce the pressure of the air by 1 mm. Take the logarithm of both sides and note that 428 UNIFIED MATHEMATICS - Iog 10 e must reduce log 760 to log 759. Find at what 8000 height the mercury column is reduced 1 cm. At what height would the pressure be reduced to 660? Are these heights ever attained ? 4. Find the barometric pressure when h = 1000, 5000, 10,000, and 15,000 feet, assuming P= 29.92, when h = 0, in which P is the numerical value of the pressure in inches of mercury and the height is h feet. Find the height for which the pressure decreases ^ of 1 inch. 5. Show that if the height of the elevation is measured in miles, the pressure in inches is given approximately by the formula A P = 29.92 e i. Note that 26,000 is nearly 5 x 5280. The constant 26,200, above, is taken for simplicity instead of 26,240. 6. To what change in height does the maximum variation of the barometer recorded on the photograph,, page 60, correspond ? 7. Compute by the progressive method of Section 4, Chapter XII, the value of e, from the series, by summing 8 terms to 8 decimal places. 8. Plot on the same diagram and compare the two graphs, of y = e 1 and y = 10*. 9. Plot the curve y = e~ z *, taking ^ inch as .1 on the hori- zontal and on the vertical axes. This curve represents what is termed the normal distribution curve, which is of funda- mental importance in all statistical work. In general, large groups of individuals may be distributed as to ability in any given quality over the area under such a curve ; the middle ab- scissa at x = represents average ability, and deviation to one side or the other represents, on one side, ability above the aver- LAWS OF GROWTH 429 age and, on the other, ability below the average. The total number of all individuals considered is represented by the area between the curve and the avaxis. An interesting graphical illustration of a normal distribu- tion curve is the crowd at a football game when a great bleacher is not filled. The central aisles are all tilled to a height representing the middle ordinate and from this out in either direction, the ordinates drop off, frequently in strikingly symmetrical manner, and corresponding quite closely to the normal distribution curve. 5. The curve of healing of a wound. Closely allied to the formulas expressing the law of organic growth, y = e kt , and the law of " organic decay," y = e~ kt , is a recently discovered law which connects algebraically by an equation and graph- ically by a curve, the surface-area of a wound, with time Sg.rm. Scofe-- Wound ' 30 15^--- _ :H:::::::: :::::::::::::::::::5 it::^t + - J j ) ::: 20 10 :: ^V ;;:; 1 c^ ^v^^ 10 5 3f - > ; "!!>! |:j ._.:* = | !S i!;|!::i:E::|:::: DayO 4 8 12 16 20 24 28 32 36 40 44 48 52 56 GO 6466 MaylS J9 23 27 31Je.4 8 12 16 20 24 28./Z.2 6 .10 14 1820 Progress of healing of a surface wound of the right leg, patient's age 31 years The observed curve oscillates about the smoother, calculated curve. expressed in days, measured from the time when the wound is aseptic or sterile. When this aseptic condition is reached, by washing and flushing continually with antiseptic solutions, two observations at an interval commonly of four days give the "index of the individual," and this index, and the two measurements of area of the wound-surface, enable the physi- cian-scientist to determine the normal progress of the wound- surface, the expected decrease in area, for this wound-surface of this individual. The area of the wound is traced carefully 430 UNIFIED MATHEMATICS on transparent paper, and then computed by using a mathe- matical machine, called a planimeter, which measures areas. The areas of the wound are plotted as ordinates with.the respective times of observation measured in days as abscissas. After each observation and computation of area the point so Sg.rm. 10 V - \ ^ ^. ^, ' I / ,/, Ht II ,. X' /. ll \ ^ ** " r -: / -^ s \ .' -N i - / "- - ~ \ \ j*-*' - s s - >, * - \> / S S s~ ^ s ? V ^, f -" - ^ o j * ' , -^ 2te ^M ^-~ I'll, -< / /'< 1 \ \ 2 i a 2 t ) 16 24 32 Sept. 23 Oct. 1 9 17 25 Nov. 2 8 12 Progress of a surface wound of the right knee Two infections in the course of healing are indicated. obtained is plotted to the same axes as the graph which gives the ideal or prophetic curve of healing. Two such ideal curves and also the actual observed curves are represented in our diagrams. When the observed area is found markedly greater than that determined by the ideal curve, the indication is that there is still infection in the wound. This is the case de- picted, as will be noted, in the smaller diagram. A rather surprising and unexplained situation occurs frequently when the wound-surface heals more rapidly than the ideal curve would indicate ; in this event secondary ulcers develop which bring the curve back to normal. This is the type which is represented by our larger diagram. This application of mathematics to medicine is largely due to Dr. Alexis Carrel of the Rockefeller Institute of Medical Research. He noted that the larger the wound-surface, the more rapidly it healed, and that the rate of healing seemed to be proportional to the area. This proportionality constant is LAWS OF GROWTH 431 not the same for all values of the surface or we would have an equation of the form, $4&"i in which Si is the area at the time that the wound is rendered sterile and observations to be plotted really begin. The actual formulas, as developed by Dr. P. Lecomte du Nolly of Base Hospital 21, Compiegne, France, are (1) ; = Si-&> giving the characteristic constant of the wound. Si is the measure of the area, first observation; S 2 is a second measurement taken after 4 days. (2) S. = S.-i[l-i'( + V4~n)], wherein S n is the area after 4 n days ; similarly, S n _i is the area after 4(n 1) days, etc. ; each ordinate is obtained from the preceding ; i is the constant as determined above. Recent experiments by Dr. du Nolly show that there is a normal value of i dependent upon the age of the individual and the size of the wound, and that the individual index as determined by two observations will doubtless reveal facts concerning the general health of the individual. The data given are taken from the Journal of Experimental Medicine, reprints kindly furnished by Major George A. Stewart of the Rockefeller Institute. The diagrams are reproduced from the issue of Feb. 1, 1918, pp. 171 and 172, article by Dr. T. Tuffier and R. Desmarres, Auxiliary Hospital 75, Paris. 6. Damped vibrations. The combination by multiplication of ordinates in the two functions, y == e~*i' and y =.sin k<, which we have seen to be fundamental in the mathematical interpre- tation of many phenomena of nature, gives a formula which also has wide application. The formula y = e~** sin k<, 432 UNIFIED MATHEMATICS expresses the law by which the decrease in intensity of the vibrations defined by y = sin k. 2 t may be determined, under certain conditions. Damped vibration curve by multiplication of ordinates Thus a pendulum swinging in the air, when the friction is proportional to the velocity, has this form of equation as the equation of motion. We have indicated on our diagram the curves / y = e~S and y = e~ ( and the damped vi- bration curve y = e~ l sin 2 TT. The student should check the values, re- drawing all the curves on double the scale of the illustration in the text. A beautiful damped vibration curve is obtained by the discharge of an electrical condenser. In our illustration the equation v = e~* 1 sin 2 irt Damped vibration produced electrically by the discharge of a condenser LAWS OF GROWTH 433 represents quite closely the curve, using the maximum ordi- nate as unity, i.e. when t = 0, and on the horizontal axis the measure of the time in seconds of one complete vibration is taken as unity ; in the electrical occurrence represented by this photograph, and the corresponding light phenomenon which produced the photograph, the action took place in about -^j of one second, and -^^ of one second is approximately the time of one vibration on this curve. PROBLEMS 1. Plot the following curves, in the order given : y = sin 2 irt r , taking two half-inches to represent t = 1 on the horizontal axis and 6.2 half-inches to represent unity on the vertical axis ; use the graphical method. y = ~, taking 6.2 half-inches to represent unity on the vertical axis, and two half-inches to represent one second on the horizontal axls - y = e~'* sin 2irf, by multiplication of ordinates. Note that by taking five half-inches to represent unity on the vertical axis each half -inch represents .2 and each twentieth of an inch represents .02. These facts are to be used when you multiply ordinates to obtain the third of these curves. For the values of the powers of e consult the table at the back of the book. 2. Given that an automobile wheel which is revolving freely at the rate of 400 revolutions per minute is allowed to come to rest by the action of the friction and air resistance ; assuming that the subsequent velocities per minute are given at the end of t minutes by the equation to determine these velocities, plot the graph of the func- tion. At what time will the number of revolutions be 434 UNIFIED MATHEMATICS reduced to approximately 200 per minute? to 100? to 50? to 3? It may be assumed that at 3 revolutions per minute the law no longer holds and that the wheel will stop at about that time. 3. Given that the horizontal displacement of a second pendulum is 4 inches, and that the horizontal displacement is. given by the equation x = 4 cos 2 irt, and that the amplitudes are decreased according to the " law of organic decay," the position being given at any instant by the equation * = 4 e^i cos 2 ^. Find the displacement of the pendulum after 10 seconds ; after 100 seconds ; after one hour. When does the pendulum have a displacement of only 1 inch ? of ^ inch ? of y 1 ^ inch ? This type of retardation is found when the friction is proportional to the velocity. 4. Given that a fly-wheel revolving freely with a velocity of 500 revolutions per minute is allowed to come to rest. If the velocity at the end of t seconds is given by the equation find the velocity at the end of 10 seconds ; at the end of 100 seconds ; at the end of 30 seconds. When will the velocity be reduced to 1 revolution per minute ? With a heavy oil as lubricator heavy fly-wheels follow approximately this law. CHAPTER XXVIII POLAR COORDINATES (See Section 3, Chapter VII) 1. Uses. For many purposes the representation of func- tions by the system of polar coordinates is desirable. Thus, effective pressure on the crank head by the piston head varies for every angle. It is convenient to give this pressure- diagram in polar coordinates ; on every radius is plotted a length representing graphically the effective turning pressure on the crank for that angle. 2. Plotting in polar coordinates. The coordinates of points which satisfy an equation given in polar coordinates are obtained precisely as in rectangular coordinates. An equation in polar coordinates involves r and 0, radius vector and vectorial angle ; by substituting in the given equation particular values of one of the variables and solving for the correspond- ing values of the other points on the curve are obtained. Illustrative problem. Plot the curve 3.42 5.00 r = 10 sin 20. 6.43 7.66 Note that when = 10, r = 10 sin 20 = 3.42, which 35 40 45 60 9.40 ing how the second loop and other loops are obtained, by 9.85 giving to values increasing by 5 intervals up to 360. Note 10 that no further computation is needed. Follow the progress 9.85 of the curve on the diagram given on the next page. 435 436 UNIFIED MATHEMATICS i 4 v<& The formulas for transformation from rectangular coordi- nates to polar coordinates should be noted : x = r cos 6, y = r sin 6. The formulas for transformation from polar to rectangular coordinates are : r= ^ ^ j?/ :R = arctan , x and sin = or cos = POLAR COORDINATES 437 PROBLEMS 1. Plot the curve r = 10 sin 6. Prove that this is a circle. Take any point on the circle, as (r, 0), and show that the co- ordinates satisfy the equation. 2. Plot the following curves : (a) r sin 6 = 5. (6) r cos = 10. (c) r sin (0-30) =10. (d) r = 10 sin 30. (e) r = 10 - 10 cos 0. (/) r = 6. 3. Plot the curve r = 2 a tan sec ; transform to rec- tangular coordinates. This curve is a " cissoid " and can be used in the "duplication. of the cube" problem. 4. Plot the curve r = 10 sec -j- 5. This is a " conchoid of Nicomedes," and can be used to effect the solution of the problem to trisect any angle. 5. Plot the polar diagram of effective pressures on the crank for different angles of ; use the data of problem 2 in the problems given under piston-rod motion.* 6. Plot the curve r = 10 10 cos 0. This curve is called a " cardioid " because of its shape. 7. Plot the curve r = 10 5 cos 0. This is called a " limaqon of Pascal." 8. Plot r = 10 20 cos 0, another type of limaQon. 9. Show that the polar equation of any conic is 2m f _ 1 e cos ' wherein 2 m is one half of the right focal chord. 10. Plot the parabola r= 10 1 cos 438 UNIFIED MATHEMATICS 11. Plot the hyperbola r= 10 1 - 2 cos B For what values of is r infinite in value ? What directions do these values give ? Are these lines from the origin then the asymptotes ? 12. Plot the spiral of Archimedes given by r = 10 6. 13. Plot the hyperbolic spiral given by 14. Plot r=sin20. 15. Plot r = sin + sin 2 0, and compare the polar with the Cartesian (x, y) representation. CHAPTER XXIX COMPLEX NUMBERS 1. Object. In the study of the number field, indicated in our first chapter, we found that in the extraction of square roots we were limited to positive numbers. Again in solving quad- ratic equations, and in the discussion of the roots of algebraic equations, we found that no number of the kind we had con- sidered could occur as the even root of a negative quantity. We can extend the number field, removing the limitation that square roots and even roots must be taken of positive quan- tities only, by creating another class of numbers, complex numbers. These numbers, after the fundamental operations with them have been properly defined, apply to our algebraic equations, x and the constants being complex numbers. In the extended number field it is possible to prove that every rational integral equation has a root and that such an equation of the nth degree has n roots. 2. Complex numbers. We define V 1, designated by ?', as a number which, multiplied by itself, equals 1 ; this re- quires, then, an extension of the meaning of multiplication and a reexamination of the fundamental processes as applied to the old numbers with this newly found number and other new numbers which follow directly from it. This discussion is given graphically in the next section. The square root of any other negative number, a, is regarded as VaV 1, or Va i. Such a number, e.g. V 7, is called a pure imag- inary. To add a pure imaginary to a real number both must be written and the combination is called a complex number. 439 440 UNIFIED MATHEMATICS Thus, x + yi and a + bi, or 3 + 2V 1 and V5 V3i. are complex numbers. Addition and multiplication are explained graphically in sections 6 and 7 below. 3. Graphical representation. Our real numbers can all be conceived graphically, as well as analytically, as derived from the unit 1. Thus, integers are obtained by the repetition of the unit ; fractions are obtained by the subdivision of the unit ; and nega- tive numbers are obtained from the negative unit, which in turn is ob- tained by reversing the direction of the positive unit. Analytically the imaginary unit repeated as a factor gives 1 ; graphically then we would desire an operation which repeated gives a reversal of direction. How is the reversing from -f 1 to 1 effected? Evidently by turning the an angle of 180 or -180. The the The imaginary unit ob- tained graphically positive unit through be regarded then as represented by V 1, or i, can middle position of this rotating unit, and the upper position is regarded as + i and the lower as i. This vertical line is taken as the axis of pure imaginaries. Thus, V 4, or 2 i, is represented two units up on this axis and V 2 is repre- sented V2 units down on this axis. A complex number, x + yi, may now be uniquely represented by the point (or, y] in the complex plane, in which the y-axis coincides with the vertical axis of pure imaginaries. The fundamentally important facts concerning these num- bers are : 1. Complex numbers are combined according to the laws of the real numbers (which we have discussed in the first chapter), noting that i 2 = 1. COMPLEX NUMBERS 441 2. The combination of any two or more complex numbers, by the operations of addition, subtraction, multiplication, division (except by zero), involution, and evolution (with certain excep- tions), ALWAYS produces a complex number. mASmm Representation of three complex numbers Two complex numbers of the form x + yi and x yi, sym- metrically pla,ced with respect to the axis of reals, are called conjugate complex numbers. Their sum and their product are real numbers. 4. Complex roots in pairs. In any rational integral algebraic equation with real coefficients, if a -f bi is a root of the A equa- tion, then a bi is also a root of the equation. The proof depends upon the fact that when a 4- bi is substituted in a^c n + a^x"' 1 -f a?x n ~ z -f a n , the resulting expression is of the form P -f Qi, in which P and Q being real numbers, P involves powers of a and the even powers of bi, and Q is obtained from expressions involving odd powers of bi. Now if then P = and Q = ; otherwise you have a real number equal to a pure imaginary. Substituting a bi for x in 442 UNIFIED MATHEMATICS changes the signs of the terms involving the odd powers of i, and does not change the sign of the even powers. Hence a bi substituted gives but P=0 and Q = 0, hence P Qi = Q. Therefore a bi is also a root of the equation if a + bi is a root. Complex roots go in pairs. Illustrative problem. 1. Find the product of 2-fV 3 by 3 V 2 and put the product in the x -f yi form. Graphical representation of the product of 2 + V3 i by 3 (2 + V3 t)(3 - \/2 i) = 6 + 3 v'3"i - 2 \/2 i - \/6 i 2 , but t 2 = 1 , giving as product, 6+V6+(3V3 -2\/2)i. Ans. 6+V6= a; 3V3-2V2 = 6. 2. Divide 3 + V 3 by 5 2V 3 and express the quotient in x + yi form. 3 + V3t t)(5 + 2V3t) _ 15+ _2V^3 5-2V3i (5 - 2 V3 t)(5 + 2V3 i) _ 9 + llv'3i _ _9_ llv/3t 37 ~ 37 37 25-12? 3. Factor x z + y 2 into complex factors, linear in x and y. x* + yz = (x + iy) (x iy). COMPLEX NUMBERS 443 PROBLEMS 1. Write the conjugate complex numbers : a. 3 + V^. cl ! + ' b. -4-21 e. 2. Rationalize the denominator in the following expressions by using the conjugate complex number as multiplier, reduc- ing the quotient obtained in this way to the form a + bi. a. 5 ~ V ~ 2 . a. J- b. 3 + V-2 5 V5 - 4 - 2 1 c. 3 3. Write the following expressions in the form a + bi: a. t- + i* + t+i. 2 2 ' n . n > 6. t -5 .(_ 3 ( -j. 3 p + 4 i 8 . 3 + 2*3-2 i c. t^+i 20 . 2 2 d. 4. Locate the points represented by the complex numbers in problem 1. 5. Square - - + -z; square - - ^-f. These are roots L 2i of a^- 1 = 0. Multiply _l + ^ by- \-^j>-i. What is 22 22 the cube of ---f^t? 1 t l + i 6. Square - - and - . Give an equation with real V2 V2 coefficients which these numbers satisfy. What are the square roots of i? 444 UNIFIED MATHEMATICS 5. Vectors. Polar representation of complex numbers. The complex number x + yi may be regarded either as determined by the point P(x, y) or by the vector OP, which by its length and direction determines the position of P. The angle which OP as a ray makes with the sc-axis is called 0, the amplitude or angle, and the length OP is called r, the modulus of the com- plex number. In other words, the polar coordinates of P are faff); r = Vx 2 -(- y 2 , cos = - , and sin 6 = ^. r r The complex number may be writ- ten in the form r(cos 6 + i sin 0), which is termed the polar form. t 0:rr i Modulus and amplitude of . a complex number Tlie modulus, r or V 2 + y 2 , is a positive number representing the length of the vector or the distance of the point x + iy from the origin. This modulus is sometimes called the stretching factor or the tensor; see section 7. 6. Addition of vectors. "\Vhen the complex number is repre- sented by a vector, the sum of two complex numbers will be represented by the diagonal of the parallelogram formed by the two vectors ; see Chapter IX, section 2. The student should verify the fact by a diagram. 7. Product of complex numbers. Given two complex num- bers, either in polar form or in rectangular form, the product of the two numbers is also a complex number ; further the modulus of the product is the product of the moduli, and the amplitude or angle of the product is the sum of the amplitudes of the factors. COMPLEX NUMBERS 445 Let r^cos 61 + i sin 0^ ; x v + yj, mod Va^ 2 + y^, ampl X = tan" 1 ^ and r 2 (cos 2 4- 1 sin 2 ) ; a^ + */2*> mod Va^ 2 + 2/2 2 > ampl 2 = tan" 1 be two complex numbers. Their product is 7 'i r 2[ cos $i cos $2 s i n $1 g i n $2 4- *'(sin $1 cos ^2 4- cos $1 s i jl $2)] #1^2 */i#2 4- * (^12/2 4- #22/i)> m d Vajj 2 ^ 2 4- 2/i 2 ?/2 2 4- ^i 2 ^ 2 + x ^y^ ampl 03 = tan s ' . The polar product may be written rtr 2 [cos (0 t 4- 2 ) + i sin (0 X + 2 )], showing that the product of the moduli i\ and r 2 is the modulus r t r 2 of the product and the amplitude is X + 2 , the sum of the amplitudes. It is left as an exercise for the student to show that the analytical expressions for modulus and amplitude establish the same facts. When any complex number is used as a multiplier, the modulus of the product is the modulus of the multiplicand stretched in the ratio of the modulus of the multiplier to unity. For this reason the modulus is sometimes termed the stretching factor. 8. De Moivre's theorem. Evidently, if t and 2 are set equal to 0, our product formula may be written : (1) [r(cos 4- i sin 0)] 2 = r 2 (cos 2 4 t sin 2 0). Evidently by mathematical induction, by simple introduction of one further factor ?-(cos + 1 sin 0) at a time, it can be shown that (2) [r(cos 4- i sin 0)] n = r"(cos nO + i sin w0). This theorem, which holds for all values of n, is called De Moivre's theorem. We have proved it only for n an integer. Taking the wth root of each member of equation (2), we have (r n ) " (cos ?i0 + i. sin nff) n = r(cos 4 i sin 0). 446 UNIFIED MATHEMATICS Let r" = k and nO = 0', which, as no limitation was imposed on 6, imposes no limitation as to value on 6', and we have [k(cos 6' + i sin 0')]" = k "(cos - + i sin -Y \ n n J i.e. our formula holds for a fractional exponent of the form -. n By raising to the mth power both sides, it can be shown to hold for any fractional exponent. For ?i = 1, [r(cos6 + izwff)~\- 1 = r(cos 9 + i sin 0) cos i sin 6 = r(cos 2 6 + sin 2 6) = r -1 (cos 0i sin 0), whence [r(cos + i sin 0)]" 1 = r^[cos (- 0) + i sin ( 0)], which establishes the formula when n = 1. By raising both sides to the nth power, n any rational number, the theorem is established for all rational exponents. The theorem can be established also for irrational values of n. PROBLEMS 1. Write the following complex numbers and their conju- gates in polar form, giving modulus and amplitude : b -4-2i g ' c . _3-V2-V^. h ' 23 ' d. l + i._ ? . 1 V3\ 2. Show that (cos 30 + 1 sin 30) 2 = cos 60 + i sin 60, by multiplication. 3. Show that (cos 30 + z sin30) 8 = z, i.e. cos 90 + i sin 90. 4. Show that - - = cos ( - 30) + i sin ( - 30). cos 30 + i sin 30 COMPLEX NUMBERS 447 5. Show that (cos 60 + i sin 60)* = cos 30 + i sin 30. 6. What is the value of (cos 30 + i sin 30)* by De Moivre's theorem ? 7. Plot, using 2 inches as 1 unit, cos 15 + i sin 15, cos 30 + i sin 30, cos 45 + i sin 45. o/jrvo 8. Plot the point B, cos -- 1- i sin - ; connect by a chord with the point A, cos + * sin ; take this length as a chord, successively seven times on the unit circle about O. This chord is the side of a regular inscribed polygon of seven sides. 9. Roots of unity. Plotting the solutions of the following equations on the complex number diagram, (See Section 3, above), x 1 = gives one point, 1 ; a? 1 = gives two points, 1 and 1 ; a^-l = gives three points, 1, - 5 + -5- *' ~ >~~^~ 7 ' u m a Ji s* 1 = gives four points, 1, 1, i and i ; of 1 = gives six points, x 8 1 = 0, or (x 4 l)(x 4 + 1) = 0, gives eight points, which may be obtained by methods of quadratic equations. For a-* + 1 = ^ + 2 x 2 + 1 - 2 a? = (& + I) 2 -(V2 ) = (a 2 + 1 - V2 z)(o; 2 + 1 + V2 x), whence, x 2 - and 448 UNIFIED MATHEMATICS a 8 1 = gives eight points, 1, 1, L i, 4. - t, - V2 V2 V2 V2 11. 11. -| - i, and 1. V2 V2 V2 V2 a; 5 1 = gives five points ; see the solution obtained in problem 4, page 97. The solutions are V5 + 1 - 2 y5 . V5 + 1 , V10-2V5 . ~ *' ~ ~ * -1 + V5 ; . d -1+V5 Plotting these points on the complex diagram gives the ver- tices of a regular pentagon. Graphically representing these points we have the following diagrams : Graphical solutions of y?-\ = 0, s> - 1 = 0, x 6 - 1 = 0, x 12 - 1 = 0, x 8 - 1 = 0, and x 5 - 1 = COMPLEX NUMBERS 449 The roots of any equation of the form x n 1 = 0, n integral, are, aside from + 1 or 1, complex numbers. Since any real number less than 1 when multiplied by itself gives a number less than 1 and since any real number greater than 1 multiplied by itself gives a number greater than 1, it follows that the roots other than 1 or 1 are complex. Further, the modulus of each of these complex roots is a real number which, taken n times as a factor, produces 1 ; hence the modulus of any nth root of unity is 1. We need then to know only the real part of any root of unity to plot it, since the root .itself, having a modulus 1, lies on the unit circle, x 2 + y 2 = 1. The nth roots of unity can be obtained graphically by finding the angles which repeated n times give 360 or integral multiples of 360. Thus for the twelfth roots of unity these angles are 0, 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, and 330. Writing the corresponding complex numbers in polar form we have the twelve twelfth roots of unity. If we went farther, taking 360, 390, 420, , we would simply repeat values already obtained. The twelfth roots of unity are, then, V3 1 . 1 V3 . . 1 V3 . , V3 1 . - --- j, ---- - i, i. -\ ----- i, and + --- 1, 2222 22 22 as on* the figure. 10. Historical note. Just as negative numbers were gener- ally accepted only after a graphical scheme of representation of these numbers was introduced by Descartes, so imaginary numbers were neglected and even rejected by mathematicians until a graphical system of representation was found. In 1797 a Norwegian, Caspar Wessel, presented the scheme of representation of complex numbers to the Danish Academy, but public recognition of his work is only recent. A French- 450 UNIFIED MATHEMATICS man, J. R. Argand, discovered the same system independently in 1806 and it appears that the German J. C. F. Gauss in 1831 again independently rediscovered this graphical method. The latter made extensive use of the diagram and since then these numbers play a vital role in the development of algebra. Quite recently the practical importance of these numbers has been more generally recognized by physicists and engineers. Applications have been made to problems in electricity, to the steam turbine by Steinmetz, and to numerous other vector problems. The term imaginary is a misnomer, as our development shows. So far as actuality is concerned, 3 + V 3 exists as a number quite as much as 3 or V3 ; all numbers are the product of intelligence reacting on the experiences of life, and in this sense all numbers are imaginary, the product of the imagi- nation. 1 1 . Mathematical unity. The complex numbers are fittingly chosen to conclude our treatment of plane analytic geometry, elementary algebra, and elementary trigonometry since, as the observant student will have noticed, we have here involved the fundamental principles of these subjects as well as theo- rems of plane geometry. We might note that while regular polygons of seven and nine sides cannot be constructed with ruler and compass, since the solutions of these equations lead to cubics which cannot be solved in terms of quadratic irrationalities, there are other polygons having a prime number of sides which can be so con- structed. Gauss, when only 19 years old, showed that the polygon of 17 sides, and, in general, the polygon of sides 2 2 " -f 1 in number, when this number is prime, can be con- structed with ruler and compass. The corresponding alge- braic fact is that x m 1 = 0, when m is a prime number equal to 2 2 " + 1, is solvable by the methods of quadratics, and the roots can be expressed in functions involving only square roots. COMPLEX NUMBERS 451 PROBLEMS 1. Solve ce 3 1 = and plot the points on the polar diagram. 2. Solve similarly y? + 1 = and plot. 3. Solve a; 5 4- 1 = 0. Using the results given for of 1 = 0, write all the solutions of cc 10 1 = 0. 4. Derive the formulas for cos 2 9 and sin 2 6, using De Moivre's theorem. HINT, cos 2 + i sin 2 = (cos 6 + i sin 0) 2 . Square the right-hand member and then equate cos 2 6 to the real part and put sin 2 equal to the coefficient of i. 5. Derive formulas for cos 30 and sin 3d by the process of problem 4. 6. Show on the diagram how to obtain the twelfth roots of unity. What equation do these numbers satisfy ? 7. Show geometrically and algebraically how you can obtain the solutions of the equation x* - I = . from the complex diagram. Use also the formulas for sin ^ 6 and cos \Q to obtain sin 15 and cos 15 from sin 30 and cos 30. 8. Solve y? 1 = 0, and plot the points on the complex diagram. 9. Solve a; 16 1 = 0. What angles are involved ? CHAPTER XXX SOLID ANALYTIC GEOMETRY: POINTS AND LINES 1. The third and fourth dimensions. We have found that on a line the position of any point may be given by a single number, x, which locates the point with reference to one fixed point on the line. The single number, commonly x, represents distance, in terms of some unit of length, from the point of reference, and direction by means of a + or sign. In a plane the position of any point may be given by a pair of numbers which locate the point with reference to two fixed lines in the plane. The two numbers, x and y commonly, represent the distances in determined order, in terms of some unit of length, from each of the two given lines of reference, and direction as before. By analogy, continuing with the proper changes, it is obvious that in space the position of any point may be given by a set of three numbers which locate the point with reference to three fixed planes in space. The three numbers, x, y, and z commonly, represent the distances in determined order, in terms of some unit of length, from the three given planes of reference, and the direction in each case is determined by the algebraic sign of the number. If the analogy could be continued we could state that the position of any point in a four-dimensional space would probably be given by a set of four numbers which locate the point with reference to four given " three-dimensional " spaces. The four numbers, x, y, z, and w commonly, would then represent the "distances" in determined order, in terms of some unit of length, from, each of the spaces of reference. Without a pre- cise definition of what we mean by " distance " of a point in 452 SOLID GEOMETRY: POINTS AND LINES 453 " four-dimensional space " from a " three-dimensional space " these analogies must be regarded as purely fanciful, and devoid of physical significance. LOCATION OF A POINT Upon or in a With reference to y means of line, one point, one variable, one dimensional. zero dimensional. X. plane, two lines, two variables, two dimensional one dimensional. (, y). space (ordinary), three planes, three variables, three dimensional. two dimensional. . O, y, z). hyperspace, four three-spaces, four variables, four dimensional. three dimensional. (x, y, z, w>). n-space, n(n 1) -spaces, n variables, n dimensional. ( 1) dimensional. (Xi, X 2 , X 3 , "-X n ). 2. Space coordinates. The position of a point in ordinary space is determined by location with respect to three inter- secting planes, called the coordi- nate planes. Just as our lines of reference were chosen perpen- dicular to each other, for conven- ience, in plane analytics, so here the planes of reference are taken mutually perpendicular, like the three sides of a box or like the front wall, the floor, and the left- hand wall of a room. The three lines of intersection of these planes with each other in pairs are called the axes of coordinates, Axes in space designated as x-axis, y-axis, and z-axis; the three planes are named xy-, xz-, and yz-planes respectively ; the point common to the three planes and to the axes is called the origin. 454 UNIFIED MATHEMATICS -& The numbers x, y, and z represent respectively distances from the yz-, xz-, and ay-planes ; direction is determined as indicated by the arrowheads upon the diagram. No general agreement has been reached as to which axis to use as the vertical axis. The system shown is called a right-handed system, since the 90 rotation of the positive ray of the a-axis into the positive y ray advances a right-handed screw along the z-axis, and similarly with the other axes, by cyclical interchange in x, y, z order. The positive directions of these three axes can be repre- sented by the thumb, first finger, and second finger of the right hand. To determine the coordi- nates of any point P in space, planes are drawn or conceived through the point parallel to the coordinate planes ; the dis- tances OL, OM, and ON cut off on the axes are given with proper sign as the coordinates (x, y, z) of the point P. Space is divided by the coordinate planes into eight divi- sions, called octants. The signs of the coordinates of any point within an octant are given in xyz order to distinguish the octants. Thus the + octant is at the right, below, and back. To every point in space corresponds one set of coordinates and -only one, and, conversely, to every set of three numbers corresponds one and only one point in space. When a point is given by its coordinates, the position is determined on the diagram by passing a plane through the a/-axis at the x of the point, parallel to the t/z-plane ; on the intersection of this plane with the ccy-plane indicate the y coordinate. The third coordinate must be represented in perspective, and the direc- O(0, 0, 0); L(3.5, 0, 0); M (0,3,0); N(0, 0, 2) ; CK3.5, 0, 2) ; R(3.5, 8, 0) ; 5^0,3,2); />(3.5, 3, 2) SOLID GEOMETRY: POINTS AND LINES 455 tion and length of these units in perspective are taken paral- lel and equal to the units represented on the third axis. Drawing by L. Makielskl. An artist's conception of a rectangular solid in space 3. Fundamental propositions of solid geometry. The follow- ing propositions of solid geometry have constant application in our further work. The student would do well to review these propositions in any elementary work on solid geometry and further to verify the reasonableness of these propositions on our figures. a. Two planes intersect in a straight line. 6. If a line is perpendicular to each of two intersecting lines, it is perpendicular to the plane of the two lines, i.e. it is perpendicular to every line in the plane of the two given lines. Thus, PQ on the diagram below is perpendicular to QL, to QM, to QD, and to every line in the a;2-plane which passes through Q. A line in the #2-plane which does not pass through Q does not intersect PQ, but the angle which it makes with PQ is defined as the angle which any parallel to it which does intersect PQ makes with PQ. This will then be a right angle. c. The angles between two pairs of parallel lines are equal or supplementary. d. If two planes are perpendicular to a third, their inter- section line is perpendicular to the third. 456 UNIFIED MATHEMATICS e. The dihedral angle between two planes is measured by the plane angle formed by two lines, one in each plane, both' perpendicular to the edge of the dihedral angle. /. If a line (PQ) is perpen- The dihedral angle, between two planes, is measured by the angle between two lines dicular to a plane (xz) and from the foot of the perpendicular a second perpendicular (QL or QD) is drawn to any line (OX or NL] in the plane (zx), then the line connecting any point (P) on the first perpendicular to the intersection point (L or Z>) is perpendicular to the line (QX or NL) in the plane. NOTE. We will refer to these propositions as 3 a, 36, 3 c, 3d, 3e, and 3/. 4. Vectorial repre- sentation. For some purposes it is convenient to think of three numbers (x, y, z) as representing the vector from the origin to the point. The length OP of the vector is called r, and the vectorial Point P(x, y, z) or P(r, SOLID GEOMETRY: POINTS AND LINES 457 angles which this vector makes with the x-, y-, and z-axes are termed direction angles and are represented by the letters, , ft, and y, respectively. By proposition 3 /, the triangles PLO, PQO, and PNO are right triangles. Hence x = r cos a = r cos z = r cos . Further, the square on the diagonal OP( = r) of our rectangu- lar prism is the sum of OQ 2 and QP 2 , and OQ 2 = ON* + NQ>, whence r 2 r 2 cos 2 a + r 2 cos 2 ft + y 2 cos 2 y, or cos 2 + cos 2 /3 + cos 2 y = 1. Evidently, also, r 2 = z 2 + y 2 + z 2 . 5. Parametric equations of a line. The equations x = r cos a, y = r cos /?, z = r cos y, when , ft, and y are fixed and r is a variable parameter, serve as the equations of the straight line OP. Cos a, cos ft, and cos y are called direction cosines of this line. Evidently any point E(a, b, c) on the line OP produced in either direction satisfies this relationship, if r is taken as the distance from to E. If E is on the other side of the origin from P, then the direction angles of the vector OE are supple- mentary to a, ft, and y and have cosines opposite in sign to cos a, cos ft, cos y. In this case r is taken as negative and it is evident that with this interpretation the coordinates of the point E satisfy the given equations. A line which does not pass through has the same direction cosines as the line parallel to it through 0, positive directions on both being the same. If such a line passes through P(XI, ?/!, z x ) in space, the corresponding parametric equations are x Xi = r cos a, y y\ = r cos ft, z z l = r cos y. 458 I M VIED MATHEMATICS In plane analytics the corresponding formulas for straight lines are x = r cos a, y = r cos /?, where cos /8 = sin a x Xi = r cos a, y y l = r cos ft. The student should check these formulas and show their rela- tions to the ordinary equations used. 6. Distance formulas and spheres. is the square of the distance from (x, y, z) to (0, 0, 0). a;2 + yZ _|_ 2 2 _ fi is an equation which is satisfied by every point on a sphere. m 5= = E The sphere : (x - h) 2 + (y - fc) 2 + (z - I)- = r- r = is the distance between two points (x l9 y lf z^) and (x*,, y.,, z 2 ). represents the square of the distance from (x, y, 2) to (h, k, I) and hence, (^ _ 7^2 +(y- fc)2 + ( 2 _ Z) 2 = r 2 is the equation of a sphere whose center is (h, k, /) and whose radius is r. SOLID GEOMETRY: POINTS AND LINES 459 7. Point of division. , _ <>; J/2 - J/3 - Z3 Step for step, and letter for letter, the proof follows that given for the point of division in a plane ; the only change is that the z- term is added. Thus, since / = 1, it follows that = } e tc. PROBLEMS 1. "\Vhat does the equation x = 3 or x 3 = represent on a line, the or-axis, i.e. when you are considering points on a line ? What does this equation represent in plane analytics ? What does this equation represent in space analytics ? 2. What does the equation x 2 = 9 represent in one dimen- 'sioual analytics? How are these two points located with reference to the origin ? What does the equation x 2 + y 2 = 9 460 UNIFIED MATHEMATICS represent in the ccy-plane ? How are the points, which lie on this locus, located with reference to the origin '.' What does the equation x 2 + y 2 + z 2 = 9 probably represent in xyz- coordinates ? 3. Where do all lines lie which make an angle of +30 with the o?-axis ? Where do all lines lie which make an angle of + 70 with the ?/-axis ? Where does a line lie which has the angle a equal to 30 and the angle ft equal to 70 ? Deter- mine the angle y, using the relation cos 2 + cos 2 ft + cos 2 y = 1. Given that a = 30 and ft = 60, what is the value of y ? If = 30, ft = 45, what is the value of y ? 4. Write the equations, in parametric form, of a line through the origin which has the direction angles a = 30, ft = 70, and the third angle as determined in the preceding problem. Write the equations when these angles are 30, 60, and 90. 5. Write the equations of lines parallel to the two lines of the preceding problem and passing through the point (-2,3, -7). 6. Write the equation of the sphere whose radius is 10 and whose center is the point (2, 3, 4). Find three other points on this sphere. 7. Find the coordinates of the points of trisection of the line joining A(2, 3, - 7) to B(2, - 3, - 4). If the line AB is extended through B by its own length, what are the coordi- nates of the point so determined ? 8. Given that the parametric equations of a line are x = 3r, y = 2r^ z = or, find 10 points upon the line by giving to r values from 4 to + 5. Determine the direction cosines of this line. Determine from your trigonometric tables the angles , ft, and y. 9. Given the parametric equations of a line x 3 = 3r, y + 5 = 2r, and z 7 = 5 r, find 10 points on the locus ; determine the direction cosines ' and the angles a, ft, and y. SOLID GEOMETRY: POINTS AND LINES 461 8. Angle between two lines. We have had occasion to note that the angle between two non-intersecting, or skew, lines in space is defined as the angle be- tween two intersecting lines which are respectively parallel to these given lines. For convenience these parallels may be taken through 0. The angle between two lines having direction cosines (i, fa, y t ) and (2, /8s, 72) respectively is ob- tained as follows : Take any two points P t and The angle between two lines in space P 2 , one on each line, having vector distances r t and r 2 . Evidently I\P?= r? -f r 2 2 - 2 ?v- 2 cos 0. But Equating the two values and canceling, noting that ri 2 = Xl * + y? + zf, and rf = xj + y? + z 2 2 , this gives cos 1 , yi = 1\ cos fa, Zi = = r 2 cos 2 , 2/2 = r 2 cos Pz> z i r z cos 72- Now and Substituting, cos cos ! cos 0-2 + cos Pi cos fa + cos y x cos y 2 , a relation which is independent, as it must be, of the particular points P l and P 2 chosen. The corresponding formula in the plane for the angle be- tween two lines is cos = cos ! cos a% + sin o^ sin 462 UNIFIED MATHEMATICS The adaptation and the proof are left to the student as an exercise. Compare with formula on page 247. ' Frequently the direction cosines of a line in space are repre- sented by I, in, n or 1 1} m^ n l7 etc. The condition that two lines be perpendicular is evidently ?i/ 2 + m l m<, + ?* 1 w 2 = 0, and the condition for parallel lines is that I, = 1 2 , m l = ra 2 , and n t = n 2 . It can be shown by using vectors that the relation . /!/ 2 4- wiiWo + i2 = 1> combined with If + mf + n^ = 1 and 1 2 - + w 2 2 + n 2 2 = 1 reduces to l t = /,, ra t = ra 2 , and % = n. 9. First- degree equation. We now show that the equation (a) Ax + By+Cz + D = Q represents a plane. A plane is, by definition, a surface which is such that the straight line joining any two points in the surface lies wholly on the surface, i.e. any other point on the line is also on the surface. Let PI(X I} y } , Zx) and P 2 (x 2 , y^, z 2 ) be any two points which satisfy equation (a). (6) .:Ax l + Byi+Cz l +D = 0. (c) ^la: 2 + By* + Cz + D = 0. Let P 3 (%3, 2/3, z 3 ) be any other point on the line joining P(XI, yi, %i) to P 2 (a- 2 , y 2 , z 2 ) and let this point divide P : P 2 into P P Jc segments such that ~ - = PS PZ ""2 Then, # 3 , y 3 , and z 3 may be written _ 3 ~~ Substituting these values in the left-hand member of equa- tion (a), we have [ B i z | fti -f- fc 2 KI -\- K% K l -|- ft" 2 which may be written, by rearrangement of terms, -A- (Ax, + By, + &!+ D) + r^y- (^ 2 + By, + Cz 2 +D) ; SOLID GEOMETRY: POINTS AND LINES 463 but this expression, by (&) and (c) above, is zero ; hence the point P 3 (x 3 , 1/3, z 3 ), any point on the straight line joining PI to P 2 > which points are on the locus of (a), is also on this locus. .-. Ax + By + Cz + D = represents a plane. The converse proposition is demonstrated in section 5 of the next chapter. A plane to pass through three given points is determined by substitution of the three points in equation (a) and solving for three of the constants, e.g. A, B, and C, in terms of the fourth. If the fourth constant chosen happens to be zero, another selection must be made. Illustrative problem. Find the plane through (4, 4, 4), (3, 0, - 5), and (0, - 3, - 7). Let Ax + By + Cz + D = represents the equation of the plane. Substituting, x 3 (1) 4^l + 4B + 4C + .D = 0. (2) 3 A - 5 C + D = 0. x 4 (8) -3#-7C+D = 0. Since it happens that only A, C, and D occur in (2), eliminate B between (1) and (3) by multiplying (1) by 3 and (3) by 4 and adding, obtaining (4) 12 A - 16 C + 1 D = 0. -4(2) 3A- 5 C+ J> = 0. Eliminate A between (4) and (2) by multiplying (3) by 4 and add- ing to (4). This gives (5) 4 C + 3 D = ; C = - fD. Substituting in (2) the value of C found gives 3 A + * D + D = 0, A = |f D. Substituting value of C in (3) gives The equation of the plane may be written, - }f Dx+ } { Dy - J Dz + D = 0, or 19x + 26y 9z + 12 = 0. Ans. 464 UNIFIED MATHEMATICS PROBLEMS 1. Find the angle between the two lines of problem 5 of the preceding list of problems. 2. Find the direction cosines proportional to 3, 2, and 5 ; find those proportional to 2, 3, and 4 ; find the angle between two lines having these direction cosines that you have found. 3. Find the equation of a plane whose intercepts on the axes are, respectively, 3, 5, and 7. 4. Find the equation of a plane through the points (3, 0, 5), (-2, 11, 7), and (0, 11, 7). 5. The parametric equations of any line through (3, 0, 5) can be written x 3 = r cos a, y = r cos /3, z 5 = r cos y. If this line is to pass through (2, 11, 7), these coordinates must satisfy these equations. Make the substitutions, re- spectively; square and add the corresponding numbers and thus obtain the value of r, giving the distance of the point (3, 0, 5) from (2, 11, 7). Find then the values of cos a, cos (3, and cos y. 6. If a rectangular box with sides parallel to the coordinate planes has the line joining (3, 0, 5) to ( 2, 11, 7) as a principal diagonal, find the lengths of the sides, the length of the diagonal, and so find the direction cosines of the line joining the two points. 7. Discuss the loci of the following equations and find three points on each locus : a. 3 x + 11 = 0. b. z 2 -9 = 0. c. z-5 = 0. d. x y 5 = 0. e. z 8. Where do points lie which are common to the loci of the two following equations : x = 3, and y = 5 ? 3-3 = and 2-2y + 10 = 0? CHAPTER XXXI SOLID ANALYTICS; FIRST-DEGREE EQUATIONS AND EQUATIONS IN TWO VARIABLES 1. Locus of an equation in three variables. Any equation involving three variables has for its locus a surface which may, in special forms of the equation, reduce to one or more lines or points. We obtain points on such a surface by giving values to two coordinates, e.g., x and y, and solving for the third, e.g., z. Thus we have found that any first-degree equa- tion represents a plane. NOTE, x 2 + y 2 + 2 = represents only a point (0, 0, 0), or a point sphere. x 2 + y 2 = represents the z-axis since everywhere on this axis x = and y = 0. 2. Intersections of loci. (See Chapter V, Section 2.) Any point which satisfies two equations involving three variables lies, in general, upon a curve which is common to the two surfaces represented. When three equations are regarded as simultaneous, points of intersection of the three surfaces are obtained. Under special relations between the three given equations, these points may lie upon a line, but, in general, three simultaneous rational integral algebraic equations determine a finite num- ber of points of intersection. Just as a family of lines through the intersection of two given lines is obtained in Chapter V, Section 4, in the form lj_ -f kL = 0, so the equation f^x, y, z) + kf 2 (x, y, z) = 465 466 UNIFIED MATHEMATICS represents a family of surfaces which pass through the inter- section curves of the two given surfaces. Thus, z 2 + j/ 2 + z 2 = 25 represents a sphere of radius 5 ; z 2 = 9 repre- sents two planes parallel to the yz-plane ; the equation X 2 + yz + Z 2 _ 25 - fc(z 2 - 9) = represents for all values of A; a surface through the intersections of the sphere and the plane. For k = 1, this surface reduces to a cylinder, 3,2 Z 2 _ 16 = 0. 3. Cylindrical surfaces. Any equation in two variables, as x and y, represents in space a cylinder whose axis is parallel to the axis designated by the third variable. Cylindrical surfaces: ELLIPTIC HYPERBOLIC PARABOLIC The curves indicated on these surfaces are cubic space curves. If an equation f(x, y) = is given in x and y, any point ( x i> 2/i) which satisfies the equation will lie upon the curve in the xy-pla,ne given by f(x, y)= 0. Considered as a point in space, the point (x 1} y 1} 0) satisfies the equation, and further it is evident that (aj l5 y 1} z), irrespective of the value of z, will also satisfy f(x, y) = 0, since the z-coordinate does not enter at all. SOLID ANALYTICS 467 Thus, (3, 4, 0) satisfies the equation x 2 + y 2 25 = and also the points (3, 4, 1) or (3, 4, 10) or (3, 4, 8,) will satisfy the equation /^2 l i*2 y^ (i But all points (xi, y 1} z), for varying values of z only, lie on a parallel to the z-axis through (x t , y^ and hence all points on the surface generated by a straight line moving parallel to the z-axis and touching the curve f(x, y) = 0, in the ary-plane, lie upon a cylinder. The curve f(x, y} = is called the directrix of the cylinder and the moving line is called the generator or element of the cylindrical surface. Similarly, when an equa- tion is given in x and z or in y and z, a cylinder is represented. A plane given by a first-degree equation in two variables, or one variable, is a special case of the preceding. 4. Straight line as the intersection of two planes. Just as the equation ^ = - represents in the plane the straight line joining PJ(XI, .Vi) to P 2 (x z , yz)> so the three equations, x Xi y yi z Zi represent in space the straight line joining There are three equal- ities which are ob- tained by leaving out in turn each of the fractions, but there are only two inde- pendent equations, as v Line joining PI to P- in space 468 UNIFIED MATHEMATICS the third equality would follow always from the first two which were given. These formulas can be obtained directly from the properties of similar triangles, or from the parametric equations of section 5 of the preceding chapter. The latter method brings out the important fact that the values x 2 x^ y z y^ and z 2 Zi are proportional to the direction cosines of the given line, and the values themselves of these cosines can be obtained, using the fact that the sum of the three squares is equal to unity. The derivation of the theorems mentioned is left as an exercise to the student. The parametric forms of the equations of a straight line may be written, fl-fli _ y_-J/i _ z-z l _ r cos a cos /8 cos y or \ a; a?! _ y y\ _ z z^ _r k cos a k cos (3 k cos y k Further, any equations which can be put in one of the two forms above represent a straight line, and the denominators of the fractions are proportional to the direction cosines of the line. The equations of the straight line in the form x - Xi _ y yi _ z Zi a b c wherein a, 6, and c are necessarily proportional to the direction cosines of the line, are called the standard or symmetrical equations of the line. In general, any curve in space is given as the intersection of two surfaces by the equations of the two surfaces. In particu- lar, the straight line is given by the equations of any two planes which pass through the line. Of the infinite number of planes, the pencil of planes, which pass through a given line, the three planes, called projecting planes of the line, which are parallel to the coordinate axes are of particular SOLID ANALYTICS 469 importance. These equations will evidently be first-degree equations in two variables. In the standard form the equality of any two members gives one of the projecting planes through the given line. Illustrative problem. Find the direction cosines of the straight line determined by the two planes (a) x + y -3z- 5 = 0. (6) 3 x - y - 5 z - 11 = 0. Find the projecting planes parallel to the coordinate axes (or .perpen- dicular to the coordinate planes) ; find the points where this line pierces the coordinate planes. Any plane through the line of intersection is given by (c) x + y-3z-5 + fc(3x-y-5z-ll)=0. Giving to k the value f, which is equivalent to multiplying (a) by 5, and (6) by 3 and adding, and simplifying you have, 5 x + 5 y - 15 z - 25 - 9 x + 3 y + 15 z + 33 = 0, or (d) 4x + 8j/ + 8=0, as the plane of projection on the xy-plane. Eliminating y, k 1, gives, ( e ) 4 x 8 z 1C = or x 2z 4 = 0, the plane of projection on the xz-plane. Eliminating x, k = , gives, (/) _4(/ + 4z-f4 = 0, which might have been obtained from (d) and (e), thepiane of projection on the yz-plane. Solving for x, in (d) and (e), x = 2(y + 1) and x = 2(z + 2). x = 2(j/+ l)=2(z + 2). x-0_y+ l_z + 2 2 1 1 The denominators 2, 1, and 1 are proportional to the direction cosines of this line. Hence cos = 2 m, cos /3 = 1 m and cos 7 = 1 m, giving, cos 2 a + cos 2 /3 + cos 2 -y = 4 m 2 + m z -f- m 2 = 1; 6m 2 =l; m=; - . V6 Either sign may be taken, but for convenience, make cos positive. oil cos a = ^ , cos p = ^ , cos y = ^ , the direction cosines. V6 V6 V6 470 UNIFIED MATHEMATICS To find where this line pierces any coordinate plane, as z = 0, solve the equation of the coordinate plane as simultaneous with the two given planes which determine the line. This gives here (4, 1,0) as the piercing point with the xy-plane. Similarly we find the intersection with any plane. A parallel line to our given line through a given point would be de- termined by two planes through the given point parallel to the two given planes which determine the line. Why ? Determine the parallel line through (1, 5, 6). 5. Normal form of the equation of a plane. (See Section 3, Chapter IX.) In the plane, the equation x cos a+y sin ap=Q, which may be written x cos a + y cosftp = 0, represents the equation of a straight line in normal form, which line is such that the perpendicular from the origin upon it has the length p and makes the angles a and ft with #-axis and y-axis. Similarly, in space, the equation x cos + y cos ft + z cos y p = represents a plane which is such that the perpendicular from the origin upon it has the length p and makes the angles , ft, and y with the ic-axis, ?/-axis, and 2-axis respectively. Evidently, if a plane is given and a perpendicular ON of length p, having direction cosines a, ft, and y, is dropt from the origin to this plane, the point Nis (p cos a, p cos ft, p cos y) and the extension of the per- ON of length p; ON' of length 2p Direction angles of ONN' : a, P, -y pendicular by the length p gives the point N' (2 p cos a, 2p cos ft, 2p cos y). Any point P(x, y, z) which is equidis- tant from 0(0, 0, 0) and N'(2pcosa, 2 p cos ft, 2 p cosy) SOLID ANALYTICS 471 lies on our plane. Writing and equating these distances, we have, a 2 + y 1 + z 2 = (x - 2 p cos )2 + (y - 2 p cos /3) 2 + (z - 2 p cos /3) 2 . Whence, cos )# + (4p cos /3)i/ + (4p cos y)z = 4 J9 2 (COS 2 + COS 2 ft + COS 2 y), giving finally x cos a + y cos /8 -f- z cos y p = as the equation. In the plane the distance from any point (a^, y^ to a line is obtained by writing the equation of the line in normal form and substituting therein for x and y, x t and y v In space the dis- tance of a point (x 1} y 1} Zj) from a plane is obtained by writing the equation of the plane in normal form and substituting therein these coordinates of the point for x, y, and z, respectively. To reduce a linear equation to normal form, you divide the equation through, after transposing all terms to the left-hand member, by the square root of the sum of the squares of the coefficients of x, y, and z, choosing the sign opposite to the sign of the constant term. The proof is not similar to the proof of the corresponding theorem in plane analytics. Parallel planes are represented by linear equations having the corresponding coefficients, of x, y, and z, equal or propor- tional. PROBLEMS 1. Put the following equations in normal form and deter- mine the distance of each plane from the origin : a. 2x-3y+4:Z-ll=0. 6. a;-f-y + z-5 = 0. c. 2 x - 3 y 11 = 0. d. z-7 = Q. Determine the direction cosines and the direction angles of the normals to each of the above planes. 472 - UNIFIED MATHEMATICS 2. Find the equations in standard form of a line from (1, 2, 5) perpendicular to the first plane in problem 1 ; through (0, 0, 0) perpendicular to the second plane in problem 1; through (2, 3, 4) perpendicular to the third plane in problem 1. Determine in each of these three problems the intersection of the perpendicular with the plane. 3. Find the piercing points with the coordinate planes of the following lines : a. 2 x 3 y + 4 2 11 = and x y + z 5 = 0. b. 2x 3y+4z-ll = 0andz 7 = 0. c. 2z-3y- 11 = and z-7 = 0. 4. Put the three lines of problem 3 in standard form. Note that in the second and third cases, since the given line lies in a plane parallel to the x?/-plane, the line makes an angle of 90 with the z-axis, i.e. cos y = 0. The equations of the sec- ond and third lines in standard form would have a zero de- nominator, and so it is better to put these equations in the form given in the third of these problems. The values of cos a and cos ft are determined here from the equation 2 x 3y 11 = 0, 2 3 giving cos a = and cos ft = V13 V13 5. Find the equations of the straight lines through the two points : a . ^, 5, - 2) and (0, 0, 7). 6. (3, 5, - 2) and (0, 0, 0). c. (3, 5, - 2) and (- 3, 5, +2). 6. Find the angle between the lines d *-2_y_z-l ' ~~ 7. Do the two lines in problem 6 intersect ? How can you determine whether any two given lines intersect ? Note that the problem is entirely analogous to the problem in plane analytics as to whether three given lines intersect, and is solved in the same manner. Write the equations of two intersecting lines. SOLID ANALYTICS 473 8. Determine the curve in which the sphere x* + f- + z 2 400 = is intersected by the plane y 9 = 0. Note that substituting y = 9 is equivalent to writing Sphere cut by a plane which gives, of course, a new surface passing through the intersection curve of the first two surfaces. 9. Find the intersection of the sphere x 2 + y 2 + z z 100 = and the cylinder x 2 + y 2 36 = 0. 10. Upon what cylinder, parallel to one of the coordinate axes, does the intersection of the plane x = 5 with the surface ar 2 + 4 y 2 = 25 z lie ? 11. Find the intersection of the line with the sphere x 2 + y- + z 2 100 = by substituting these values in the equation of the sphere and solving for r. Note that since the right-hand coefficients, 2, 3, and 5, are not the 474 UNIFIED MATHEMATICS direction cosines of this line, but only proportional to them, the values of r obtained are not the distances from (1, 2, 3) to the points of intersection with the sphere, but are propor- tional to these distances. The points of intersection are obtained by substituting the values of r found back in the equations of the line and solving for (x, y, z). Hyperboloid of one sheet Hyperboioid of two sheets CHAPTER XXXII SOLID ANALYTICS: QUADRIC SURFACES 1. General equation. In plane coordinates, any equation of the second degree represents a conic section. Similarly, in space coordinates, any equation of the second degree repre- sents a quadric surface. The types of quadric surfaces, limited in number, are closely allied to the types of conic sec- tions. In plane analytics, it is shown that the general equation of the second degree, containing the " cross-term " xy, intro- duces no new curves, only the same curves, represented by the different types of equations in which no xy-term appears. It is likewise true in space that the general equation contain- ing any or all of the " cross-terms," yz, xz, and xy, presents no surfaces different from those which may be represented by the general equation containing no cross-term. Methods of transformation of coordinates quite similar to those discussed in Chapter XXIV apply to space coordinates, but the limitations of a first course, preclude any discussion of the methods and results. Any surface given by an equation of the second degree is 475 476 UNIFIED MATHEMATICS cut by any plane in some form (including, of course, limiting forms) of conic section. The coordinate planes very evidently cut any quadric surface in a conic, since the curve of inter- section in the coordinate plane is given by an equation of the second degree in the two variables of that plane. The trans- formations mentioned above are desirable for the general proof, but another method is indicated below. SPHERE Ellipsoids : PROLATE OBLATE . GENERAL SPHEROID SPHEROID ELLIPSOID 2. Ellipsoids. The equation of a sphere has been given as An ellipsoid is given by the equation a 2 6 2 This surface is related to the three spheres, (X - _ 7)2 = very much as the ellipse is related to its auxiliary circles. The parametric equations of the above ellipsoid are x h = a cos a, y Jc = b cos (3, z I = c cos y. SOLID ANALYTICS: QUADRIC SURFACES 477 The elimination of a, (3, and y, employing cos 2 a + cos 2 ft + cos 2 y = 1, gives the equation of the ellipsoid in the standard form above. The quantities a, b, and c represent the semi-axes of the ellipsoid. If two of these denominators are equal to each Ellipsoid of revolution, with x-axis as axis of revolution other, the ellipsoid is an ellipsoid of revolution about an axis parallel to the axis corresponding to the term with the odd denominator. Thus, 25 16 16 is an ellipsoid of revolution, obtained by revolving the curve -- (- = 1 about the x-axis. The derivation of the formula of the ellipsoid of revolution is as follows, PN* + NM 2 =PM*, but QM=PM, radii of the circle QPR about M, with lettering as indicated on diagram given above. Now for all points on this circle the ^-coordinate is the same, 25 which is a relation true for every point on the circle obtained 478 UNIFIED MATHEMATICS by rotating the point Q on the ellipse - + ^ = 1 about its 25 16 axis. But Q is a,ny point on the ellipse, and hence P may be any point on the surface obtained by revolving the ellipse about its axis. Hence, for every point on this surface, ,* + ** = 16(1- or *! 4-^.4.11 = 1 25 16 16 Any ellipsoid of revolution obtained by revolving an ellipse about its major axis is called a prolate spheroid, and is shaped like a football ; an oblate spheroid is obtained by rotating an ellipse about its minor axis, and is shaped like a circular cushion or the earth. PROBLEMS 1. Find the equation of the Iphere having the center at the origin and passing through the point (2, 5, 6). Give the seven points which lie on this sphere and are symmetrically situated to the given point with respect to the coordinate planes. 2. Find the, equation of the preceding sphere if the center is at (3, 2, 12). Find by using conditions of symmetry with respect to planes through the center parallel to the coordinate planes seven further points on this sphere. 3. Write the equation of the ellipsoid having the center at the origin and semi-axes equal to 2, 3, and 5 respectively (x, y, and z order). Find three points on this ellipsoid. Write the equations of three circles which lie on this surface. Write the equations of the traces on the coordinate planes, i.e. the inter- sections with these planes. Draw the graph. 4. If a football is 10 inches long with a diameter of 8 inches, write the equation of the surface, assuming it to be an ellipsoid. Draw the graph to scale. SOLID ANALYTICS: QUADRIC SURFACES 479 5. Assuming that an air cushion 18 inches in diameter and 6 inches high is an ellipsoid, write the equation of the surface. Draw the graph to scale. 6. Find the six principal foci of the ellipsoid in problem 3. These are the foci of the traces on the coordinate planes. 3. Hyperboloids. By rotating the hyperbola about either axis, a hyperboloid of revolution is obtained. Rotation about the principal axis, the #-axis here, gives a surface of two separated parts, called a hyperboloid of revolution of two sheets. The equation is, _ a 2 6 2 6 2 ~ The method of derivation, which we outline, is general, and being applied to the surface obtained by revolving any curve, y =/( a O about the cc-axis, will give the equation of the surface in the form y 2 + z 2 = [/(z)] 2 . . aj2 yZ Given -- 7- = 1> revolved about the -axis. a 2 o 2 Hyperboloid of two sheets, of revolution The curves are slightly distorted. 480 UNIFIED MATHEMATICS Any point P(x, y, z) on this surface is obtained by the rotation of a point Q(x, y, 0) about the x-axis. The point Q generates a circle in a plane parallel to the yz- plane, in which x has everywhere the value given by OM. The equation of this circle is Hyperboloid of two sheets y-axis as principal axis. This radius r is evidently a function of x, being defined by the original equation given ; hence, r z = (ordinate on the hy- perbola) 2 = b z ( - 1 V V 2 / Hence the point P(x, y, z) satisfies the equation or, by rearrangement of terms, the hyperboloid of revolu- tion of tico sheets. Note that precisely this surface would have been obtained by revolving a; 2 - s= 1 in the orz-pla.ne about the x-axis. a 2 6 2 Similarly, the hyperboloid of revolution of one sheet is obtained by revolving a hyperbola about its conjugate axis. The pre- ceding hyperbola revolved about the y-axis gives SOLID ANALYTICS: QUADRIC SURFACES 481 __ a 2 & 2 a 2 The student will note that the axis of rotation in each case is given by the odd term. Corresponding to these surfaces of revolution are the general hyperboloids, a 2 6 2 hyperboloid of two sheets, and ^_^ 2 + ?_ 2 = l a 2 6 2 c 2 hyperboloid of one sheet, which represent in each case a surface having a principal axis parallel to the axis of the odd term, e.g. the first has the cc-axis as principal axis and the second has the y-axis as principal axis. Chang- ing the principal axis to another of the coordinate axes inter- changes two of the algebraic signs in the equation. 4. Paraboloids. By revolving the parabola y z = 4 ax about its axis the surface \f + z"- = 4 ax is obtained. This is called a paraboloid of revolution, or a circular parabo- loid, and is the type of surface which is fundamental in theater and auditorium construction. The derivation of the equation is left as an exercise for the student. The elliptic paraboloid is given by the equation Hyperboloid of revolution, one sheet y-axis as principal axis. and sections parallel to the icy-plane are ellipses. 482 UNIFIED MATHEMATICS Elliptic paraboloid Hyperbolic paraboloid ; --5^ The corresponding " IL " standard forms with J _ - jj II -5-"--" A '4-V, / ::::! I 1 ,- 1 - - rrk *<:: --] -- respectively, as prmci- V s:' A~ i * / / ^ ---- + V^ ~T* S ^=k z 1 ; r 2 z 2 ^r 4 - ::/:: -, + -^ = 4 .V> \ - - - - tt 2 C 2 / \ / \ c: oo :: ... : s C* : :: _ t- i J_IT / - [- / 3 7 t ; t y " ^Z Q % j :::jL:: a 2 6 2 S ' r^" ' 'M# JTT^JBJI: gives the most compli- cated of the quadratic E nip tic p ar al ill f ::::::::::i joloid of revc surfaces, the hyperbolic paraboloid, a saddle- lution shaped surface which SOLID ANALYTICS: QUADRIC SURFACES 483 is here represented by a photograph of a model of such a surface. The hyperbolic paraboloid may be generated by the motion of a given parabola, x 2 4 a 2 z = 0, moving parallel to the xz- plane and having its vertex moving on the parabola 5. Cones. If any straight line is revolved about another straight line in its plane as an axis, a cone of revolution is generated. Limit- ing forms are the cylinder, when the re vol viiig line is par- allel to the axis, a plane when the re- volving line is per- pendicular to the axis, and a straight line when the revolv- ing line coincides Cone generated by a straight line rotating about an axis in the same plane with its axis. Let v = -x revolve about OX. The cone of revolution geu- crated has the equation The cone is itself a limiting form of the hyperboloids, as will be noted below. The general equation of the cone, whose axis is the y-axis and whose vertex is the origin, is 6. Conic sections. The method which we will here outline to prove that every plane section of a cone may be given by 484 UNIFIED MATHEMATICS Conic sections : ELLIPSE I'AKA HULA HTFKKBOLJ an equation of the second degree applies to a cone having an ellipse, parabola, or hyperbola as a base as well as to the cir- cular base, which is taken for convenience. '*" '/" Z~ Given the cone ^ = 0. a- IP b~ Evidently, any one of the planes given by x = k cuts this |||||i||ii^iiNiiiiiiiiiiiiiiiimtimimuu< ~3J^ --it} - a\mx + fc) 2 a 2 z 2 = 0. Evidently, PM = z = z', since PM is drawn in one of two per- pendicular planes, and perpendicular to the other. The elliptic section depicted in its own plane Further, BM=yf = stituting, we have, 6V 2 1 -f-m 2 - a 2 m 2 )z' 2 - a 2 z ' -|- m 2 , whence x = 2 a*km x / Sub- m 2 Vl + m 2 m 2 )- x - a =fc 2 - aV 2 = 0, = 0. 486 UNIFIED MATHEMATICS But this is an equation of the second degree in x' and z'. Fur- ther, the coefficient of x' 2 is (b z a 2 m 2 ) and of z' 2 is a 2 (l-f m 2 ); hence the curve is an ellipse if m 2 > , a parabola if m 2 = , 7i2 ^ ^ and a hyperbola if ra 2 < a 2 When the cutting plane has the form y = mx + nz -f k, the proof is more complicated but not essentially different. Every section of a cone may be represented by an equation of the second degree in two variables. Ellipse, parabola, hyperbola, and two straight lines as intersections of a cone by a plane PROBLEMS 1. Name and discuss the following surfaces : a. x*+ y*+ z 2 100 =0. b. a 2 + 2?/ 2 + 3z 2 -100 =0. c. a 2 + 2y 2 -4z 2 -100 =0. d. x* + 2y* -100 =0. e. x 2 + 2y*-z* =0. /. x 2 -2?/ 2 -4z 2 -100 =0. g. x 2 -100 =0. h. x*-2 2 SOLID ANALYTICS: QUADRIC SURFACES 487 i. o; 2 -2.y 2 . =0. j. x- + 2y 2 =0. k. z 2 + 2 1/ 2 + 4 z 2 =0. I z 2 + 2y 2 + 4z 2 + 100 =0. 2. How would the addition of a term, 10 x, affect the locus of each of the preceding twelve expressions ? Discuss the change in each locus produced by changing the sign of x 2 in each expression from + to . 3. Find the equation of the cone obtained by revolving the line in the a^/-plane y = 4 x 10 about the o>-axis ; about the y-axis ; about the z-axis. 4. Find the equation of the paraboloid obtained by revolv- ing the parabola z 2 = 8 x about the #-axis ; find the equation of the surface obtained by revolving this surface about the z-axis. Why has the latter surface not received particular discussion ? 5. Find the equation of the cone obtained by revolving the line y = 4 x 10 about the line y = 6. Note that this differs from the problems which we have considered in the text only by a change of origin, or a transformation of the type 6. Find the locus of a point which is equidistant from the point (4, 0, 0) and from the plane x + 4 = 0. What is the surface ? 7. Find the locus of a point the sum of whose distances from the points (4, 0, 0) and ( 4, 0, ) is constant and equal to 10. What is the surface ? 8. Find the locus of a point the difference of whose distances from two points (4, 0, 0) and ( 4, 0, 0) is constant and equal to 6. 9. How would you find in space coordinates the distance from a point to a line ? Apply to finding the distance from PAJ. i.u v x 2 y 3 z 8 (1, 3, 5) to the line = - = . 488 UNIFIED MATHEMATICS 7. Limiting forms. The limiting forms corresponding to the ellipsoid are given by equations of the type 3.2 y2 2 2 (x-h? (y-ky + (inJI 2 = o, o 2 6 2 c 2 the first of which represents the point (0, 0, 0) and the second the point (h, Jc, I). The method of approaching this limit is best indicated by writing the equation as o2 + &2 + c2~ As k approaches 0, the semi-axes Vfca 2 , Vfcft 2 , and Vfcc 2 ap- proach as a limit. In a similar way the hyperboloid equations approach, as limits, the equations representing cones asymptotic to the given hyperboloids. The limiting forms of the paraboloids are equations in two variables, and reduce to two planes, or to cylinders. The limiting forms of equations in two variables represent- ing cylinders correspond, with proper and more or less evident changes, to the limiting forms of the corresponding equations in plane analytics. Thus, any equation of the second degree f(x, y, z) = 0, whether in one or two or three variables, of which the left- hand member can be factored into two real linear factors in the variables, represents two planes which constitute also a type of quadric surface. 8. Applications. The applications of the conic sections which have been given in plane analytics are strictly applica- tions of surfaces, or solids having these surfaces as boundaries. Thus, a bridge having a parabolic arch uses a solid having a parabolic cylinder as bounding surface. SOLID ANALYTICS: QUADRIC SURFACES 489 The equation of the paraboloid in the Hill Auditorium, with the foot as unit of length, is y 2 + z 2 = 70.02 x ; the skylight in the ceiling of the Hill Auditorium is bounded by an elliptical cylinder, 2>2 y2 + - = 1, dimensions in feet. 76 2 50 2 Any of the automobile reflectors are paraboloids, in general, of revolution. Hyperboloids are used as revolving cones in the manufacture of iron pipes ; these pipes are passed between two revolving cones whose axes are inclined at 90 to straighten the pipes. 9. Circular sections. Given the ellipsoid represented by a 2 the question arises as to what planes cut this surface in circular sections. The method which we have given above, under section 7, for determining the nature of the curve cut out of the cone by the plane y = mx + A; applies to this problem. In the ellipsoid above planes y = k cut the surface in ellipses. The develop- ment as given shows that any plane y = mx cuts this surface in a curve given by the intersection, also, of the surface, z 2 _ * ~ ~ ' and the given plane, or a curve, =1 a 2 (l + m 2 ) & 2 (1 + m 2 ) c 2 (?/ "~ /Hi' [ tJ ^^ and the line f* A as axes of z = I x= reference, both lying in the plane of the section. This curve can be written, c 2 (6 2 - U 2 m 2 )a;' 2 + a 2 6 2 (l + m 2 )z' 2 = aWtl + m 2 ). 490 UNIFIED MATHEMATICS Equating the coefficients of x' 2 and z' 2 gives ra = a 2 (c 2 - 6 2 ) 6x/a 2 -c 2 The two planes, y = _ ^ x, and all planes parallel to them, cut this surface in circular sections. For these to be real planes c must be intermediate in value between a and 6. If c is not intermediate between a and 6, then planes either of the form y = raz or y nx will make real circular sections. The method applies to elliptic cylinders, to elliptic cones, and to hyperboloids, as well as to the ellipsoid. A simpler method, assuming c as the intermediate value, is to cc 2 y 2 find in the ellipse \- ^- = 1 a diameter of length 2 c ; this Ct diameter with the z-axis determines the plane of a circular section. 10. Tangent planes and tangent lines. The formula Ax& + By$ + G(x + ajj) + F(y + 7/ t ) + C = 0, which gives the tangent to at the point PI(XI, yi) on the curve applies in space analytics, with the addition of the corresponding z 2 and z terms, to give the tangent plane to the quadric surface at a point PI(X I} y l} Zi) on the surface. The tangent plane at PI(X I} y 1} Zj) to the surface A& + By 2 +Cz 2 +2 Gx + 2'Fy + 2Ez + K=Q is given by the equation, + G(x SOLID ANALYTICS: QUADRIC SURFACES 491 When the point PI(X^ y l} 2 t ) is not on the quadric surface, this equation represents not the tangent plane but the polar plane of the point (x lf y ly Zj) with respect to the surface. For any point outside of the surface, tangent planes to the surface have their points of tangency situated upon a plane, the polar plane of the point PI(X I} yi, i). A more complete discussion of the polar plane would reveal many other points of similarity between the polar plane as related to its quadric surface and the polar line as related to its conic. The intersection of a tangent plane at a point PI(X I} y l} z^ on the surface with any other plane through P t gives a tan- gent line to the surface at P^ 11. Ruled surfaces. Generating lines. Any surface which can be generated by the motion of a straight line moving according to some law is termed a ruled surface. Evidently, by its method of generation, such a surface has straight-line elements, called rectilinear generators, which lie wholly upon the surface. Certain of the quadric surfaces are ruled surfaces. Evi- dently all the cylinders, the cones, and the pairs of planes belong in this class. The ellipsoid, being confined to a finite portion of space, does not have right-line elements lying wholly upon the surface ; nor do the elliptic paraboloid and the hyperboloid of two sheets have right-line elements. The hyperbolic paraboloid and the hyperboloid of one sheet do have rectilinear generators. We will find the equa- tion of the families of lines which lie wholly upon one of the surfaces in question ; the method will apply to the other ruled quadric surfaces. Any point upon the hyperboloid o;2 _ 172 z 2 a 2 6 2 c 2 ~ very evidently satisfies the equation *? _ = 1 _ ?- a 2 6 2 ~ c 2 ' 492 UNIFIED MATHEMATICS which may be written, This indicates that any point which satisfies the pair of linear equations / v \ -J.tfi-!\ a b \ c x + 1 = a b will satisfy the equation of our surface since it will make the product represented by the left-hand member of our equation, 18,17,16,15 14 13 12 11 10 9 8 7 6 5 4 3.2 19.20212223 24 25 26 27 28 293031.32.1 30,31,32 123 29,28 27 26 25 9 10 11,12,13,14 19 18 17,16,15 The right-line generators on this hyperboloid of revolution are formed by connecting corresponding points on two circular sections An elliptic section is also indicated. in the second form above, equal to the product of the factors, representing the right-hand member. But every point which SOLID ANALYTICS: QUADRIC SURFACES 493 satisfies the pair of equations for any given value of Tc lies upon a straight line, the intersection of the two planes given by the linear equations. Hence, every point upon this line lies upon the given surface for any value of k. It can be shown that no two lines of this family of lines, i.e. no two lines given by two values of k, intersect. Another family of lines also lies upon this surface. The equations of this second family of lines, with the parameter fc, are as follows : a b k Every member of this family of lines can be shown to inter- sect every member of the preceding family and no member of its own family. PROBLEMS 1. Find the equations of the rectilinear generators of the following surfaces : a. x 2 y 2 z 2 = 0. 6. x * + y 2 - z 2 =16. c. x 2 y z 4 z 0. d. x*-y* =0. 2. Find the circular sections of the following surfaces : 2 ,/2 ,.2 * il 6 -\ ft , __ jZ ^_ . _ I 25 16 ( .) b. a 2 25 y 2 16 z- 9 = 0. c. re 2 + 4 y2 = 9z. d. afl + 4y 2 = 9. e. 100 + 36~ 16 = 1. 494 UNIFIED MATHEMATICS 3. Write the equations of the tangent planes to each of the surfaces in the preceding problem at the point (a^, y lt z^) in each case upon the given surface. 4. In problem 1 a, above, take ki = 1 and fe 2 = 2 and show- that these two lines of the same family of rectilinear generators do not intersect. Write the second family of rectilinear gen- erators of the same surface and show that, taking k = 1 (any other value would do), this line does intersect a given line (&! = 1) of the first set. How could you make this proof general ? TABLES PAGE Constants with their logarithms 496 Squares and cubes of integers, 1 to 100 497 Four-place logarithms of numbers, 100 to 999 498 Four-place logarithms of numbers, 1000 to 1999 500 Four-place logarithms of sines and cosines, 10' intervals .... 502 Four-place logarithms of tangents and cotangents, 10' intervals . . 504 Four-place logarithms of sines, to 9, and of cosines, 81 to 90, by minutes 506 Four-place logarithms of tangents, to 9, and of cotangents, 81 to 90, by minutes 507 Four-place natural sines and cosines, 10' intervals 508 Four-place natural tangents and cotangents, 10' intervals .... 510 Radian measure of angles 612 Minutes as decimals of one degree ; & and e~ z tables 513 The accumulation of 1 at the end of n years 514 The present value of 1 due in n years 515 The accumulation of an annuity of 1 per annum at the end of n years 616 The present value of an annuity of 1 per annum for n years . . . 517 The annual sinking fund which will accumulate to 1 , or, by addition of i, the annuity which 1 will purchase 618 495 496 UNIFIED MATHEMATICS Constants with their logarithms. NI-MBER LOGARITHM Base of natural logarithms e = 2.71828183 0.4342945 Modulus of common logarithms . . . u = 0.43429448 9.0377843-10 Circumference of a circle in degrees . . = 360 2.5563025 Circumference of a circle in minutes . = 21600 4.3344538 Circumference of a circle in seconds . =1296000 6.1126050 Radian expressed in degrees . . . . = 57.29578 1.7581226 Radian expressed in minutes .... =3437.7468 3.5362739 Radian expressed in seconds .... =206264.806 5.3144251 Ratio of a circumference to diameter . TT = 3.14156592 0.4971499 IT = 3.14159265358979323846264338328 VOLUMES AXD WEIGHTS Cubic inches in 1 gallon (U S.) . . . =231 2.3636 Gallons in 1 cubic foot = 7.48 .8739 Cubic inches in 1 bushel =2150.4 3.3325 Pounds per cubic foot of water (4 C.) . =62.43 1.79.">1 Pounds per cubic foot of air (0 C.) . . = 0.0807 8.9069-10 Cubic feet in 1 cubic meter . . . = 35.32 1.5480 Cubic meters in 1 cubic yard .... =0.76 9.8808-10 Cubic inches in 1 liter = 61.03 1.7855 Liters in 1 gallon (U. S.) =3.786 -5783 Pounds in 1 kilogram = 2.2 or 2.205 .3434 Metric ton in pounds = 2205 3.3434 Volume of sphere, f wr* = 4.1888r3 4.1888 -6221 LENGTHS AND AREAS Inches in 1 meter (by Act of Congress) = 39.37 1.5952 Feet in 1 rod, 16.5 ; yards in 1 rod . . = 5.5 Square feet in 1 acre = 43560 4.6391 160 square rods = 1 acre ; 640 acres = 1 square mile ; 3.281 feet = 1.094 yards = 1 meter. TABLES 497 Squares and cubes of integers, 1 to 100. Square roots and cube roots of 1 to 100. Reciprocals of 1 to 100. ss D D g W H a o % o t> a D H H n o La 02 D O $x & O Utf H o or 02 O yO D O Utf w" n n2 n 3 VH V7i 1 n n 2 n 3 Vn \'n it n n i 1 l 1.000 1.000 1.0000 51 2,601 132,651 7.141 3.708 .0196 i 4 8 2.414 1.260 .5000 52 2,704 140,608 7.211 3.733 .0192 I 9 27 1.732 1.442 .3333 53 2,809 148,877 7.280 3.756 .0189 4 16 64 2.000 1.587 .2500 54 2,916 157,464 7.348 3.780 .0185 5 25 125 2.236 1.710 .2000 55 3,025 166,375 7.416 3.803 .0182 6 36 216 2.449 1.817 .1667 56 3,136 175,616 7.483 3.826 .0179 7 49 343 2.646 1.913 .1429 57 3,249 185,193 7.550 3.849 .0175 8 64 512 2.828 2.000 .1250 58 3,364 195,112 7.616 3.871 .0172 9 81 729 3.000 2.080 .1111 59 3,481 205,379 7.681 3.893 .0170 10 100 1,000 3.162 2.154 .1000 60 3,600 216,000 7.746 3.915 .0167 11 121 1,331 3.317 2.224 .0909 61 3,721 226,981 7.810 3.936 .0164 12 144 1,728 3.464 2.289 .0833 62 3,844 238,328 7.874 3.958 .0161 13 169 2,197 3.606 2.351 .0769 63 3,969 250,047 7.937 3.979 .0159 14 196 2,744 3.742 2.410 .0714 64 4,096 262,144 8.000 4.000 .0156 1.5 225 3,375 3.873 2.466 .0667 65 4,225 274,625 8.062 4.021 .0154 1C, 256 4,096 4.000 2.520 .0625 66 4,356 287,496 8.124 4.041 .0152 17 289 4,913 4.123 2.571 .0588 67 4,489 300,763 8.185 4.062 .0149 IS 324 5,832 4.243 2.621 .0556 68 4,624 314,432 8.246 4.082 .0147 lit 361 6.859 4,359 2.668 .0526 69 4,761 328,509 8.307 4.102 .0145 20 400 8,000 4.472 2.714 .0500 70 4,900 343,000 8.367 4.121 .0143 21 441 9,261 4.583 2.759 .0476 71 5,041 357,911 8.426 4.141 .0141 22 484 10,648 4,690 2.802 .0455 72 5,184 373,248 8.485 4.160 .0139 23 529 12,167 4.796 2.844 .0436 73 5,329 389,017 8.544 4.179 .0137 "it 576 13,824 4.899 2.884 .0417 74 5,476 405,224 8.602 4.198 .0135 2.5 625 15,625 5.000 2.924 .0400 75 5,625 421,875 8.660 4.217 .0133 26 676 17,576 5.099 2.962 .0385 76 5,776 438,976 8.718 4.236 .0132 27 729 19,683 5.196 3.000 .0370 77 5,929 456,533 8.775 4.254 .0130 is 784 21,952 5.292 3.037 .0357 78 ' 6,084 474,552 8.832 4.273 .0128 29 841 24,389 5.385 3.072 .0345 79 6,241 493,039 8.888 4.291 .0127 30 900 27,000 5.477 3.107 .0333 80 6,400 512,000 8.944 4.309 '.0125 31 961 29,791 5.568 3.141 .0323 81 6,561 531,441 9.000 4.327 .0123 32 1,024 :^,7r,s 5.657 3.175 .0313 82 6,724 551,368 9.055 4.344 .0122 33 1,089 35,937 5.745 3.208 .0303 83 6,889 571,787 9.110 4.362 .0120 34 1,156 39,304 5.831 3.240 .0294 84 7,056 592,704 9.165 4.380 .0119 3.5 1,225 42,875 5.916 3.271 .0286 85 7,225 614,125 9.220 4.397 .0118 315 1,296 46,656 6.000 3.302 .0278 86 7,396 636,056 9.274 4.414 .0116 37 1,369 50,653 6.083 3.332 .0270 87 7,569 658,503 9.327 4.431 .0115 3S 1,111 :>I>7L> 6.164 3.362 .0263 88 7,744 681,472 9.381 4.448 .0114 39 1,521 6.245 3.391 .0256 89 7,921 704,969 9.434 4.465 .0112 40 1,600 64,000 6.325 3.420 .0250 90 8,100 729,000 9.487 4.481 .0111 41 1,681 68,921 6.403 3.448 .0244 91 8,281 753,571 9.539 4.498 .0110 42 1,764 74,088 6.481 3.476 .0238 92 8,464 77s.oss 9.592 4.514 .0109 43 1,849 79,507 6.557 3.503 .0233 93 8,649 804,357 9.644 4.531 .0108 44 1,930 85,184 6.633 3.530 .0227 94 8,836 830,584 9.695 4.547 .0106 4.5 2,025 91,125 6.708 3.557 .0222 95 9,025 857,375 9.747 4.563 .0105 4(i 2,116 97,336 6.782 3.583 .0217 96 9,216 884,736 9.798 4.579 .0104 47 2,209 103,823 6.856 3.609 .0213 97 9,409 912,673 9.849 4.595 .0103 4S 2,304 110,592 6.928 3.634 .0208 98 9,604 941,192 9.899 4.610 .0102 49 2,401 117,649 7.000 3.659 .0204 99 9,801 970,299 9.950 4.626 .0101 .50 2,500 125,000 7.071 3.684 .0200 100 10,000 1,000,000 10.000 4.642 .0100 X n* n 3 Vn ^ 1 n n* n Vn V^ 1 n n 498 UNI FIED MATHEMATICS Logarithms of numbers from 100 to 549. 1 2 3 4 5 6 7 8 9 To avoid inter- polation in the 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 first ten lines. 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 following pages. 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 22 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 1 2.2 2 4.4 3 6.6 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 4 8.8 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 5 11.0 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 6 13.2 7154 18 2553 2577 2601 2025 2648 2672 2695 2718 2742 2765 8 17^6 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 9 19.8 21 20 19 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 1 2.1 2.0 1.9 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 2 4.2 4.0 3.8 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 3 6.3 6.0 5.7 4 84 80 76 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 5 10!5 lo!o 9.5 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 6 12.6 12.0 11.4 7 14.7 14.0 13.3 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 8 16.8 16.0 15.2 9 18.9 18.0 17.1 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 18 17 16 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 1 1.8 1.7 1.6 2 36 34 32 29 462.4 4639 4654 4669 4683 4698 4713 4728 4742 4757 3 5^4 5.1 4.8 4 7.2 6.8 6.4 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 5 9.0 8.5 8.0 6 10 8 10 2 9.6 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 8 14.'4 13^6 12.'8 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 9 16.2 15.3 14.4 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 15 14 13 1 1.5 1.4 1.3 1 2 3 4 5 6 7 8 9 2 3.0 2.8 2.6 3 4.5 4.2 3.9 4 6.0 5.6 5.2 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 5 7.5 7.0 6.5 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 6 9.0 8.4 7.8 37 5682 5694 5705 5717 '5729 5740 5752 5763 5775 5786 7 10.5 9.8 9.1 8 12 112 104 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 9 13^5 12!6 11.7 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 11 12 1 1.1 1 1.2 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 2 2.2 2 2.4 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 3 3.3 3 3.6 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 4 4.4 4 4.8 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 5 5.5 5 6.0 666 6 72 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 7 7.7 7 8.4 8 8.8 8 9.6 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 9 9.9 9 10.8 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 H 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 j? 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 3 '2 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 e 3 11 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 3 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 1 1 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 1 J 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 * s?" TABLES 499 Logarithms of numbers from 550 to 999. 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 w 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 , 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 1 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 4 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 I 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 1 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 .2" 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 ^ 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 i 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 M 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 1 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 10 3 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 jj 1 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 I 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 a 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 V 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 9 a 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 1 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 3 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 9949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 3 5 6 8 80 81 82 83 84 9031 9036 9042 9047 9085 9090 9096 9101 9138 9143 9149 9154 9191 9196 9201 9206 9243 9248 9253 9258 9053 9058 9106 9112 9159 9165 9212 9217 9263 9269 9063 9069 9074 9079 9117 9122 9128 9133 9170 9175 9180 9186 9222 9227 9232 9238 9274 9279 9284 9289 85 86 87 88 89 9294 9299 9304 9309 9345 9350 9355 9360 9395 9400 9405 9410 9445 9450 9455 9460 9494 9499 9504 9509 9315 9320 9365 9370 9415 9420 9465 9469 9513 9518 9325 9330 9335 9340 9375 9380 9385 9390 9425 9430 9435 9440 9474 9479 9484 9489 9523 9528 9533 9538 90 91 92 93 94 9542 9547 9552 9557 9590 9595 9600 9605 9638 9643 9647 9652 9685 9689 9694 9699 9731 9736 9741 9745 9562 9566 9609 9614 9657 9661 9703 9708 9750 9754 9571 9576 9581 9586 9619 9624 9628 9633 9666 9671 9675 9680 9713 9717 9722 9727 9759 9763 9768 9773 95 9 9777 9782 9786 9791 9823 9827 9832 9836 9868 9872 9877 9881 9912 9917 9921 9926 9956 9961 9965 9969 9795 9800 9841 9845 9886 9890 9930 9934 9974 9978 9805 9809 9814 9818 9850 9854 9859 9863 9894 9899 9903 9908 9939 9943 9948 9952 9983 9987 9991 9996 G 8 500 UNIFIED MATHEMATICS Logarithms of numbers between 1000 and 1499. 1 3 3 4 5 6 7 8 9 1M 0000 0004 0009 0013 0017 0022 0026 0030 0035 0039 101 0043 0048 0052 0056 0060 0065 0069 0073 0077 0082 103 0086 0090 0095 0099 0103 0107 0111 0116 0120 0124 103 0128 0133 0137 0141 0145 0149 0154 0158 0162 0166 104 0170 0175 0179 0183 0187 0191 0195 0199 0204 0208 105 0212 0216 0220 0224 0228 0233 0237 0241 0245 0249 106 0253 0257 0261 0265 0269 0273 0278 0282 0286 0290 107 0294 0298 0302 0306 0310 0314 0318 0322 0326 0330 108 0334 0338 0342 0346 0350 0354 0358 0362 0366 0370 109 0374 0378 0382 0386 0390 0394 0398 0402 0406 0410 110 0414 0418 0422 0426 0430 0434 0438 0441 0445 0449 111 0453 0457 0461 0465 0469 0473 0477 0481 0484 0488 113 0492 0496 0500 0504 0508 0512 0515 0519 0523 0527 113 0531 0535 0538 0542 0546 0550 0554 0558 0561 0565 114 0569 0573 0577 0580 0584 0588 0592 0596 0599 0603 115 0607 0611 0615 0618 0622 0626 0630 0633 0637 0641 116 0645 0648 0652 0656 0660 0663 0667 0671 0674 0678 117 0682 0686 0689 0693 0697 0700 0704 0708 0711 0715 118 0719 0722 0726 0730 0734 0737 0741 0745 0748 0752 119 0755 0759 0763 0766 0770 0774 0777 0781 0785 0788 130 0792 0795 0799 0803 0806 0810 0813 0817 0821 0824 131 0828 0831 0835 0839 0842 0846 0849 0853 0856 0860 133 0864 0867 0871 0874 0878 0881 0885 0888 0892 0896 133 0899 0903 0906 0910 0913 0917 0920 0924 0927 0931 134 0934 0938 0941 0945 0948 0952 0955 0959 0962 0966 1 3 3 4 5 6 7 8 9 135 0969 0973 0976 0980 0983 0986 0990 0993 0997 1000 136 1004 1007 1011 1014 1017 1021 1024 1028 1031 1035 137 1038 1041 1045 1048 1052 1055 1059 1062 1065 1069 138 1072 1075 1079 1082 1086 1089 1092 1096 1099 1103 139 1106 1109 1113 1116 1119 1123 1126 1129 1133 1136 130 1139 1143 1146 1149 1153 1156 1159 1163 1166 1169 131 1173 1176 1179 1183 1186 1189 1193 1196 1199 1202 133 1206 1209 1212 1216 1219 1222 1225 1229 1232 1235 133 1239 1242 1245 1248 1252 1255 1258 1261 1265 1268 134 1271 1274 1278 1281 1284 1287 1290 1294 1297 1300 135 1303 1307 1310 1313 1316 1319 1323 1326 1329 1332 136 1335 1339 1342 1345 1348 1351 1355 1358 1361 1364 137 1367 1370 1374 1377 1380 1383 1386 1389 1392 1396 138 1399 1402 1405 1408 1411 1414 1418 1421 1424 1427 139 1430 1433 1436 1440 1443 1446 144'9 1452 1455 1458 140 1461 1464 1467 1471 1474 1477 1480 1483 1486 1489 141 1492 1495 1498 1501 1504 1508 1511 1514 1517 1520 143 1523 1526 1529 1532 1535 1538 1541 1544 1547 1550 143 1553 1556 1559 1562 1565 1569 1572 1575 1578 1581 144 1584 1587 1590 1593 1596 1599 1602 1605 1608 1611 145 1614 1617 1620 1623 1626 1629 1632 1635 1638 1641 146 1644 1647 1649 1652 1655 1658 1661 1664 1667 1670 147 1673 1676 1679 1682 1685 1688 1691 1694 1697 1700 148 1703 1706 1708 1711 1714 1717 1720 1723 1726 1729 149 1732 1735 1738 1741 1744 1746 1749 1752 1755 1758 1 3 3 4 5 6 7 8 9 TABLES 501 Logarithms of numbers between 1500 and 1999. 1 2 3 4 5 6 7 8 9 . 150 1761 1764 1767 1770 1772 1775 1778 1781 1784 1787 151 1790 1793 1796 1798 1801 1804 1807 1810 1813 1816 159 1818 1821 1824 1827 1830 1833 1836 1838 1841 1844 153 1847 1850 1853 1855 1858 1861 1864 1867 1870 1872 154 1875 1878 1881 1884 1886 1889 1892 1895 1898 1901 155 1903 1906 1909 1912 1915 1917 1920 1923 1926 1928 156 1931 1934 1937 1940 1942 1945 1948 1951 1953 1956 157 1959 1962 1965 1967 1970 J973 1976 1978 1981 1984 158 1987 1989 1992 1995 1998 2000 2003 2006 2009 2011 159 2014 2017 2019 2022 2025 2028 2030 2033 2036 2038 160 2041 2044 2047 2049 2052 2055 2057 2060 2063 2066 161 2068 2071 2074 2076 2079 2082 2084 2087 2090 2092 162 2095 2098 2101 2103 2106 2109 2111 2114 2117 2119 163 2122 2125 2127 2130 2133 2135 2138 2140 2143 2146 164 2148 2151 2154 2156 2159 2162 2164 2167 2170 2172 165 2175 2177 2180 2183 2185 2188 2191 2193 2196 2198 166 2201 2204 2206 2209 2212 2214 2217 2219 2222 2225 167 2227 2230 2232 2235 2238 2240 2243 2245 2248 2251 168 2253 2256 2258 2261 2263 2266 2269 2271 2274 2276 169 2279 2281 2284 2287 2289 2292 2294 2297 2299 2302 170 2304 2307 2310 2312 2315 2317 2320 2322 2325 2327 171 2330 2333 2335 2338 2340 2343 2345 2348 2350 2353 173 2355 2358 2360 2363 2365 2368 2370 2373 2375 2378 173 2380 2383 2385 2388 2390 2393 2395 2398 2400 2403 174 2405 2408 2410 2413 2415 2418 2420 2423 2425 2428 1 2 3 4 5 6 7 8 9 175 2430 2433 2435 2438 2440 2443 2445 2448 2450 2453 176 2455 2458 2460 2463 2465 2467 2470 2472 2475 2477 177 2480 2482 2485 2487 2490 2492 2494 2497 2499 2502 178 2504 2507 2509 2512 2514 2516 2519 2521 2524 2526 179 2529 2531 2533 2536 2538 2541 2543 2545 2548 2550 180 2553 2555 2558 2560 2562 2565 2567 2570 2572 2574 181 2577 257!) 2582 2584 2586 2589 2591 2594 2596 2598 182 2601 2603 2605 2608 2610 2613 2615 2617 2620 2622 183 2625 2627 2629 2632 2634 2636 2639 2641 2643 2646 184 2648 2651 2653 2655 2658 2660 2662 2665 2667 2669 185 2672 2674 2676 2679 2681 2683 2686 2688 2690 2693 186 2695 2697 2700 2702 2704 2707 2709 2711 2714 2716 187 2718 2721 2723 2725 2728 2730 2732 2735 2737 2739 188 2742 2744 2746 2749 2751 2753 2755 2758 2760 2762 189 2765 2767 2769 2772 2774 2776 2778 2781 2783 2785 190 2788 2790 2792 2794 2797 2799 2801 2804 2806 2808 191 2810 2813 2815 2817 2819 2822 2824 2826 L'S2S 2831 IK 2833 2835 2838 2840 2842 2844 2847 2849 2851 2853 193 2856 2858 2860 2862 2865 2867 2869 2871 2874 2876 194 2878 2880 2882 2885 2887 2889 2891 2894 2896 2898 195 2900 2903 2905 2907 2909 2911 2914 2916 2918 2920 196 2923 2925 2927 2929 2931 2934 2936 2938 2940 2942 197 2945 2947 2949 2951 2953 2956 2958 2960 2983 2964 198 2967 2969 2971 2973 2975 2978 2980 2982 2984 2986 199 2989 2991 2993 2995 2997 2999 3002 3004 3006 3008 1 2 3 4 5 6 7 8 9 502 UNIFIED MATHEMATICS Log sin A from to 45 - 10' 20' 30' 40' 50' 60' A d. 7.4637 7648 9408 *0658 *1627 *2419 89 Do not interpolate, but 1 8.2419 2 5428 3088 3668 5776 6097 4179 6397 4637 6677 5050 6940 5428 7188 88 87 use the special table 3 7188 7423 7645 7857 8059 8251 8436 86 which gives these values 4 8436 8613 8783 8946 9104 9256 9403 85 by minutes. 5 9403 9545 9682 9816 9945 *0070 *0192 84 6 9.0192 0311 0426 0539 0648 0755 0859 83 7 0859 0961 1060 1157 1252 1345 1436 82 91 8 1436 1525 1612 1697 1781 1863 1943 81 SO 9 1943 2022 2100 '2176 2251 2324 2397 80 73 P. P. 10 9.2397 2468 2538 2606 2674 2740 2806 79 68 11 2806 2870 2934 2997 3058 3119 3179 78 62 92 90 88 86 84 12 3179 3238 3296 3353 3410 3466 3521 77 57 1 9.2 9.0 8.8 8.6 8.4 13 3521 3575 3629 3682 3734 3786 3837 76 53 2 18.4 18.0 17.6 17.2 16.8 14 3837 3887 3937 3986 4035 4083 4130 75 49 3 4 27.6 36.8 27.0 26.4 36.0 35.2 25.8 25.2 34.4 33.6 5 46.0 45.0 44.0 43.0 42.0 15 4130 4177 4223 4269 4314 4359 4403 74 46 6 55.2 54.0 52.8 51.6 50.4 16 4403 17 4659 4447 4491 4700 4741 4533 4781 4576 4821 4618 4861 4659 4900 73 72 43 40 7 8 9 64.4 73.6 82.8 63.0 61.6 72.0 70.4 81.0 79.2 60.2 58.8 68.8 67.2 77.4 75.6 18 4900 4939 4977 5015 5052 5090 5126 71 38 19 5126 5163 5199 5235 5270 5306 5341 70 36 82 80 78 76 74 1 8.2 8.0 7.8 7.6 7.4 2 16.4 16.0 15.6 15.2 14.8 20 9.5341 5375 5409 5443 5477 5510 5543 69 34 3 24.6 24.0 23.4 22.8 22.2 21 5543 22 5736 23 5919 5576 5609 5767 5798 5948 5978 5641 5828 6007 5673 5859 6036 5704 5889 6065 5736 5919 6093 68 67 66 32 31 29 4 5 6 7 32.8 41.0 49.2 57.4 32.0 31.2 40.0 39.0 48.0 46.8 56.0 54.6 30.4 29.6 38.0 37.0 45.6 44.4 53.2 51.8 24 6093 6121 6149 6177 6205 6232 6259 65 28 8 65.6 64.0 62.4 60.8 59.2 9 73.8 72.0 70.2 68.4 66.6 30' 72 70 68 66 64 1 7.2 7.0 6.8 6.6 6.4 25 6259 26 6418 27 6570 6286 6313 6444 6470 6595 6620 6340 6495 6644 6366 6521 6668 6392 6546 6692 6418 6570 6716 64 63 62 27 25 24 2 3 4 5 14.4 21.6 28.8 36.0 14.0 13.6 21.0 20.4 28.0 27.2 35.0 34.0 13.2 12.8 19.8 19.2 26.4 25.6 33.0 32.0 28 6716 6740 6763 6787 6810 6833 6856 61 23 6 43.2 42.0 40.8 39.6 38.4 29 6856 6878 6901 6923 6946 6968 6990 60 22 7 8 50.4 57.6 49.0 47.6 56.0 54.4 46.2 44.8 52.8 51.2 9 64.8 63.0 61.2 59.4 57.6 30 9.6990 31 7118 32 7242 33 7361 34 7476 7012 7033 7139 7160 7262 7282 7380 7400 7494 7513 7055 7181 7302 7419 7531 7076 7201 7322 7438 7550 7097 7222 7342 7457 7568 7118 7242 7361 7476 7586 59 58 57 56 55 21 21 20 18 1 2 3 4 5 62 6.2 12.4 18.6 24.8 31.0 60 58 6.0 5.8 12.0 11.6 18.0 17.4 24.0 23.2 30.0 29.0 56 54 5.6 5.4 11.2 10.8 16.8 16.2 22.4 21.6 28.0 27.0 35 7586 36 7692 37 7795 38 7893 39 7989 7604 7622 7710 7727 7811 7828 7910 7926 8004 8020 7640 7744 7844 7941 8035 7657 7761 7861 7957 8050 7778 7877 7973 8066 7692 7795 7893 7989 8081 54 53 52 51 50 18 17 id 16 15 . 6 7 8 9 1 37.2 43.4 49.6 55.8 52 5.2 36.0 34.8 42.0 40.6 48.0 46.4 54.0 52.2 50 48 5.0 4.8 33.6 32.4 39.2 37.8 44.8 43.2 50.4 48.6 46 44 4.6 4.4 2 10.4 10.0 9.6 9.2 8.8 40 9.8081 41 8169 42 8255 43 8338 44 8418 8096 8111 8184 8198 8269 8283 8351 8365 8431 8444 8125 8213 8297 8378 8457 8140 8227 8311 8391 8469 8155 8241 8324 8405 8482 8169 8255 8338 8418 8495 49 48 47 46 45 15 14 14 13 13 3 4 5 f. 7 8 9 15.6 20.8 26.0 31.2 36.4 41.6 46.8 15.0 14.4 20.0 19.2 25.0 24.0 30.0 28.8 35.0 33.6 40.0 38.4 45.0 43.2 13.8 13.2 18.4 17.6 23.0 22.0 27.6 26.4 32.2 30.8 36.8 35.2 41.4 39.6 60' 50' 40' 30' 20' 10' A d. P.P. Log cos A from 45 to 90. TABLES 503 Log sin A from 45 to 90. p. p. A 0' 10' 20 7 30' 40' 50' 60' A" (1 p.p. 45 8495 8507 8520 8532 8545 8557 8569 44 12 12 46 8569 8582 8594 8606 8618 8629 8641 43 12 42 40 38 36 47 8641 8653 8665 8676 8688 8699 8711 12 2 2^4 1 4.2 4.0 3.8 3.6 48 8711 8722 8733 8745 8756 8767 8778 41 11 3 3.6 2 84 8.0 7.6 7.2 49 8778 8789 8800 8810 8821 8832 8843 40 1 1 4 4.8 3 12.6 12.0 11.4 10.8 5 6.0 4 16.8 16.0 15.2 14.4 6 7.2 521.020.0 19.0 18.0 501 (.8843 8853 8864 8874 8884 8895 8905 39 10 7 8.4 625.2 24.022.821.6 51 8905 8915 8925 8935 8945 8955 8965 10 8 9.6 729.428.026.625.2 833.632.030.428.8 937.836.034.2 32.4 52 53 54 8965 9023 9080 8975 9033 9089 8985 9042 9098 8995 9052 9107 9004 9061 9116 9014 9070 9125 9023 9080 9134 I 9 10 9 9 10.8 34 32 30 28 55 56 9134 9186 9142 9194 9151 9203 9160 9211 9169 9219 9177 9228 9186 9236 34 33 9 8 1 3.4 3.2 3.0 2.8 2 6.8 6.4 6.0 5.6 3 10.2 9.6 9.0 8.4 4 13.6 12.8 12.011.2 57 58 59 9236 9284 9331 9244 9292 9338 9252 9300 9346 9260 9308 9353 9268 9315 9361 9276 9323 9368 9284 9331 9375 32 31 30 8 8 7 5 17.0 16.0 15.014.0 6 20.4 19.2 18.0 16.8 7 23.822.4 21.0 19.6 601 t.9375 8393 9390 9397 9404 9411 9418 ''!( 7 8 27.2 25.6 24.0 22.4 61 9418 9425 9432 9439 9446 9453 9459 H 6 930.628.827.025.2 62 9459 9466 9473 9479 9486 9492 9499 7 63 9499 9505 9512 9518 9524 9530 9537 'Hi 7 64 9537 9543 9549 9555 9561 9567 9573 M 6 26 24 22 20 3 1 2.6 2.4 2.2 2.0 65 9573 9579 9584 9590 9596 9602 9607 J4 5 5 2 5.2 4.8 4.4 4.0 6G 9607 9613 9618 9624 9629 9635 9640 23 5 a 3 7.8 7.2 6.6 6.0 67 9640 9646 9651 9656 9661 9667 9672 >2 5 .3 4 10.4 9.6 8.8 8.0 5 13.0 12.0 11.0 10.0 6 15.6 14.4 13.2 12.0 68 69 9672 9702 9677 9706 9682 9711 9687 9716 9692 9721 9697 9725 9702 9730 : !i 5 5 +* 1 7 18.2 16.8 15.4 14.0 s 820.8 19.2 17.6 16.0 30' - 923.421.6 19.8 18.0 a 70 9.9730 9734 9739 9743 9748 9752 9757 19 5 71 9757 9761 9765 9770 9774 9778 9782 IS 4 1 19 18 17 16 72 9782 9786 9790 9794 9798 9802 9806 17 4 y 1 19 18 17 16 73 9806 9810 9814 9817 9821 9825 9828 it; 3 2 3.8 3.6 3.4 3.2 74 9828 9832 9836 9839 9843 9846 9849 15 3 1 3 5.7 5.4 5.1 4.8 A 4 76 72 68 6.4 5 9^5 9.0 8.5 8.0 75 9849 9853 9856 9859 9863 9866 9869 14 3 *3 6 11.4 10.8 10.2 9.6 76 9869 9872 9875 9878 9881 9884 9887 13 3 B 7 13.3 12.6 11.9 11.2 77 9887 9890 9893 9896 9899 9901 9904 12 3 8 15.2 14.4 13.6 12.8 9 17.1 16.2 15.3 14.4 78 79 9904 9919 9907 9922 9909 9924 9912 9927 9914 9929 9917 9931 9919 9934 11 10 2 3 a 3 80 9.9934 9936 993S 9940 9942 9944 9946 ,, 2 1 15 14 13 12 81 9946 9948 9950 9952 9954 9956 9958 8 2 1 1.5 1.4 1.3 1.2 82 OOfiS 0060 9961 9963 9964 9966 9908 7 2 a 2 3.0 2.8 2.6 2.4 3 4.5 4.2 3.9 3.6 4 6.0 5.6 5.2 4.8 83 84 0068 9976 9969 9977 9971 9979 9972 9980 9973 9981 0078 9982 9976 9983 1 5 1 1 5 7.5 7.0 6.5 6.0 6 9.0 8.4 7.8 7.2 7 10.5 9.8 9.1 8.4 8120112104 96 85 86 9983 9989 9985 9990 9986 *w ** 1 4.8 4.7 4.6 4.5 4.4 2 9.6 9.4 9.2 9.0 8.8 30 tan 9.5611 5650 5689 5727 5766 5804 5842 cot 69 M 314.4 14.1 13.813.513.2 log cot 0.4.389 4350 4311 4273 4234 4196 4158 tan log 39 419.2 18.818.418.017.6 21 tan 9.5842 log cot 0.4158 5879 5917 4121 4083 5954 4046 5991 4009 6028 3972 6064 cot 3936 tan 68 log 37 37 524.023.523.022.522.0 628.828.227.627.026.4 733.632.932.231.530.8 22 tan 9.6064 6100 6136 6172 6208 6243 6279 cot 67 36 838.437.636.836.035.2 log cot 0.3936 3900 3864 3828 3792 3757 3721 tan log 38 943.242.341.440.539.6 60' 50' 40' 30' 10' P.P. Logarithms of tangents and cotangents, 67 to 90. TABLES 505 Logarithms of tangents and cotangents, 23 to 46. P.P. 10' 20' 30' 40' 50' 60' A" d. P.P. 23 tan 9.6279 6314 6348 6383 6417 6452 6486 cot 66 34 log cot 0.3721 3686 3652 3617 3583 3548 3514 tan log 34 tA" ton Q ,: 1^1; R^on RX.HI RX.V7 Ron r.i;-..i W7 ,.t * 43 1 4.3 42 4.2 it tan .o-*o oozu 000,5 log cot 0.3514 3480 3447 DOS/ 3413 OIKU OOO-l 3380 3346 DOS/ COl DO .5.5 33 13 tan log 33 1 33 3.3 32 3.2 286 8 4 2 6.6 6.4 3 12.9 12.6 25 tan 9. 6687 6720 6752 6785 6817 6850 6882 cot 64 33 i 9.9 9.6 4 17.2 16.8 log cot 0.3313 3280 3248 3215 3183 3150 31 18 tan log 33 i 13.2 12.8 5 21.5 21.0 26 tan 9.6882 6914 6946 6977 7009 7040 7072 cot 63 s 32 ."i 16.5 16.0 6 25.8 7 30 1 25.2 29 4 log cot 0.3118 3086 3054 3023 2991 2960 2928 tan log 32 7 19.8 23 1 19.2 22 4 8 34 4 33 6 27 tan 9. 7072 7103 7134 7165 7196 7226 7257 cot 62 31 8 26.4 25.6 938.7 37.8 log cot 0.2928 2897 2866 2835 2804 2774 2743 tan log 31 1 29.7 28.8 28 tan 9.7257 7287 7317 7348 7378 7408 7438 cot 61 30 log cot 0.2743 2713 2683 2652 2622 2592 2562 tan log 30 31 30 41 40 29 tan 9.7438 7467 7497 7526 7556 7585 7614 cot 60 20 31 1. 1 4.1 282 4.0 80 log cot 0.2562 2533 2503 2474 2444 2415 2386 tan log 29 1 2 .1 6.2 o.u 6.0 3 12.3 4 16.4 12.0 16.0 30 tan 9.7614 7644 7673 7701 7730 7759 7788 cot 59 29 i 4 9.3 12.4 9.0 12.0 520.5 20.0 log cot 0.2386 2356 2327 2299 2270 2241 2212 tan log 29 B i, 15.5 18 6 15.0 18 6 ?4.6 7 ?8 7 24.0 28 31 tan 9. 7788 7816 7845 7873 7902 7930 79.> cot 58 28 7 21.7 21.0 8 32 8 32 log cot 0.2212 2184 2155 2127 2098 2070 2042 tan log 28 8 24.8 24.0 9MJ 36.0 32 tan 9. 7958 7986 8014 8042 8070 8097 8125 cot 57 28 9 27.9 27.0 log cot 0.2042 2014 1986 1958 1930 1903 1875 tan log 28 39 38 33 tan 9. 81 25 8153 8180 8208 8235 8263 8290 cot 56 27 29 28 1 3.9 2 78 3.8 7 6 log cot 0.1875 1847 34 tan 9.8290 8317 1820 8344 1792 8371 1765 1737 1710 tan log 27 8398 8425 8452 cot 55 27 1 2 2.9 5.8 2.8 5.6 3 11> 4 15.6 5 19 5 11 A 11.4 19 log cot 0.1710 1683 1656 1629 1602 1575 1548 tan log 27 I 4 5 8.7 11.6 14.5 8.4 11.2 14.0 i> 2. : 5 A 7 27.3 8312 22^8 26.6 304 35 tan 9.8452 8479 8506 30' 8533 8*559 8586 861 3 cot 54 27 6 17.4 7 20.3 8 2:5.2 16.8 19.6 22.4 9 35.1 34.2 log cot 0.1548 1521 36 tan 9.S613 8639 1494 8666 1467 8692 1441 1414 8718 8745 1387 tan log 27 8771 cot 53 26 9 26.1 25.2 log cot 0.1387 1361 1334 1308 1282 1255 1229 tan log 26 37 36 37 tan 9.8771 8797 8824 8850 8876 8902 8928 cot 52 26 27 26 1 3.7 2 7.4 3.6 7.2 log cot 0.1229 1203 1176 1150 1124 1098 1072 tan log 26 1 1 2.7 5.4 2.6 5.2 3 11.1 10.8 1 8.1 7.8 4 14.8 14.4 38 tan 9.8928 8954 8980 9006 9032 9058 9084 cot 51 26 I 10.8 10.4 5 18.5 r, -22 -2 7 25 9 18.0 21.6 25 2 log cot 0.1072 1046 39 tan 9.9084 9110 1020 9135 0994 9161 0968 0942 9187 9212 091 6 tan log 2(i 9238 cot 50 s 26 5 a 7 13.5 16.2 IS 'I 13.0 15.6 18 2 28^8 log cot 0.0916 0890 0865 0839 0813 0788 0762 tan log 26 21.6 20.8 9 33.3 32.4 I 24.3 23.4 40 tan 9.9238 9264 9289 9315 9341 9366 9392 cot 49 26 log cot 0.0762 0736 0711 0685 0659 0634 0608 tan log 26 41 tan 9.9392 9417 9443 9468 9494 9519 9544 cot 48 25 25 24 35 34 log cot 0.0(i08 0583 0557 0532 0506 0481 0456 tan log 25 ] 2 5 2 4 1 3.5 3.4 42 tan 9.9544 9570 9595 9621 9646 9671 9697 cot 47 s 25 50 "is 2 7.0 3 10.5 6.8 10.2 log cot 0.0456 0430 0405 0379 0354 0329 0303 tan log 25 ."5 7.5 4 14.0 13.6 1 10.0 9.6 6 17.5 17.0 43 tan 9.9697 9722 9747 9772 9798 9823 9848 cot 46 25 5 12.5 15 12.0 14 4 6 21.0 7 24 5 20.4 23 8 log cot 0.0303 0278 0253 0228 0202 0177 0152 tan log 25 7 \7'.S lf.8 8 28^0 27^2 44' tan 9.9S48 9874 '.IS'. Ill 9924 9949 9975 0000 cot 45 25 8 20.0 19.2 931.5 30.6 log cot 0.0152 0126 0101 0076 0051 0025 0000 tan log 25 22.5 21.6 45 tan 0.0000 0025 0051 0076 0101 0126 0152 cot 44 25 log cot 0.0000*9975*9949 *9924 *9899*9874*9848 tan log 25 P.P. 60' 50' 40' 30' 20' 10' 0' A" d. P.P. Logarithms of tangents and cotangents, 44 to 67. 506 UNIFIED MATHEMATICS Log sin by minutes from 0" to 9^. 6.4637 7648 9408 *0658 *1627 *2419 *3088 *3668 *4180 *4637 50 10 7.4637 5051 5429 5777 6099 6398 6678 6942 7190 7425 7648 40 20 7648 7859 8061 8255 8439 8617 S7S7 8951 *109 *261 *408 30 30 7.9408 551 GS! 822 952 *078 *200 *319 *435 *548 *658 20 40 8.0658 765 870 972 *072 *169 *265 *358 *450 *539 *627 10 50 8.1627 713 797 880 961 *041 *119 *196 *271 *346 *419 89 1 8.2419 490 561 630 699 766 832 898 962 *025 *088 50 10 8.3088 150 210 270 329 388 445 502 558 613 66840 20 668 722 775 828 880 931 982 *032 *082 *131 *179 30 30 8.4179 227 275 322 368 414 459 504 549 593 637 20 40 637 680 723 765 807 848 890 930 971 *011 *050 10 50 8.5050 090 129 167 206 243 281 318 355 392 428 088 2 8.5428 464 500 535 571 605 640 674 708 742 776 50 10 776 809 842 875 907 939 .972 *003 *035 *066 *097 40 20 8.6097 128 159 189 220 250 279 309 339 368 397 30 30 397 426 454 483 511 539 567 595 622 650 677 20 40 677 704 731 758 784 810 837 863 889 914 940 10 50 940 965 991 *016 *041 *066 *090 *115 *140 *164 *188 87 3 8.7188 212 236 260 283 307 330 354 377 400 423 50 10 423 445 468 491 513 535 557 580 602 623 645 40 20 645 667 688 710 731 752 773 794 815 836 857 30 30 857 877 898 918 939 959 979 999 *019 *039 *059 20 40 8.8059 078 098 117 137 156 175 194 213 232 251 10 50 251 270 289 307 326 345 363 381 400 418 436 086 4 8.8436 454 472 490 508 525 543 560 578 595 613 50 10 613 '630 647 665 682 699 716 733 749 766 783 40 20 783 799 816 833 849 865 882 898 914 930 946 30 30 946 962 978 994 *010 *026 *042 *057 *073 *089 *104 20 40 8.9104 119 135 150 166 181 196 211 226 241 256 10 50 256 271 286 301 315. 330 345 359 374 389* 403 085 5 8.9403 417 432 446 460 475 489 503 517 531 545 50 10 545 559 573 587 601 614 628 642 655 669 682 40 20 682 696 709 723 736 750 763 776 789 803 816 30 30 816 829 842 855 868 881 894 907 919 932 945 20 40 945 958 970 983 996 *008 *021 *033 *046 *058 *070 10 50 9.0070 083 095 107 120 132 144 156 168 180 192 84 6 9.0192 204 216 228 240 252 264 276 287 299 311 50 10 311 323 334 346 357 369 380 392 403 415 426 40 20 426 438 449 460 472 483 494 505 516 527 539 30 30 539 550 561 572 583 594 605 616 626 637 648 20 40 648 659 670 680 691 702 712 723 734 744 755 10 50 755 765 776 786 797 807 818 828 838 849 859 83 7 9.0859 869 879 890 900 910 920 930 940 951 961 50 10 961 971 981 991 *001 *011 *020 *030 *040 *050 *06040 20 9.1060 070 080 089 099 109 118 128 138 147 157 30 30 157 167 176 186 195 205 214 224 233 242 252 20 40 252 261 271 280 289 299 308 317 326 336 345 10 50 345 354 363 372 381 390 399 409 418 427 436 082 8 9.1436 445 453 462 471 480 489 498 507 516 525 50 10 525 533 542 551 560 568 577 586 594 603 612 40 20 612 620 629 637 646 655 663 672 680 689 697 30 30 697 705 714 722 731 739 747 756 764 772 781 20 40 781 789 797 806 814 822 830 838 847 855 863 10 50 863 . 871 879 887 895 903 911 919 927 935 943 81 10' 3' V Log cos by minutes from 81 to 90. TABLES 507 Log tan by minutes from (T to 9. 10' 6.4637 10 7.4637 5051 7648 9408 *0658 5429 5777 6099 *1627 6398 *2419 6678 *3088 6942 *3668 7190 *4180 7425 *4637 50 7648 40 20 7648 7860 8062 8255 8439 8617 8787 8951 *109 *261 *409 30 30 7.9409 00 1 689 823 952 *078 *200 *319 *435 *548 *658 20 40 8.0658 765 870 972 *072 *170 *265 *359 *450 *540 *627 10 50 8.1627 713 798 880 962 *041 *120 *196 *272 *346 *419 89 1 8.2419 491 562 631 700 767 833 899 963 *026 *089 50 in 8.3069 150 211 271 330 389 446 503 559 614 669 40 20 669 723 776 829 881 932 983 *033 *083 132 181 30 30 8.4181 229 276 323 370 416 461 506 551 595 638 20 40 638 682 725 767 809 851 892 933 973 *013 *053 10 50 8.5053 092 131 170 208 246 283 321 358 394 431 088 2 8.5431 467 503 538 573 608 643 677 711 745 779 50 10 779 812 845 878 911 943 975 *007 *038 *070 *101 40 20 8.6101 i:?2 163 193 223 254 283 313 343 372 401 30 30 401 430 459 487 515 544 571 599 627 654 682 20 40 682 709 736 762 789 815 842 868 894 920 945 10 50 945 971 996 *021 *046 *071 *096 *121 *145 *170 *194 87 :{ 8.7194 218 242 266 290 313 337 360 383 406 429 50 10 429 452 475 497 520 542 565 587 609 631 652 40 20 652 674 096 717 739 760 781 802 823 844 865 30 30 865 886 906 927 947 967 988 *008 *028 *048 *067 20 40 8.8067 087 107 126 146 165 185 204 223 242 261 10 50 261 280 299 317 336 355 373 392 410 428 446 86 4 8.8446 465 483 501 518 536 554 572 589 607 624 50 10 624 642 659 676 694 711 728 745 762 778 795 40 20 795 812 829 845 862 878 895 911 927 944 960 30 30 960 976 992 *008 *024 *040 *056 *071 *087 *103 *118 20 40 8.9118 134 150 165 180 196 211 226 241 256 272 10 50 272 287 302 316 331 346 361 376 390 405 420 85 5 8.9420 434 449 463 477 492 506 520 534 549 563 50 10 563 577 591 605 619 633 646 660 674 688 701 40 20 701 7i:> 729 742 756 769 782 796 809 823 836 30 30 836 849 862 875 888 901 915 927 940 953 966 20 40 966 979 992 *005 *017 *030 *O43 *05o *068 *080 *093 10 50 9.0093 105 118 130 143 155 167 180 192 204 216 084 (i 9.0216 228 240 253 265 277 289 300 312 324 336 50 10 336 348 360 371 383 395 407 418 430 441 453 40 20 453 464 476 487 499 510 521 533 544 555 567 30 30 567 578 688 600 611 622 633 645 656 667 678 20 40 678 688 099 710 721 732 743 754 764 775 786 10 50 786 796 807 818 828 839 849 860 871 881 891 083 1 9.0891 902 912 923 933 943 954 964 974 984 995 50 10 995 *005 *015 *025 *035 *045 *055 *066 *076 *086 *096 40 20 9.1096 106 Lie 125 188 145 155 165 175 185 194 30 :) I'M 204 214 223 233 243 252 262 272 281 291 20 40 291 300 310 319 329 338 348 357 367 376 385 10 50 385 395 404 413 423 432 441 450 460 469 478 82 s 9.1478 487 496 505 515 524 533 542 551 560 569 50 10 569 687 696 605 613 622 631 640 649 658 40 20 658 867 675 684 698 702 710 719 728 736 745 30 30 745 754 762 771 779 788 797 805 814 822 831 20 40 831 839 848 856 864 873 881 890 898 906 915 10 50 915 923 931 940 948 956 964 973 981 989 997 81 III Log cot by minutes from 81 to 90. 508 UNIFIED MATHEMATICS Numerical values of the sine function, to 45. 10' 20' 30' 40' 60' 60' P.P. e 0.0000 0029 0058 0087 0116 0145 0175 89 29 i 0175 0204 0233 0262 0291 0320 0349 88 29 2 0349 0378 0407 0436 0465 0494 0523 87 29 30 29 3 0523 0552 0581 0610 0640 0669 0698 86 29 1 30 29 4 0698 0727 0756 0785 0814 0843 0872 85 29 2 6.0 5.8 3 9.0 8.7 5 0.0872 0901 0929 0958 0987 1016 1045 84 29 4 12.0 11.6 5 15.0 14.5 6 1045 1074 1103 1132 1161 1190 1219 83 29 6 18.0 17.4 7 1219 1248 1276 1305 1334 1363 1392 82 29 7 21.0 20.3 8 1392 1421 1449 1478 1507 1536 1564 81 29 8 24.0 23.2 n 97 Q 2fi 1 9 1564 1593 1622 1650 1679 1708 1736 80 29 10 0.1736 1765 1794 1822 1851 1880 1908 N 79 29 28 27 11 1908 1937 1965 1994 2022 2051 2079 78 28 12 2079 2108 2136 2164 2193 2221 2250 77 28 1 2.8 2.7 2 56 54 13 2250 2278 2306 2334 2363 2391 2419 76 28 3 84 8.1 14 2419 2447 2476 2504 2532 2560 2588 75 28 4 11.2 10.8 5 14.0 13.5 6 16 8 16 2 15 0.2588 2616 2644 2672 2700 2728 2756 74 28 7 19.6 18.9 16 2756 2784 2812 2840 2868 2896 2924 73 28 8 22.4 21.3 17 2924 2952 2979 3007 3035 3062 3090 72 28 9 25.2 24.3 18 3090 3118 3145 3173 3201 3228 3256 71 28 19 3256 3283 3311 3338 3365 3393 3420 70 27 26 25 20 0.3420 3448 3475 3502 3529 3557 3584 69 27 1 2.6 2.5 21 3584 3611 3638 3665 3692 3719 3746 68 27 2 5.2 5.0 22 3746 3773 3800 3827 3854 3881 3907 67 27 3 7.8 7.5 4 10 4 10 23 3907 3934 3961 3987 4014 4041 4067 66 27 5 13 12.5 24 4067 4094 4120 4147 4173 4200 4226 65 26 6 15.6 15.0 7 18.2 17.5 30' 8 20.8 20.0 9 23.4 22.5 25 0.4226 4253 4279 4305 4331 4358 4384 64 26 26 4384 4410 4436 4462 4488 4514 4540 63 26 27 4540 4566 4592 4617 4643 4669 4695 62 26 24 23 28 4695 4720 4746 4772 4797 4823 4848 61 26 1 2.4 2.3 29 4848 4874 4899 4924 4950 4975 5000 60 25 2 4.8 4.6 3 7.2 6.9 4 96 9.2 30 0.5000 5025 5050 5075 5100 5125 5150 59 25 5 12.0 11.5 31 5150 5175 5200 5225 5250 5275 5299 58 25 6 14.4 13.8 32 5299 5324 5348 5373 5398 5422 5446 57 24 7 16.8 16.1 33 5446 5471 5495 5519 5544 5568 5592 56 24 8 19.2 18.4 9 21.6 20.7 34 5592 5616 5640 5664 5688 5712 5736 55 24 35 0.5736 5760 5783 5807 5831 5854 5878 54 24 22 21 20 36 5878 5901 5925 5948 5972 5995 6018 53 23 37 6018 6041 6065 6088 6111 6134 6157 52 23 1 2.2 2.1 2.0 2 44 42 40 38 6157 6180 6202 6225 6248 6271 6293 51 23 3 6^6 6.3 6.0 39 6293 6316 6338 6361 6383 6406 6428 50 22 4 8.8 8.4 8.0 5 11.0 10.5 10.0 6 13.2 12.6 12.0 40 0.6428 6450 6472 6494 6517 6539 6561 49 22 7 15.4 14.7 14.0 41 6561 6583 6604 6626 6648 6670 6691 48 22 8 17.6 16.8 16.0 42 6691 6713 6734 6756- 6777 6799 6820 47 22 9 19.8 18.9 18.0 43 6820 6841 6862 6884 6905 6926 6947 46 21 44 6947 6967 6988 7009 7030 7050 7071 45 21 60' 40' 30' 20' 10' d. P.P. Numerical values of the cosine function, 45 to 90. TABLES 509 Numerical values of the sine function, 45 to 90. (K 10' 20' 30' 40' 50' 60' (I P. P. 45 0.7071 7092 7112 7133 7153 7173 7193 44 20 M 7193 7214 7234 7254 7274 7294 7314 43 20 21 20 1 18 47 7314 7:3 7353 7373 7392 7412 7431 4',' 20 1 21 20 19 18 4S 7431 7451 7470 7490 7509 7528 7547 41 19 2 4.2 4.0 3.8 3.6 49 7547 7566 7585 7604 7623 7642 7660 40 19 1 6.3 6.0 5.7 5.4 1 8.4 8.0 7.6 7.2 5 10.5 10.0 9.5 9.0 51) 0.7660 7679 7698 7716 7735 7753 7771 3 18 a 12.6 12.0 11.4 10.8 51 n 5:1 7771 7880 7986 7790 7898 8004 7808 7916 8021 7826 7934 8039 7844 7951 8056 7862 7969 8073 7880 7986 8090 :is 37 ::<; 18 18 17 7 s ! 14 7 14.0 13.3 12.6 16.8 16.0 15.2 14.4 18.9 18.0 17.1 16.2 54 8090 8107 8124 8141 8158 8175 8192 35 17 17 16 15 14 H 0.8192 8208 8225 8241 8258 8274 8290 34 16 1 1.7 1.6 1.5 1.4 5*i 8290 S307 8323 8339 8355 8371 8387 :w 16 1 3.4 3.2 3.0 2.8 57 8387 8403 8418 8434 8450 8465 '8480 3'i 16 :i 4 5.1 4.7 4.5 4.2 68 64 60 56 M 8480 8496 8511 8526 8542 8557 8572 31 15 5 8.5 S.O 7.5 7.0 m 8572 8587 8601 8616 B831 8646 8660 30 15 i 10.2 9.6 90 8.4 7 11.9 11.2 10.5 9.8 8 13.6 12.8 12.0 11.2 60 0.8660 8675 8689 8704 8718 8732 8746 ','9 14 i 15.3 14.4 13.5 12.6 61 8746 8760 8774 8783 8802 8816 8829 >s 14 n 8829 ssi:i 8857 8870 8884 8897 8910 iJ 14 <; 13 13 12 11 10 H 8988 9001 9013 9026 9038 9051 9063 25 12 i 1.3 1.2 1.1 1.0 i 2.6 2.4 2.2 2.0 :i 3.9 3.6 3.3 3.0 H 0.9063 9075 9088 9100 9112 9124 9135 >4 12 -I 5.2 4.8 4.4 4.0 N 9135 9147 9159 9171 9182 9194 9205 23 12 5 6.5 6.0 55 5.0 7 6H 020:> 9272 9216 9283 9228 9293 9239 9304 9250 9315 9261 O.S2.-> 9272 9336 n 21 11 11 a 7 7.8 7.2 6.6 6.0 9.1 8.4 7.7 7.0 104 96 88 80 6 9336 9346 9356 9367 9377 9387 9397 io 10 1 11.7 10.8 9.9 9.0 30 7 70 0.9397 9407 9117 9426 9436 9446 9455 19 10 71 9455 '. 14 (,:> 9474 9483 9492 9502 9511 IS 9 n 9511 9520 9528 9537 9546 9555 9563 17 9 73 9563 9572 9.-.SO 9588 9596 9605 9613 16 8 74 9613 9621 9628 9636 9644 9652 9659 15 8 75 0.9659 9667 9674 9681 9689 9696 9703 14 7 7<; 9703 9710 9717 9724 !I7:50 9737 9744 13 7 77 9744 9750 9757 9763 9769 9775 9781 n 6 7S 9781 9787 9793 9799 9805 9811 9816 n 6 n 9816 9822 9827 9833 9838 9843 9848 10 5 Interpolate men- tally, using the mul- 89 0.9848 9853 9858 9863 9868 9872 9877 9 5 tiplication table. SI 9877 9881 9886 9890 9894 9899 9903 8 4 S' 9903 Wl>7 9911 9914 9918 9922 9925 7 4 S3 rasa 9932 9936 9939 9942 994f> 6 3 S4 9945 9948 9951 9954 9957 9959 9962 & 3 S5 0.9962 9964 9967 9969 9971 9974 9976 4 2 H6 9976 9978 9980 9981 9983 9985 9986 1 2 S7 M86 9988 9989 9990 9992 9993 9994 1 1 HS '.!''!.-, 9996 9997 9997 9998 9998 1 1 S9 MNM 9999 9999 *0000 *0000 *0000 *0000 60' 50' 40' 30' W (1 P.P. Numerical values of the cosine function, to 45. 510 UNIFIED MATHEMATICS Numerical values of the tangent function, to 45. 0' 10' 20' 30' 40' 50' 60' d. P. P. 0.0000 0029 0058 0087 0116 0145 0175 89 29 29 30 31 32 33 1 0175 0204 0233 0262 0291 0320 0349 88 29 1 2.9 3.0 3.1 3.2 3.3 2 0349 0378 0407 0437 0466 0495 0524 87 29 2 5.8 6.0 6.2 6.4 6.6 3 0524 0553 0582 0612 0641 0670 0699 86 29 3 8.7 9.0 9.3 9.6 9.9 4 0699 0729 0758 0787 0816 0846 0875 85 29 4 11.6 12.0 12.4 12.8 13.2 5 14.5 15.0 15.5 16.0 16.5 6 17.4 18.0 18.6 19.2 19.8 5 6 0.0875 1051 0904 1080 0934 1110 0963 1139 0992 1169 1022 1198 1051 1228 84 83 29 30 7 20.3 21.0 21.7 22.4 23.1 8 23.2 24.0 24.8 25.6 26.4 9 26.1 27.0 27.9 28.8 29.7 7 1228 1257 1287 1317 1346 1376 1405 82 30 8 1405 1435 1465 1495 1524 1554 1584 81 30 34 35 36 37 38 9 1584 1614 1644 1673 1703 1733 1763 80 30 1 3.4 3.5 3.6 3.7 3.8 2 6.8 7.0 7.2 7.4 7.6 10 11 0.1763 1944 1793 1974 1823 2004 1853 2035 1883 2065 1914 2095 1944 2126 79 78 30 30 3 10.2 10.5 10.8 11.1 11.4 4 13.6 14.0 14.4 14.8 15.2 5 17 17 5 18 18 5 19 12 2126 2156 2186 2217 2247 2278 2309 77 30 6 20.4 21.0 21.6 22.2 22.8 13 2309 2339 2370 2401 2432 2462 2493 76 31 7 23.8 24.5 25.2 25.0 26.6 14 2493 2524 2555 2586 2617 2648 2679 75 31 8 27.2 28.0 28.8 29.6 30.4 9 30.6 31.5 32.4 33.3 34.2 15 0.2679 2711 2742 2773 2805 2836 2867 74 31 39 40 41 42 43 16 17 2867 3057 2899 3089 2931 3121 3962 3153 2994 3185 3026 3217 3057 3249 73 72 32 32 1 3.9 4.0 4.1 4.2 4.3 2 78 80 82 84 86 18 3249 3281 3314 3346 3378 3411 3443 71 32 3 11.7 12'.0 12.3 12.6 12.9 19 3443 3476 3508 3541 3574 3607 3640 70 33 4 15.6 16.0 16.4 16.8 17.2 5 19.5 20.0 20.5 21.0 21.5 6 23.4 24.0 24.6 25.2 25.8 20 0.3640 3673 3706 3739 3772 3805 3839 69 33 7 27.3 28.0 28.7 29.4 30.1 21 3839 3872 3906 3939 3973 4006 4040 68 34 8 31.2 32.0 32.8 33.6 34.4 22 4040 4074 4108 4142 4176 4210 4245 67 34 9 35.1 36.0 36.9 37.8 38.7 23 4245 4279 4314 4348 4383 4417 4452 66 34 24 4452 4487 4522 4557 4592 4628 4663 65 35 44 45 46 47 48 1 4.4 4.5 4.6 4.7 4.8 30' 2 8.8 9.0 9.2 9.4 9.6 3 13.2 13.5 13.8 14.1 14.4 25 0.4663 4699 4734 4770 4806 4841 4877 64 36 4 17.6 18.0 18.4 18.8 19.2 5 22 22 5 23 23 5 24.0 26 4877 4913 4950 4986 5022 5059 5095 63 36 6 26.4 27.0 27.6 28.2 28.8 27 5095 5132 5169 5206 5243 5280 5317 62 37 7 30.8 31.5 32.2 32.9 33.6 28 29 5317 5543 5354 5581 5392 5619 5430 5658 5467 5696 5505 5735 5543 5774 61 60 38 38 8 35.2 36.0 36.8 37.6 38.4 9 39.6 40.5 41.4 42.3 43.2 49 50 51 52 53 30 0.5774 5812 5851 5890 5930 5969 6009 59 39 1 4.9 5.0 5.1 5.2 5.3 31 6009 6048 6088 6128 6168 6208 6249 58 40 2 9^8 lo!o 10'2 10^4 lo!6 32 6249 6289 6330 6371 6412 6453 6494 57 41 3 14.7 15.0 15.3 15.6 15.9 33 6494 6536 6577 6619 6661 6703 6745 56 42 4 19.6 20.0 20.4 20.8 21.2 34 6745 6787 6830 6873 6916 6959 7002 55 43 5 24.5 25.0 25.5 26.0 26.5 6 29.4 30.0 30.6 31.2 31.8 7 34.3 35.0 35:7 36.4 37.1 35 36 0.7002 7265 7046 7310 7089 7355 7133 7400 7177 7445 7221 7490 7265 7536 54 53 44 45 8 39.2 40.0 40.8 41.6 42.4 9 44.1 45.0 45.9 46.8 47.7 37 7536 7581 7627 7673 7720 7766 7813 52 46 38 7813 7860 7907 7954 8002 8050 8098 51 48 54 55 56 57 58 39 8098 8146 8195 8243 8292 8342 8391 50 49 1 5.4 5.5 5.6 5.7 5.8 2 10.8 11.0 11.2 11.4 11.6 3 16.2 16.5 16.8 17.1 17.4 40 0.8391 8441 8491 8541 8591 8642 8693 49 50 4 21.6 22.0 22.4 22.8 23.2 41 42 43 44 8693 9004 9325 9657 8744 9057 9380 9713 8796 9110 9435 9770 8847 9163 9490 9827 8899 9217 9545 9884 8952 9271 9601 9942 9004 9325 9657 *0000 48 47 46 45 52 54 55 57 5 27.0 27.5 28.0 28.5 29.0 6 32.4 33.0 33.6 34.2 34.8 7 37.8 38.5 39.2 39.9 40.6 8.43.2 44.0 44.8 45.6 46.4 9 48.6 49.5 50.4 51.3 52.2 60' 50' 40' 30' 20' IV 0' d. P.P. Numerical values of the cotangent function, 45 to 90. TABLES 511 Numerical values of the tangent function, 45 to 90. 10' 20' 30' 40' 50' 60' d. P.P. 45 1.000 1.006 1.012 1.018 1.024 1.030 1.03644 6 6 7 8 9 10 11 46 1.036 1.042 l.OIS 1.054 1.060 1.066 1.07243 6 1060708091011 47 1.072 1.079 1.085 1.091 1.098 1.104 1.11142 6 2 1.2 1.4 1.8 1.8 2.0 2.2 48 1.111 1.117 1.124 1.130 1.137 1.144 1.15041 6 3 1.8 2.1 2.4 2.7 3.0 3.3 4 1.150 1.157 1.104 1.171 1.178 1.185 1.19240 7 4 2.4 2.8 3.2 3.6 4.0 4.4 5 3.0 3.5 4.0 4.5 5.0 5.5 6 3.6 4.2 4.8 5.4 6.0 6.6 50 1.192 1.199 1.206 51 1.235 1.242 1.250 52 1.280 1.288 1.295 1.213 1.257 1.303 1.220 1.265 1.311 1.228 1.23539 7 1.272 1.28038 8 1.319 1.32737 8 7 4.2 4.9 5.6 6.3 7.0 7.7 8 4.8 5.6 6.4 7.2 8.0 8.8 9 5.4 6.3 7.2 8.1 9.0 9.9 53 1.327 1.335 1.343 1.351 1.360 1.368 1.37636 8 54 1.376 1.385 1.393 1.402 1.411 1.419 1.42835 9 12 13 14 15 16 1 1.2 1.3 1.4 1.5 1.6 2 2.4 2.6 2.8 3.0 3.2 55 1.428 1.437 1.446 1.455 1.464 1.473 1.48334 9 3 3.6 3.9 4.2 4.5 4.8 56 1.IS3 1.492 1.501 i.an 1.520 1.530 1.5403310 4 4.8 5.2 5.6 6.0 6.4 57 1.540 1.550 1.560 1.570 1.580 1.500 1.6003210 5 6.0 6.5 7.0 7.5 8.0 6 72 78 84 90 96 58 1.600 1.611 1.021 1.632 1.643 1.653 1.66431 11 7 8^4 9.1 9.8 10.5 11.2 59 1.664 1.675 1.686 1.698 1.709 1.720 1.7323011 8 9.6 10.4 11.2 12.0 12.8 9 10.8 11.7 12.6 13.5 14.4 60 1.732 1.744 1.756 1.767 1.780 1.792 1.804 29 12 n|w 1O *>ll 91 61 1.804 1.816 1.829 1.842 1.855 1.868 1.881 28 13 19 1 ' V . 1 62 1.881 1.894 1.907 1.921 1.935 1.949 1.963 27 14 1 1.7 1.8 1.9 2.0 2.1 63 1.963 1.977 1.991 2.006 2.020 2.035 2.050 26 14 2 3.4 3.6 3.8 4.0 4.2 3 51 54 57 60 63 6ft 2.050 2.066 2.081 2.097 2.112 2.128 2.145 25 16 4 6^8 7^2 7^6 8.0 8.4 5 8.5 9.0 9.5 10.0 10.5 6102108114120 12 6 65 2.145 2.161 2.177 2.194 2.211 2.229 2.246 24 17 7 11.9 12.6 13.3 14.014.7 66 2.246 2.264 2.282 2.300 2.318 2.337 2.356 23 18 8 13.6 14.4 15.2 16.0 16.8 67 2.356 2.375 2.394 2.414 2.434 2.455 2.475 22 20 9 15.3 16.2 17.1 18.0 18.9 68 2.475 2.496 2.517 2.539 2.560 2.583 2.6052122 69 2.605 2.628 2.651 2.675 2.699 2.723 2.7472024 22 23 2* 25 26 1 2.2 2.3 2.4 2.5 26 30' 2 4.4 4.6 4.8 5.0 5.2 3 6.6 6.9 7.2 7.5 7.8 76 2.747 2.773 2.798 71 2.904 2.932 2.000 2.824 2.989 2.850 3.018 2.877 2.9041926 3.047 3.078 18 29 4 8.8 9.2 9.6 10.0 10.4 5 11.011.5 12.012.5 13.0 6 13.2 13.8 14.4 15.0 15.6 72 3.078 3.108 3.140 3.172 3.204 3.237 3.271 17 32 7 15.4 16.1 16.8 17.5 18.2 73 3.271 3.305 3.340 3.376 3.412 3.450 3.487 16 36 8 17.6 18.4 19.2 20.020.8 74 3.487 3.526 3.566 3.606 3.647 3.689 3.7321541 9 19.820.721.622.523.4 27 28 59 62 63 75 3.732 3.776 3.821 3.867 3.914 3.962 4.0111446 76 4.011 4.061 4.113 77 4.331 4.390 4.449 4.165 4.511 4.219 4.574 4.275 4.331 13 53 4.638 4.7051262 1 2.7 2.8 5.9 6.2 6.3 2 5.4 5.6 11.8 12.4 12.6 3 8.1 8.4 17.7 18.6 18 9 78 4.705 4.773 4.843 4.915 4.989 5.066 5.1451173 4 10.8 11.2 23.624.825.2 7t 5.145 5.226 5.309 5.396 5.485 5.576 5.671 10 88 5 13.5 14.029.531.031.5 6 16.2 16.8 35.4 37.2 37.8 7 18.9 19.641.3 43.444.1 80 5.671 5.769 5.871 5.976 6.084 6.197 6.314 9 8 21. 6 22.4 47.2 49.650.4 81 6.314 6.435 6.561 6.691 6.827 6.968 7.115 8 9 24.3 25.2 53.1 55.8 56.7 H'J 7.115 7.269 7.429 7.596 7.770 7.953 8.144 7 83 8.144 8.345 8.556 8.777 9.010 9.255 9.514 6 64 66 68 70 72 84 9.514 9.78810.078 10.385 10.712 11.059 11.430 5 1 6.4 6.6 6.8 7.0 7.2 2 12.8 13.2 13.6 14.0 14.4 f 3 19.2 19.8 20.421.021.6 8511.43011.82612.251 12.706 13.197 13.72714.301 4 425.626.427.228.028.8 86 14.301 14.1)24 15.<;05 ,H7 1 d. nsi 20.206 21.470 HM2s.o:i 31.242 34.368 16.350 22.904 iiS.ISS 17.169 24.542 42.964 18.075 10. OS1 :i 26.43228.636 2 49.10457.290 1 5 32.0 33.0 34.0 35.0 36.0 6 38.4 39.6 40.8 42.0 43.2 7 44.8 46.2 47.6 49.0 50.4 8 51 .2 52.8 54.4 56 57 6 8957.29068.75085.940 114.59 171.89 343.77 infinit. 957.659.461.263.064.8 <>o 50' 40' 30' 20' 10' P.P. Numerical values of the cotangent function, to 45. 512 rNIFIKI) MATHEMATICS Radian measure of angles, D to ISO" 1 or Length of arc in unit circle for angle Q- to 180 D at center. A" RADIANS A RADIANS RADIANS A RADIANS 1 2 3 0.017 0.035 0.052 4C 47 48 0.803 0.820 0.838 91 92 93 1.588 1.606 1.623 136 137 138 2.374 2.391 2.409 4 5 6 0.070 0.087 0.105 49 50 51 0.855 0.873 0.890 94 95 96 1.641 1.658 1.676 139 140 141 2.426 2.443 2.461 7 8 9 0.122 0.140 0.157 52 53 54 0.908 0.925 0.942 97 98 99 1.693 1.710 1.728 142 143 144 2.478 2.496 2.513 10 11 12 0.175 0.192 0.203 55 56 57 0.960 0.977 0.995 100 101 102 1.745 1.763 1.780 145 146 147 2.531 2.548 2.566 13 14 15 0.227 0.244 0.262 58 59 60 1.012 1.031 1.047 103 104 105 1.798 1.815 1.833 148 149 150 2.583 2.601 2.618 16 17 18 0.279 0.297 0.314 61 62 63 1.065 1.082 1.100 106 107 108 1.850 1.868 1.885 151 152 153 2.635 2.653 2.670 19 20 21 0.332 0.349 0.367 64 65 66 1.117 1.134 1.152 109 110 111 1.902 1.920 1.937 154 1.55 156 2.688 2.705 2.723 22 23 24 0.384 0.401 0.419 67 68 69 1.169 1.187 1.204 112 113 114 1.955 1.972 1.990 157 158 159 2.740 2.758 2.775 25 26 27 0.436 0.454 0.471 70 71 72 1.222 1.239 1.257 115 116 117 2.007 2.025 2.042 160 161 162 2.793 2.810 2.827 28 29 30 0.189 0.506 0.524 73 74 75 1.274 1.292 1.309 118 119 120 2.059 2.077 2.094 163 164 165 2.845 2.862 2.880 31 32 33 0.541 0.559 0.576 76 77 78 1.326 1.344 1.361 121 122 123 2.112 2.129 2.147 166 167 168 2.897 2.915 2.932 34 35 36 0.593 0.611 0.628 79 80 81 1.379 1.396 1.414 124 125 126 2.164 2.182 2.199 169 170 171 2.950 2.967 2.985 37 38 39 0.646 0.663 0.681 82 83 84 1.431 1.449 1.466 127 128 129 2.217 2.234 2.251 172 173 174 3.002 3.019 3.037 40 41 42 0.698 0.716 0.733 85 86 87 1.484 1.501 1.518 130 131 132 2.269 2.286 2.304 175 176 177 3.054 3.072 3.089 43 44 45 0.750 0.768 0.785 88 89 90 1.536 1.553 1.571 133 134 135 2.321 2.339 2.356 178 179 180 3.107 3.124 3.142 TABLES 513 Minutes as Decimals of Growth Function, e* One Degree or Seconds Decay Function, e~* as Decimals of One e* OR tf ~~* OB e- Minute X t log. i logc t Value log lo Value log* 1 .017 31 .517 0.0 oo 1.000 0.000 1.000 0.000 2 .033 32 .533 0.1 -2.303 1.105 0.043 0.905 9.957 3 .050 33 .550 0.2 -1.610 1.221 0.087 0.819 9.913 4 .067 34 .567 0.3 -1.204 1.350 0.130 0.741 9.870 5 .083 35 .583 0.4 -0.916 1.492 0.174 0.670 9.826 6 .100 36 .600 0.5 - 0.693 1.649 0.217 0.607 9.783 7 .117 37 .617 0.6 -0.511 1.822 0.261 0.549 9.739 8 .133 38 .633 0.7 -0.357 2.014 0.304 0.497 9.696 9 .150 39 .650 0.8 -0.223 2.226 0.347 0.449 9.653 10 .167 40 .667 0.9 -0.105 2.460 0.391 0.407 9.609 11 .183 41 .683 1.0 0.000 2.718 0.434 0.368 9.566 12 .200 42 .700 1.1 0.095 3.004 0.478 0.333 9.522 13 .217 43 .717 1.2 0.182 3.320 0.521 0.301 9.479 14 .233 44 .733 1.3 0.262 3.769 0.565 0.273 9.435 15 .250 45 .750 1.4 0.336 4.055 0.608 0.247 9.392 16 .267 46 .767 1.5 0.405 4.482 0.651 0.223 9.349 17 .283 47 .783 1.6 0.470 4.953 0.695 0.202 9.305 18 .300 48 .800 1.7 0.531 5.474 0.738 0.183 9.262 19 .317 49 .817 1.8 0.588 6.050 0.782 Q.165 9.218 20 .333 50 .833 1.9 0.642 6.686 0.825 0.150 9.175 21 .350 51 .850 2.0 0.693 7.389 0.869 0.135 9.131 22 .367 52 .867 2.1 0.742 8.166 0.912 0.122 9.088 23 .383 53 .883 2.2 0.788 9.025 0.955 0.111 9.045 24 .400 54 .900 2.3 0.833 9.974 0.999 0.100 0.001 25 .417 55 .917 2.4 0.875 11.02 1.023 0.091 8.958 26 .433 56 .933 2.5 0.916 12.18 1.086 0.082 8.914 27 .450 57 .950 2.6 0.956 13.46 1.129 0.074 8.871 28 .467 58 .967 2.7 0.993 14.88 1.173 0.067 8.827 29 .483 59 .983 2.8 1.030 16.44 1.216 0.061 8.784 30 .500 60 1.000 2.9 1.065 18.17 1.259 0.055 8.741 3.0 1.099 20.09 1.303 0.050 8.697 3.1 1.132 22.20 1.346 0.045 8.654 3.2 1.163 24.53 1.390 0.041 8.610 3.3 1.193 27.11 1.433 0.037 8.567 3.4 1.224 29.96 1.477 0.033 8.523 3.5 1.253 33.12 1.520 0.030 8.480 4.0 1.386 54.60 1.737 0.018 8.263 4.5 1.504 90.02 1.954 0.0111 8.046 5.0 1.609 148.4 2.171 0.0067 7.829 6.0 1.792 403.4 2.606 0.0025 7.394 7.0 1.946 1096.6 3.040 0.0009 6.960 8.0 2.079 2981.0 3.474 0.0003 6.526 9.0 2.197 8103.1 3.909 0.0001 6.091 10.0 2.303 22026. 4.343 0,0000 5.657 514 UXIFIED MATHEMATICS The accumulation of 1 at the end of n years. r n = (1 + i) M . Years. li%. *%. **%. 3%. 4%. 5%. 6%. Years. 1 1.0150 1.0200 1.0250 1.0300 1.0400 1.0500 1.0600 1 2 1.0302 1.0404 1.0506 1.0609 1.0816 1.1025 1.1236 2 3 1.0457 1.0612 1.0769 1.0927 1.1249 1.1576 1.1910 1 4 1.0614 1.0824 1.1038 1.1255 1.1000 1.2155 1.2625 4 5 1.0773 1.1041 1.1314 1.1593 1.2167 1.2763 1.3382 5 6 1.0934 1.1262 1.1597 1.1941 1.2653 1.3401 1.4185 6 7 1.1098 1.1487 1.1887 1.2299 1.3159 1.4071 1.5036 7 8 1.1265 1.1717 1.2184 1.2668 1.3686 1.4775 1.5938 8 9 1.1434 1.1951 1.2489 1.3048 1.4233 1.5513 1.6895 9 10 1.1605 1.2190 1.2801 1.3439 1.4802 1.6289 1.7908 10 11 1.1779 1.2434 1.3121 1.3842 1.5395 1.7103 1.8983 11 12 1.1956 1.2682 1.3449 1.4258 1.6010 1.7959 2.0122 12 13 1.2136 1.2936 1.3785 1.4685 1.6651 1.8856 2.1329 13 14 1.2318 1.3195 1.4130 1.5126 1.7317 1.9799 2.2609 14 15 1.2502 1.3459 1.4483 1.5580 1.8009 2.0789 2.3966 15 16 1.2690 1.3728 1.4845 1.6047 1.8730 2.1829 2.5404 16 17 1.2880 1.4002 1.5216 1.6528 1.9473 2.2920 2.6928 17 18 1.3073 1.4282 1.5597 1.7024 2.0258 / 2.4066 2.8543 18 19 1.3270 1.4568 1.5987 1.7535 2.1068 2.5270 3.0256 19 20 1.3469 1.4859 1.6386 1.8061 2.1911 2.6533 3.2071 20 21 1.3671 1.5157 1.6796 1.8603 2.2788 2.7860 3.3996 21 22 1.3876 1.5460 1.7216 1.9161 2.3699 2.9253 3.6035 22 23 1.4084 1.5769 1.7646 1.9736 2.4647 3.0715 3.8197 23 24 1.4295 1.6084 1.8087 2.0328 2.5633 3.2251 4.0489 24 25 1.4509 1.6406 1.8539 2.0938 2.6658 3.3864 4.2919 25 26 1.4727 1.6734 1.9003 2.1566 2.7725 3.5557 4.5494 26 27 1.4948 1.7069 1.9478 2.2213 2.8834 3.7335 4.8223 27 28 1.5172 1.7410 1.9965 2.2879 2.9987 3.9201 5.1117 28 29 1.5400 1.7758 2.0464 2.3566 3.1187 4.1161 5.4184 29 30 1.5631 1.8114 2.0976 2.4273 3.2434 4.3219 5.7435 30 31 1.5865 1.8476 2.1500 2.5001 3.3731 4.5380 6.0881 31 32 1.6103 1.8845 2.2038 2.5751 3.5081 4.7649 6.4534 32 33 1.6345 1.9222 .2.2589 2.6523 3.6484 5.0032 6.8406 33 34 1.6590 1.9607 2.3153 2.7319 3.7943 5.2533 7.2510 34 35 1.6839 1.9999 2.3732 2.8139 3.9461 5.5160 7.6861 35 36 1.7091 2.0399 2.4325 2.8983 4.1039 5.7918 8.1473 36 37 1.7348 2.0807 2.4933 2.9852 4.2681 6.0814 8.6361 37 38 1.7608 2.1223 2.5557 3.0748 4.4388 6.3855 9.1543 38 39 1.7872 2.1647 2.6196 3.1670 4.6164 6.7048 9.7035 39 40 1.8140 2.2080 2.6851 3.2620 4.8010 7.0400 10.2857 40 50 2.1052 2.6916 3.4371 4.3839 7.1067 11.4674 18.4202 50 60 2.4432 3.2810 4.3998 5.8916 10.5196 18.6792 32.9877 60 70 2.8355 3.9996 5.6321 7.9178 15.5716 30.4264 59.0759 70 80 3.2907 4.8754 7.2096 10.6409 23.0498 49.6514 105.7960 80 90 3.8190 5.9431 9.2289 14.3005 34.1193 80.7304 189.4645 90 100 4.4321 7.2447 11.8137 19.2186 50.5050 131.5013 339.3021 100 Years. li%. 2%. z*%. 3%. 4%. 5%. 6%. Years. TABLES 515 The present value of 1 due in n years. v n (1 + \}~ n . Years. 11%. 2%. 1%. 3%. 4%. *%. 61 Years. 1 0.9852 0.9804 0.9750 0.9709 0.9615 0.9524 0.9434 1 2 0.9707 0.9012 0.9518 0.9426 0.9246 0.9070 0.8900 2 3 0.9563 0.9423 0.9286 0.9151 0.8890 0.8638 0.8396 3 4 0.9422 0.9238 0.9060 0.8885 0.8548 0.8227 0.7921 4 5 0.9283 0.9057 0.8839 0.8626 0.8219 0.7835 0.7473 5 6 0.9145 0.8880 0.8623 0.8375 0.7903 0.7462 0.7050 6 7 0.9010 0.8706 0.8413 0.8131 0.7599 0.7107 0.6651 1 8 0.8877 0.8535 0.8207 0.7894 0.7307 0.6768 0.6274 8 9 0.8746 0.8368 0.8007 0.7664 0.7026 0.6446 0.5919 9 10 0.8617 0.8203 0.7812 0.7441 0.6756 0.6139 0.5584 10 11 0.8489 0.8043 0.7621 0.7224 0.6496 0.5847 0.5268 11 12 0.8364 0.7885 0.7436 0.7014 0.6246 0.5568 0.4970 12 13 0.8240 0.7730 0.7254 0.6810 0.6006 0.5303 0.4688 13 14 0.8118 0.7579 0.7077 0.6611 0.5775 0.5051 0.4423 14 15 0.7999 0.7430 0.6905 0.6419 0.5553 0.4810 0.4173 15 10 0.7880 0.7284 0.6736 0.6232 0.5339 0.4581 0.3936 16 17 0.7764 0.7142 0.6572 0.6050 0.5134 0.4363 0.3714 17 18 0.7649 0.7002 0.6412 0.5874 0.4936 0.4155 0.3503 18 It 0.7536 0.6864 0.6255 0.5703 0.4746 0.3957 0.3305 19 '.Ml 0.7425 0.6730 0.6103 0.5537 0.4564 0.3769 0.3118 20 21 0.7315 0.6598 0.5954 0.5375 0.4388 0.3589 0.2942 21 22 0.7207 0.6468 0.5809 0.5219 0.4220 0.3418 0.2775 22 23 0.7100 0.6342 0.5667 0.5067 0.4057 0.3256 0.2618 23 24 0.6995 0.6217 0.5529 0.4919 0.3901 0.3101 0.2470 24 26 0.6892 0.6095 0.5394 0.4776 0.3751 0.2953 0.2330 25 26 0.6790 0.5976 0.5262 0.4637 0.3607 0.2812 0.2198 26 27 0.6690 0.5859 0.6134 0.4502 0.3468 0.2678 0.2074 27 28 0.6591 0.5744 0.5009 0.4371 0.3335 0.2551 0.1956 28 29 0.6494 0.5631 0.4887 0.4243 0.3207 0.2429 0.1846 29 30 0.6398 0.5521 0.4767 0.4120 0.3083 0.2314 0.1741 30 31 0.6303 0.5412 0.4651 0.4000 0.2965 0.2204 0.1643 31 32 0.6210 0.5306 0.4538 0.3883 0.2851 0.2099 (1.1. >.'>( I 32 :n 0.6118 0.5202 0.4427 0.3770 0.2741 0.1999 0.1462 33 34 0.6028 0.5100 0.4316 0.3660 0.2636 0.1904 0.1379 34 35 0.5939 0.5000 n. tin t 0.3554 0.2534 0.1813 0.1301 35 M 0.5851 0.4902 0.4111 0.3450 0.2437 0.1727 0.1227 36 37 0.5764 0.4806 0.4011 0.3350 0.2343 0.1644 0.1158 37 38 0.5679 0.4712 0.3913 0.3252 0.2253 0.1566 0.1092 38 39 0.5595 0.4620 0.3817 0.3158 0.2166 0.1491 0.1031 39 40 0.5513 0.4529 0.3724 0.3066 0.2083 0.1420 0.0972 40 50 0.4750 0.3715 0.2909 0.2281 0.1407 0.0872 0.0543 50 60 0.4093 0.3048 0.2273 0.1697 0.0951 0.0535 0.0303 60 70 0.3627 0.2500 0.1776 0.1263 0.0642 0.0329 0.0169 70 80 0.3039 0.2061 0.1387 0.0940 0.0434 0.0202 0.0095 80 90 0.2619 0.1683 0.1084 0.0699 0.0293 0.0124 0.0053 90 100 0.2256 0.1380 0.0846 0.0520 0.0198 0.0076 0.0029 100 Years. 11%. 2%. 21%. %. 4%. % e%. Years. 516 UNIFIED MATHEMATICS The accumulation of an annuity of 1 per annum at the end of n years. _(l + i)-l sn i Years. 1J%. 2%. 2}%. 3%. 4%. %. %. Years. 1 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1 0000 1 2 2.0150 2.0200 2.0250 2.0300 2.0400 2.0500 2.0600 2 3 3.0452 3.0604 3.0756 3.0909 3.1216 3.1525 3.1836 3 4 4.0909 4.1216 4.1525 4.1836 4.2465 4.3101 4 3746 4 5 5.1523 5.2040 5.2563 5.3091 5.4163 5.5256 5.6371 5 6 6.2296 6.3081 6.3877 6.4684 6.6330 6.8019 6.9753 6 7 7.3230 7.4343 7.5474 7.6625 7.8983 8.1420 8.3938 7 8 8.4328 8.5830 8.7361 8.8923 9.2142 9.5491 9.8975 8 9 9.5593 9.7546 9.9545 10.1591 10.5828 11.0266 11.4913 t 10 10.7027 10.9497 11.2034 11.4638 12.0061 12.5779 13.1808 10 11 11.8633 12.1687 12.4835 12.8078 13.4864 14.2068 14.9716 11 12 13.0412 13.4121 13.7956 14.1920 15.0258 15.9171 16.8699 12 13 14.2368 14.6803 15.1404 15.6178 16.6268 17.7130 18.8821 13 14 15.4504 15.9739 16.5190 17.0863 18.2919 19.5986 21.0151 14 15 16.6821 17.2934 17.9319 18.5989 20.0236 21.5786 23.2760 15 16 17.9324 18.6393 19.3802 20.1569 21.8245 23.6575 25.6725 16 17 19.2014 20.0121 20.8647 21.7616 23.6975 25.8404 28.2129 17 18 20.4894 21.4123 22.3863 23.4144 25.6454 28.1324 30.9057 18 19 21.7967 22.8406 23.9460 25.1169 27.6712 30.5390 33.7600 19 20 23.1237 24.2974 25.5447 26.8704 29.7781 33.0660 36.7856 20 21 24.4705 25.7833 27.1833 28.6765 31.9692 35.7193 39.9927 21 ifi, 25.8376 27.2990 28.8629 30.5368 34.2480 38.5052 43.3923 22 23 27.2251 28.8450 30.5844 32.4529 36.6179 41.4305 46.9958 23 24 28.6335 30.4219 32.3490 34.4265 39.0826 44.5020 50.8156 24 25 30.0630 32.0303 34.1578 36.4593 41.6459 47.7271 54.8645 25 26 31.5140 33.6709 36.0117 38.5530 44.3117 51.1135 59.1564 26 27 32.9867 35.3443 37.9120 40.7096 47.0842 54.6691 63.7058 27 28 34.4815 37.0512 39.8598 42.9309 49.9676 58.4026 68.5281 28 29 35.9987 38.7922 41.8563 45.2189 52.9663 62.3227 73.6398 29 30 37.5387 40.5681 43.9027 47.5754 56.0849 66.4389 79.0582 30 31 39.1018 42.3794 46.0003 50.0027 59.3283 70.7608 84.8017 31 32 40.6883 44.2270 48.1503 52.5028 62.7015 75.2988 90.8898 32 33 42.2986 46.1116 50.3540 55.0778 66.2095 80.0638 97.3432 33 34 43.9331 48.0338 52.6129 57.7302 69.8579 85.0670 104.1838 34 35 45.5921 49.9945 54.9282 60.4620 73.6522 90.3203 111.4348 35 36 47.2760 51.9944 57.3014 63.2759 77.5983 95.8363 119.1209 36 37 48.9851 54.0343 59.7339 66.1742 81.7022 101.6281 127.2681 37 38 50.7199 56.1149 62.2273 69.1594 85.9703 107.7095 135.9042 38 39 52.4807 58.2372 64.7830 72.2342 90.4092 114.0950 145.0585 39 40 54.2679 60.4020 67.4026 75.4013 95.0255 120.7998 154.7620 40 50 73.6828 84.5794 97.4843 112.7969 152.6671 209.3480 290.3359 50 60 96.2147 114.0515 135.9916 163.0534 237.9907 353.5837 533.1282 60 70 122.3638 149.9779 185.2841 230.5941 364.2905 588.5285 967.9322 70 80 152.7109 193.7720 248.3827 321.3630 551.2450 971.2288 1746.5999 80 90 187.9299 247.1567 329.1543 443.3489 827.9833 1594.6073 3141.0752 90 100 228.8030 312.2323 432.5487 607.2877 1237.6237 2610.0252 5638.3681 100 Years 11%. 2%. **% 8% 4%. 5%. 6%. Years. TABLES The present value of an annuity of 1 for n years, 1 - v n 517 Years. U%. 2%. "i r| I 3%. 4%. 5%. 6%. Years. 1 0.9852 0.9804' 0.9756 0.9709 0.9615 0.9524 0.9434 1 2 1.9559 1.9416 1.9274 1.9135 1.8861 1.8594 1.8334 2 3 2.9122 2.8839 2.8560 2.8286 2.7751 2.7232 2.6730 3 4 3.8544 3.8077 3.7620 3.7171 3.6299 3.5460 3.4651 4 5 4.7827 4.7135 4.6458 4.5797 4.4518 4.3295 4.2124 5 6 5.6972 5.6014 5.5081 5.4172 5.2421 5.0757 4.9173 6 7 6.5082 6.4720 6.3494 6.2303 6.0021 5.7864 5.5824 7 8 7.4859 7.3255 7.1701 7.0197 6.7327 6.4632 6.2098 8 g 8.3808 8.1622 7.9709 7.7861 7.4353 7.1078 6.8017 9 10 9.2222 8:9826 8.7521 8.5302 8.1109 7.7217 7.3601 10 11 10.0711 9.7868 9.5142 9.2526 8.7605 8.3064 7.8869 11 12 10.9075 10.5753 10.2578 9.9540 9.3851 8.8633 8.3838 12 13 11.7315 11.3484 10.9832 10.6350 9.9856 9.3936 8.8527 13 14 12.5434 12.1062 11.6909 11.2961 10.5631 9.8986 9.2950 14 15 13.3432 12.8493 12.3814 11.9379 11.1184 10.3797 9.7122 15 16 14.1313 13.5777 13.0550 12.5611 11.6523 10.8378 10.1059 16 17 14.0076 14.2919 13.7122 13.1661 12.1657 11.2741 10.4773 17 18 15.6726 14.9920 14.3534 13.7535 12.6593 11.6896 10.8276 18 19 16.4262 15.6786 14.9789 14.3238 13.1340 12.0853 11.1581 19 20 17.1686 16.3514 15.5892 14.8775 13.5903 12.4622 11.4699 20 21 17.9001 17.0112 16.1845 15.4150 14.0292 12.8212 11.7641 21 22 18.6208 17.6580 16.7654 15.9369 14.4511 13.1630 12.0416 22 23 19.3309 18.2922 17.3321 16.4436 14.8568 13.4886 12.3034 23 24 20.0301 18.9139 17.8850 16.9355 15.2470 13.7986 12.5504 24 25 20.7196 19.5235 18.4244 17.4131 15.6221 14.0940 12.7834 25 26 21.3986 20.1210 18.9.506 17.8768 15.9828 14.3752 13.0032 26 27 22.0676 20.7069 10 4040 18.3270 16.3296 14.6430 13.2105 27 28 22.7267 21.2813 19.9649 18.7641 16.6631 14.8981 13.4062 28 29 23.3761 21.8444 20.4535 19.1885 16.9837 15.1411 . 13.5907 29 30 24.0158 22.3965 20.9303 19.6004 17.2920 15.3725 13.7648 30 31 24.6461 22.9377 21.3954 20.0004 17.5885 15.5928 13.9291 31 32 25.2671 23.4683 21.8492 20.3888 17.8736 15.8027 14.0840 32 33 25.8790 23.0886 22.2919 20.7638 18.1476 16.0025 14.2302 33 34 26.4817 24.4986 22.7238 21.1318 18.4112 16.1929 14.3681 34 35 27.0756 24.9986 23.1452 21.4872 18.6646 16.3742 14.4982 35 36 27.6607 25.4888 23.5563 21.8323 18.9083 16.5469 14.6210 36 37 28.2371 25.9695 23.9573 22.1672 19.1426 16.7113 14.7368 37 38 28.8051 26.4406 24.3486 22.4925 19.3679 16 8679 14.8460 38 39 "i :it>ti'i 26.9026 24.7303 22.8082 19.5845 17!0170 14.9491 39 40 20.0158 27.8566 25.1028 23.1148 19.7928 17.1591 15.0463 40 50 34.9997 31.4236 28.3623 25.7298 21.4822 18.2559 15.7619 50 60 30.3803 34.7609 30.9087 27.6756 22.6235 18.0203 16.1614 60 70 43.1549 37.4987 32.8979 29.1234 23.3945 19.3427 16.3845 70 80 46.4073 39.7445 34.4518 30.2008 23.91.54 19.5965 16.5091 80 90 49.2099 41.5869 35.6658 31.0024 24.2673 19.7523 16.5787 90 100 51.6247 43.0983 36.6141 31.5989 24.5050 19.8479 16.6175 100 Years. 1%. 21%. 3%. 4%. 5%. 6' , . Years. 518 UNIFIED MATHEMATICS The annual sinking fund which will accumulate to 1 at the end of n years. 1 s > wl C 1 " M)"- - JL w * ,/ k? IC&JLU a O.U.U. C. 5] M*W