LIBRARY OF THE University of California. Class Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/computationmensuOOIambrich COMPUTATION AND MENSURATION THE MACMILLAN COMPANY NEW YORK • BOSTON • CHICAGO ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO COMPUTATION AND MENSURATION BY P. A. LAMBERT, M.A. PROFESSOR OF MATHEMATICS, LEHIGH UNIVERSITY NciM fork THE MACMILLAN COMPANY 1907 AU rights reserved % Copyright, 1907, By the MACMILLAN COMPANY. Set up and electrotyped. Published September, 1907. J. S. Cushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE The transition from secondary school to college is disastrous for many students. This is due largely to the fact that the student has not been taught to make inde- pendent use of what he has learned. The transition should be accompanied by training in the application of the knowledge gained in the secondary school. Such training serves a threefold purpose — it makes the student realize that he has acquired increased power, it reviews the work of the secondary school in an interest- ing manner, and it gives an outlook into the work of the college. Such a transition course in mathematics the author believes may be based on this text on Computation and Mensuration. This transition course would naturally come at the end of the secondary school course or at the beginning of the college course. * The student is expected to refer constantly to his texts on Algebra, Geometry, and Trigonometry. Formulas derived in all elementary text-books are neither proved nor tabulated in this text. The aim here is to build on the foundation already laid. 165869 CONTENTS CHAPTER I Approximate Computation ARTICLE PAGK 1. Approximate Numbers 1 2. Direct Measurement 3 3. Indirect Measurement 3 4. Approximate Multiplication 5 5. Approximate Division 9 CHAPTER II Graphic Computation 6. Graphic Representation of Numbers 12 7. Addition and Subtraction of Segments 13 8. Multiplication and Division of Seginents .... 14 J^. Square Hoot of a Segment 16 10. Graphic Computation of Areas 17 11. Graphic Solution of Triangles 18 • CHAPTER III The Method of Coordinates 12. The Plane Coordinates of a Point 20 13. Area of a Triangle in Coordinates of Vertices ... 21 14. Areas of Any Rectilinear Figures 25 CHAPTER IV Volumes of Solids bounded by Planes 15. The Prismatoid 27 16. Special Cases of Prismatoids . . . . . , . .29 Vlll CONTENTS CHAPTER V Computation and Use of Trigonometric Functions ARTICLE PAGE ,17. On Angles 32 18. Trigonometric Functions 34 19. Computation of Trigonometric Functions . . . .36 20. On Vectors 39 CHAPTER VI Computation and Use of Logarithms 21. Nature of Logarithms . 22. Computation of Common Logarithms 23. Arrangement of Tables of Logarithms 24. Computation by Means of Logarithms 25. The Compound Interest Formula . 26. The Slide Rule .... 44 45 50 54 59 62 CHAPTER VII On Limits ^ 27. The Infinite Decreasing Geometric Progression ... 65 28. The Length of a Curved Line 66 29. The Computation of tt 67 30. An Important Limit 69 31. Length and Area of Involute of Circle 70 CHAPTER VIII Graphic Algebra 32. The Graph of an Equation 73 33. Equations of Lines 75 34. Graphic Solution of Equations 76 35. Inequalities treated Graphically . • . . . .78 CONTENTS IX CHAPTER IX Areas bounded by Curves ARTICLE PAGE 36. Exact Areas . . . . " 80 37. Approximate Equation of a Curve 82 38. Approximate Areas 83 CHAPTER X Volumes of Solids 39. Exact Volumes 87 40. Approximate Volumes 88 41. Applicability of Prismoidal Formula 90 COMPUTATION AND MENSURATION CHAPTER I APPROXIMATE COMPUTATION Art. 1. — Approximate Numbers Frequently numbers occurring in computation are known only approximately. If the length of a line is to be measured, it may be absurd, either from the nature of the problem to be solved or the means of measurement employed, to attempt to de- termine the length of the line beyond hundredths of a foot. The square root of a number not a perfect square can- not be expressed exactly in numbers, but its approximate value to any desired number of decimal places may be found. For example, the approximate values of V5 are as follows; to one decimal place 2.2''" to two decimal places 2.24" to three decimal places 2.236"'' to four decimal places 2.2361" to five decimal places 2.23607" to six decimal places 2.236068" to seven decimal places 2. 2360680" * *In these approximate values of Vo the figure in the last decimal place is always the nearest integer. B 1 2 COMPUTATION AND MENSURATION In each approximate value the sign + or — written after the last figure indicates that the error must be respectively added or subtracted. The error in every case is not greater than 5 units of the decimal place next following the last decimal place of the approximation. The approximate value of V5 to eight decimal places cannot be determined until the figure in the ninth decimal place has been computed. The ratio of the circumference of a circle to its diam- eter cannot be expressed exactly in numbers, but its value has been computed to 707 decimal places. This ratio is denoted by tt, and its value to 15 decimal places is 7r= 3.141592653589793. The value of tt most commonly used is 3.1416. Problem l. Find the limit of error when ^^- is used for TT. Pkoblem 2. Find the limit of error when ^^J is used for TT. Problem 3. Find the limits of error of the sums of V26, V27, V28, V29, each computed approximately to four decimal places*. Problem 4. The yard is defined by law as g^^ meter.* Find the number of meters in a yard correct to 4 decimals. Problem 5. The pound is defined by law as 2:204^2 kilo- gram.* Find the number of kilograms in a pound correct to 4 decimals. * By decision oi United States Superintendent of Weights and Meas- ures,^ on April 5, 1903, with the approval of the Secretary of the Treasury. APPROXIMATE COMPUTATION 3 Art. 2. — Direct Measurement The direct measurement of the length of a line is effected by actually applying to the line the standard unit of length. However carefully several direct measure- ments are made, the results are generally not the same. By universal agreement, the arithmetic mean of the re- sults of several direct measurements, made with the same care, is accepted as the best result to be obtained from these measurements. Problem 6. Eight measurements of a line give the values 186.4, 186.3, 186.2, 186.3, 186.2, 185.9, and 186.4 inches. Find the best value of the length of the line based on these measurements. Problem 7. Four measurements of an angle give 62° 25' 10". ; 62° 25' 1".5 ; 62° 25' 23". 3, and 62° 25' 30". 0. Find the best result determined by these measurements. Problem 8. Four measurements of a base line give 1472.34, 1471.99, 1472.25, and 1472.14 feet. Find the best approximate length of the line to be obtained from these measurements. Art. 3. — Indirect Measurement Example l. Suppose A, B, C to be three stations form- ing the vertices of a right triangle, and let it be required to determine the length of the hypotenuse AB when an obstruction between A and B makes the direct measure- ment of this length impracticable. If by direct measurement ^C= 1275.43 feet and 4 COMPUTATION And mensuration BC= 1526.56 feet, AB may be computed by means of the proposition in Geometry — The square on the hypotenuse equals the sum of the squares on the other two sides, or expressed as a formula, AB'^ = AO^ + BlP. The number expressing the length of AB is found by extracting a square root and may be carried out to any desired number of decimal places; but since the measured lines AC and BC are determined only to two decimals, it would be absurd to attempt to determine by computation the length of AB beyond two decimals. Example 2. Let it be required to determine the cir- cumference of a circle whose diameter is found by direct measurement to be 425.63 feet. The circumference of this circle is tt x 425.63 feet. Since the diameter of the circle is given to two decimals, it is impossible to determine by computation the circum- ference beyond two decimal places. It is therefore absurd to use more decimal places in the value of tt than are neces- sary to determine the product tt x 425.63 to two decimal places. What is needed is a method of approximate com- putation which will determine the circumference correct to two decimal places with the least amount of computation. Mensuration is chiefly concerned with problems of indirect measurement. Problem 9. In a right triangle the sides about the right angle measure 875.27 feet and 56b A5 feet. Com- pute the length of the hypotenuse. APPROXIMATE COMPUTATION 5 Problem lO. The diameter of a circle measures 365.18 feet. Compute the length of the circumference. Problem ll. The circumference of a circle measures 1000 feet. Compute the diameter. Art. 4. — Approximate Multiplication Example. Let it be required to compute the product TT >< 425.63 correct to two decimal places. The computation is carried out to three decimal places in order that the second decimal place may be corrected. The first step is to determine how many decimal places in the value of ir must be used to give three decimal places in the product. The computation consists in multiplying the approx- imate value of TT by each figure of 425.63 and adding the partial products. These partial products are to contain three decimal places each. • In order that the partial product of the approximate value of TT by the units' figure of 425.63 shall contain three decimal places, the value of tt must be written to three decimal places. For each integral place to the left of units' place in 425.63 the value of ir must be extended one decimal place to the right in order that the corresponding partial product shall contain three decimal places. In this example, therefore, the value of ir must be written to five decimal places, and in order that the partial product by the figure in the hundredths' place in 425.63 may be corrected in the third decimal place the value of TT is written to six decimal places. 6 COMPUTATION AND MENSURATION The dot placed over the figure in the fifth decimal place of TT indicates that this figure is the last figure to the right in the value of tt to be used to form the partial product corresponding to the first figure to the left of 425.63. As the figure of the multiplier for which a partial prod- uct is to be computed moves one place to the right, the last figure of the multiplicand to be used moves one place to the left. There is considerable advantage in forming the partial products for the figures of 425.63 from right to left. The last figures of the partial products are corrected for the rejected part of the value of tt. The signs + and — in- dicate the • direction of the errors committed by these corrections. The sign + indicates that the error is to be added, the sign — that the error is to be subtracted. The numerical value of the error cannot exceed J of a unit of the third decimal place. The computation now appears as follows : 3.141593 425.63 1256.637 + 62.832- 15.708- 1.885- .094+ 1337.156 The error of this result must lie between — IJ and + 1 units of the third decimal place. It is therefore uncertain if the figure in the second decimal place is 5 or 6. APPROXIMATE COMPUTATION 7 It is necessary in this particular problem to carry out the approximate multiplication to 4 decimals as follows : 3.1415927 425.63 1256.6371- 62.8319- 15.7080- 1.8850- .0942 + 1337.1562 The error of this result must lie between — 2 and + J units of the fourth decimal place. Hence the product correct to two places of decimals is 1337.16. The following arrangement of this computation is per- haps to be preferred. Place the units' figure of 425.63 under the third decimal place of the value of tt and then write in inverse order the figures of 425.53 under the figures of the approximate value of TT. This brings each figure of the multiplier in line with that figure of the multiplicand which is used to begin the corresponding partial product. The above ap- proximate multiplications appear as follows : 3.141593 3.1415927 36.524 36.524 1256 637-^ 12566 371- 62 832- 628 319- 15 708- 157 080- 1885- 18 850- 94+ 942 + 1337.156 1337.1562 8 COMPUTATION ANB MENSURATION Problem 12. Find the circumference of a shaft whose radius is 4.32 inches. Pkoblem 13. If a yard is 1.904 meters, find the num- ber of meters in 23.463 yards. Problem 14. Find the interest on 11525.75 for one year at 6|%, correct to f 0.01. Problem 15. The distance of the moon from the earth is 59.97 times the earth's radius. If the earth's radius is 3962.824 miles, find the distance to the moon correct within 1 mile. Problem 16. Find the area of the circle whose radius is 16:27 feet. Problem 17. Find the volume of the sphere whose radius is 3.53 feet. Problem 18. Find the volume of the cone whose base is a circle, radius 2.35 feet, altitude 5.75 feet. Problem 19. Find the volume of the rectangular par- allelopiped whose dimensions are 8.53 feet, 6.27 feet, and 4.65 feet. Problem 20. Find the length of the diagonal of the parallelopiped in Problem 19. Problem 21. Compute 168 V3 to two decimal places. Problem 22. Compute tt \/2 to three decimal places. Problem 23. Compute tt^ to four decimal places. APPROXIMATE COMPUTATION 9 Art. 5. — Approximate Division Example. Let it be required to compute correct to two places of decimals the diameter of the circle whose circumference measures 587.35 feet. In order that the figure in the second decimal place of the quotient may be controlled, the quotient of 587.35 divided by tt is computed to three places of decimals. By inspection the quotient has three integral places. Hence six figures of the quotient are to be computed. The nature of the problem shows that this computation requires six figures of the divisor and six figures in the dividend. The successive partial products are formed as in the process of approximate multiplication in the preceding article. The computation is arranged as follows : 587.350 3.14159+ 314 159 + 186.959 273 191 251 327 + 21864 18 850- 3 014 2 827- 187 157 + 30 28 + 2 The signs + and — placed after the partial products indicate the direction of the error of the partial products UNIVERSITY 10 COMPUTATION AND MENSURATION The error in the sum of the partial products must be between -j- 2 and — 1 units of the last decimal place of the dividend. Hence the last remainder lies between and -h 3 of these units, and the error of the quotient lies between and + 1 units of the third decimal place. Hence the quotient correct to two decimal places is 186.96. There is some advantage in arranging the figures of the quotient in such a manner that the partial products begin with the product of the figure of the quotient and the figure of the divisor directly over it. The computation then takes this form, 587.350 3.14159 + 314 159+ 959.681 273 191 251 327 + 21 864 18 850- 3 014 2 .827- 187 157 + 30 28 + 2 Problem 24. Find the radius of the circle whose cir- cumference is 425.76 feet. Problem 25. Find the area of the circle whose circum- ference is 628.32 feet. Problem 26. Find the circumference of the circle whose area is 3848.45 square feet. APPROXIMATE COMPUTATION 11 Problem 27. Find the diameter of a wheel which makes 373 revolutions in a mile. Problem 28. How many revolutions per mile are made by a locomotive drive wheel 4.5 feet in diameter ? Problem 29. Compute — - to four decimal places. Problem 30. Find correct to four decimal places the number of degrees in an angle of 1 radian. Problem 31. The moon revolves about the earth in 28 days, 7 hours, 43 minutes, 11.5 seconds. What is the average angle passed over in a day ? CHAPTER II GKAPHIO COMPUTATION Art. 6. — Graphic Eepresextatiox of Numbers On a straiglit line mark a series of equidistant points. Select one of these points and call it zero. Starting from zero point call the successive points in one direction + 1, + 2, + 3, + 4, 4- 5, ••• , the successive points in the oppo- site direction — 1, — 2, — 3, — 4, — 5, ••• . There is thus attached to each integral number, positive or negative, one point of the line. Insert between the points attached to two consecutive integral numbers, 3 and 4, for example, nine equidistant points and call these points 3.1, 3.2, 3.3, 3.4,-8.5, 3.6, 3.7, 3.8, 3.9. There is thus attached to each number contain- ing not more than one decimal place one point of the line. -8-7-6-5-4-3-2-1 9 1 ^ 3 4 5 6 7 8 H i 1 1 — i — t — I — I 1 1 1 1 1 — i 1 1 H , Fig. 1. Theoretically, this operation of inserting between the points attached to two consecutive numbers nine equidis- tant points may be repeated indefinitely, but practically little is gained by attempting to attach points to numbers containing more than one decimal place. Conversely, if any point is taken at random on the straight line, the number attached to the point can be de- 12 GUAPHIC COMPUTATION 13 teriiiined to one decimal place with considerable certainty, but the attempt to determine the number beyond one dec- imal place would be accompanied by great uncertainty. If the line segment joining the points attached to two consecutive whole numbers is taken as the unit of length, the number attached to any point of the line measures the length of the line segment from the zero point to this point. The algebraic sign of the number indicates the direction in which the line segment extends from the origin. The straight line with numbers attached to successive equidistant points now becomes a scale for measuring the lengths of line segments. Art. 7. — Addition and Subtraction of Segments A line segment is a definite part of a straight line and extending in a definite direction along the line. A line segment has an initial point and a terminal point. The segment extends from the initial point to the terminal point. A line segment is transferred from one straight line to another by means of dividers. Two line segments are equal when one can be trans- ferred into the other. When line segments are to be added or subtracted they are transferred to the same straight line. To add two line segments having the same sign, lay off the two segments on a straight line both in the same direction, making the terminal point of the first segment the initial point of the second. The line segment which 14 COMPUTATION AND MENSURATION now extends from the initial point of the first segment to the terminal point of the second is the sum of the two. To add two line segments having opposite signs, lay off the two segments on a straight line in opposite directions, making the terminal point of the first segment the initial point of the second. The line segment which now extends from the initial point of the first segment to the terminal point of the second is their sum. To subtract two line segments, change the sign of the segment to be subtracted and add to the other segment. To add or subtract numbers graphically, find from the scale of segments the line segments corresponding to the given numbers, then find the sum or difference of these line segments and from the scale determine the corre- sponding number. Art. 8, — Multiplication and Division of Segments To find the product of two line segments, define the product by the proportion product : multiplicand = multiplier : unity. The product of two line segments is therefore the line segment which is the fourth proportional to the two given line segments and the unit line segment. This fourth proportional is constructed by drawing' two straight lines through a common point making any con- venient angle, the positive direction on each line being indicated by the arrowhead, and laying oft' from the common point the unit segment and multiplier on the first GRAPHIC COMPUTATION 15 straight line, the multiplicand on the second straight line. Now join the terminal points of the unit segment and the multiplicand by a straight line and draw through the terminal point of the multiplier a parallel to this straight line. The point of intersection of this parallel with the second straight line is the terminal point, the common point the initial point of the product segment. In the figure 1^=^, hence 00= OA . OB. To find the quotient of two line segments, define the quotient by the proportion quotient : 1 = dividend : divisor. The quotient of two line segments is, therefore, the line segment which is the fourth proportional to the unit line segment and the two segments which are respectively the dividend and the divisor. This fourth proportional is constructed by drawing two straight lines through a common point, making any con- venient angle, laying off from the common point the dividend and divisor on the first straight line, the unit segment on the second straight line. Now join the ter- minal points of the unit segment and the divisor by a straight line and draw through the terminal point of the 16 COMPUTATION AND MENSURATION dividend a parallel to this straight line. The point where this parallel intersects the second straight line is the ter- minal point, the common point the initial point of the quotient segment. In the figure 00 OA , __ ^^ OA — = —, hence 0(7=—. To find graphically the product or the quotient of two numbers, find by means of the scale of segments the line segments corresponding to the given numbers. Then find the product or quotient of these line seg- ments and from the scale the number corresponding to the product or quotient segment. Problem 32. Find graphically the product of 8.6 and 7.3. Problem 33. Find graphically the quotient of 47.5 divided by 11.7. Art. 9. — Square Root of a Segment If the product of two equal line segments equals the given line segment, one of the equal segments is called the square root of the given line segment. GRAPHIC COMPUTATION 17 To construct the line segment which is the square root of the given line segment, find in any manner two line segments whose product equals the given line segment. On the sum of these two line segments as a diameter, con- struct a semicircumference. The perpendicular to the diameter at the point of meeting of the two line segments Fig. 4. and terminating in the circumference is the line segment required. Repeated applications of this operation make it possible to extract a root which is any power of 2 by the use of the straight line and circle. Problem 34. Find graphically V24. Problem 35. Add graphically Vl9, V2i, and V28. Art. 10. — Graphic Computation of Areas Construct as accurately as practicable the figure whose area is to be found on cross-section paper, taking the dis- tance between consecutive lines to represent the unit of length in the measurement of the figure. The square of the cross-section paper now represents the unit of area, and the area of the figure is determined by counting the squares wholly within the boundary line of the figure and 18 COMPUTATION AND MENSURATION estimating the squares partially within the boundary line. The accuracy of the result depends upon the skill in estimating the sum of the partial squares. Problem 36. Find graphically the area of the triangle whose sides are 8, 12, 15 feet. Problem 37. Find graphically the area of the circle whose diameter is 9 inches. Problem 38. Find graphically the area of the equi- lateral triangle whose side is 7 feet. Art. 11. — Graphic Solution of Triangles The data necessary for the solution of a triangle is sufficient for the construction of the triangle. If the triangle is constructed with the given parts as accurately as practicable by the aid of straight edge, pro- tractor, and dividers, by means of these same instruments the values of the unknown parts of the triangle can be found with considerable accuracy. Problem 39. Two sides of a triangle are 7.8 feet and 12.5 feet, their included angle is 28° W, Find the re- maining three parts graphically. Problem 40. Two sides of a triangle are 9.5 feet and 16.3 feet. The angle opposite the first side is 16° 15^ Find the other parts graphically. Problem 41. One side of a triangle is 10.6 feet, the adjacent angles are Qb° lb' and 80° 10^ Find graphically the other two sides. GRAPHIC COMPUTATION 19 Peoblem 42. The sides of a triangle are 18 feet, 25 feet, and 29 feet. Find the angles graphically. Problem 43. Two sides of a triangle are 4.6 feet and 12.8 feet. The angle opposite the former sides is 60° 30'. Find the remaining parts graphically. CHAPTER III THE METHOD OP OOORLINATES Art. 12. — The Plane Coordixates of a Poixt In a given plane draw two straight lines perpendicular to each other. On each of these lines, with their inter- section at zero point, construct a linear scale, denoting the Y positive direction (-3,2) (3.2) on each line by an arrowhead. Call these two lines of reference the X- -^ X axis and the y-axis. Select any point P in the plane, and through this point draw a line paral- FiG. 5. lei to the I^-axis to meet the X-axis. The number which expresses the distance and direction of this meeting point from the zero point is called the abscissa of the point P and is denoted by x. Through the point P draw a parallel to the X-axis to meet the F-axis. The number which expresses, the dis- tance and direction of this meeting point from the zero point is called the ordinate of the point and is denoted by y. The abscissa of a point may be defined as the number which expresses the distance and direction of the point 20 THE METHOD OF COORDINATES 21 from the y-axis measured on a parallel to the X-axis ; and the ordinate may be defined as the number which ex- presses the distance and direction of the point from the X-axis measured on a parallel to the y-axis. The abscissa of a point determines a straight line par- allel to the y-axis in which the point must be situated. The ordinate of a point determines a straight line par- allel to the X-axis in which the point must be situated. The abscissa and ordinate of a point together determine the point in the plane and are called the coordinates of the point. The point whose coordinates are a: = 3, ?/ = — 2 is called the point (3, - 2). Problem 44. Where is the point situated whose ab- scissa is 4? Problem 45. Where is the point situated whose ordi- nate is — 5 ? Problem 46. Locate the points (3, 5), (—4, 6), (0, 8). Problem 47. Construct the triangle whose vertices are the points (10, 4), (6, - 8), (15, 12). Problem 48. Locate the points (5, 8), (10, 7) and find the distance between them. Problem 49. Find the area of the triangle whose ver- tices are (3, 7), (8, 5), (13, 15). Art. 13. — Area of Triangle in Coordinates of Vertices Suppose the vertices of the triangle to be the points A (^r ^i)' A (^2' ^2)' ^3 C^3' ^3)- 22 COMPUTATION AND MENSURATION Two cases will be considered, Case I, when all the coor- dinates of the vertices are positive, Case II, when some of these coordinates are negative. It must be remembered that line segments parallel to the X-axis are positive when they extend in the positive direction of the X-axis, and line segments parallel to the y-axis are positive when they extend in the positive direction of the Z-axis. In the proposition The area of a trapezoid is half the sum of the parallel bases by the altitude, the bases and altitude are positive line segments. Y Fig. 6. Case I. The area of the triangle P^P^P^ equals the sum of the areas of the trapezoids P^D^P^P^^ and P^B^D^P^ minus the area of the trapezoid P^P^P^P^. Hence 2 Area triangle P^P^P^ = B^P^ x QB^P^ -f- B^P^) -F i>3Z>2 X (DgPg + D2P2) - -^1^2 X (i>iPi + AA)- Now x^ = OB^, x^ = OB^, x^ = ODg, Vi = ^i^v Vi = ^2^T Vz = A^a THE METHOD OF COORBINATES 23 Hence DiD^ =^s~~ ^v -^s^2 = ^2 ~ ^3^ -^1^2 = ^2 ~ ^v ^^^ 2 Area P^P^P^ = (2-3 - x^) (y^ + 7/^) + (^2 - ^3) C2/3 +^2) - (^2 - ^1) (^1 + ^2)- Multiplying out and simplifying, 2 Area P^P^P^ = x^y^^ + ^^^2^/3 + ^3^1 - ^xVz - ^2^/1 - ^3^2- This expression for double the area of a triangle in terms of the coordinates of its vertices is readily repro- duced by arranging the coordinates of the vertices ^1^1 in order in two vertical columns and repeating at ^2^2 the end of the column the coordinates of the first ^^ point. The three positive terms of the area are the ^ ^ products of each of the three abscissas by the next lower ordinate, and the three negative terms of the area are the products of each of the three ordinates by the next lower abscissa. The area of the triangle is positive when the perimeter is supposed to be described in such a manner that the area lies to the left of the boundary line. 24 COMPUTATION AND MENSURATION Case II. The area of the triangle P^P^P^ equals the area of the trapezoid P^E^E^P^ minus the sum of the areas of the triangles P^E^P^ and P^E^Py Hence 2 Area triangle P.P^P, = (P^E,^ + ^3^3) x A^a - P^E^ X E^P, - P^E, X P,E,, Now x^ = OZ>i, x^= —B^ 0, x^ = 02>3, ^1 = A^r ^2 = - A^2' 3/3 = - ^3^3- Hence P^E^= - y^ ^y^^ P^E^ = - «/3 + Vv D^D^ = -x^-h x^, E^P^ = —x^ + x^, P^E^ = ^3 - ^v and 2 Area P^P^P^ = (2 ^1 - ^2 - 2/3) ( ^ ^1 + ^3) - (- 3/2 + 2/1) (- ^2 + ^1) - (- ^3 + ^1) (^3 - ^i)- This reduces to 2 Area P^P^P^ = ^1^2 + ^2^/3 + ^3^1 - ^1^2 - V'i^z - Vz^v the same result as was found for Case I. This expression for double the area of a triangle in terms of the coordinates of its vertices is entirely general. Problem 50. Find the area of the triangle whose vertices are the points (12, — 5), ( — 8, 7), (10, 15), the coordinates being measured in inches. Problem 51. Find the area of the triangle the co- ordinates of whose vertices measured in chains are (15.75, 4.26), (18.25, 20.63), (21.43, 16.52). Problem 52. Show that the points (1, 4), (3, 2), (—3, 8) lie in a straight line. THE METHOD OF COORDINATES 25 Art. 14. — Areas of Any Eectilinear Figures Let PiCx^^i). ^(^2^2)' ^3(^3^3)' -^4(^4^4) be the vertices of a quadrilateral. The straight line P1P3 divides this quadrilateral into the triangles P^P^P^ and 2 Area P^P^P^ = x^y^ + 2:2^3 + x^^ - y^x^^ - y^x^ - y^x^. 2 Area P^P^P^ = x^y^ + x^^ + x^y^ - y^x^ - y^x^ - y^Xy Hence 2 Area P^PJP^P^ = x^y^-\- x^^ + x^^ + x^y^ - y^x^ - ^2^3 - ^3^4 - ^4^1- ^1^1 ^22/2 ^4^4 This expression for double the area of a quadrilateral is readily reproduced by writing the coordinates of the ver- tices in regular order in two vertical columns, the co- ordinates of the first vertex being repeated at the end of the cqlumns.; The positive terms of the expression for the double area are the products of each abscissa by the next lower ordinate ; the negative terms are the products of each ordinate by the next lower abscissa. 26 COMPUTATION And mensuration This same method gives the area in terms of the co- ordinates of its vertices of any rectilinear figure. Peoblem 53. The coordinates measured in chains of the angular points of a quadrilateral field are (10.25, 0), (3.21, 7.35), ( - 8.75, 0), (2.37, - 9.13). Construct the figure and find the area of the field in acres. Pkoblem 54. Compute the area and construct the figure of the polygon the coordinates of whose angular points measured in feet are ( - 27, 0), (20, 3.75), (16.5, 4), (15, 7.5), {^.Q^ Problems 55, 56, and 57. The coordinates measured in chains of the successive corners of three tracts of land bounded by straight lines are as follows. Compute the areas and construct the figures. Problem 55 Problem 56 Problem 57 X y X y X y -8.12 13.31 6.30 -30 90 9.31 - 16.21 24.09 7.21 140 25.11 -9.61 14.11 12.40 80 130 30. 26. 13.10 50 90 23. 26.23 15. 54 80 22.1 15.4 20. 20.37 100 40 7.23 16.48 10. 16.20 60 20 11. 1.01 16. 10 14.30 5 70 CHAPTER IV VOLUMES OP SOLIDS BOUNDED BY PLANES Art. 15. — The Prismatoid The polyhedron two of whose faces are any two poly- gons in parallel planes and whose other faces are composed of triangles formed by so joining the vertices of these par- allel polygons that each line in order forms a triangle with the preceding line and one side of either parallel polygon is called a prismatoid. The parallel polygons are called the bases of the pris- matoid. The section of the prismatoid formed by a plane midway between the bases is called the midsection. The perpendicular distance between the bases is called the altitude of the prismatoid. Call the altitude of the prismatoid A, the areas of the bases h^ and ^2^ the area of the midsection m. Let be any point in the midsection. Draw lines from to the vertices of the midsection, and to the vertices of the pris- matoid. The lines drawn from to the vertices of the midsection divide the midsection into triangles. The planes determined by and the edges of the prismatoid divide the prismatoid into two pyramid's 0-ABC and 27 28 COMPUTATION AND MENSURATION 0-BEFa, and a series of tetrahedrons 0-BDE, 0-BEF, 0-BFa, 0-BCa,0-OBa, O-ACB, and O-ABB. The volume of the pyramid 0-ABC is ^ h -b^ ; the vol- ume of the pyramid 0-BEFGr is J A • h^. Fig. 9. The triangle BHK is similar to the triangle BBE siud HK is ^ BE. Hence area of BBE is 4 times area of BHK. The tetrahedrons 0-BBE and 0-BHK have the same altitude and therefore are proportional to their bases. Hence volume 0-BBE=4: times volume 0-BHK. Tetrahedron O-^jMT = tetrahedron B-HOK. The vol- ume of B-OffK=ih' OHK. Hence volume 0-BBE = ^h'OHK. VOLUMES OF SOLIDS BOUNDED BY PLANES 29 In like manner it is proved that the volume of 0-BEF = ^h volume of O-EFG = | A volume of 0-BaC=^h volume of 0-CDa^^h volume of 0-ACD =^h volume of O-ABB = | h OKL, OLM. OMK ONP. OPQ. OQH. The sum of the volumes of all these tetrahedra is |^ A • m. It follows that the volume of the prismatoid is J A-(5j + ^2 + 4 m), that is, the volume of the prismatoid is i of the altitude times the sum of the areas of the two bases and four times the area of the midsection. This result is called the prismoidal formula. Art. 16. — Special Cases of Prismatoids When the bases are equal polygons, with their corre- sponding sides parallel, the prismatoid is a prism. When the bases are similar polygons, with their corre- sponding sides parallel, the prismatoid is a frustum of a pyramid. When one base becomes a point, the prismatoid is a pyramid. When one base becomes a line, the prismatoid is a wedge. Wlien one face is a rectangle and the bases are trape- zoids perpendicular to the rectangular face, the prismatoid is called a prismoid. This is the general shape of a railway cutting or embankment. 30 COMPUTATION AND MENSURATION In case of a cutting or embankment, when one face is not a plane, the error in computing the volume is dimin- ished by measuring the areas of 2n-\-l equidistant cross sections A^^ A^^ A^, J.^, ••• , A^^^ ^2»+i ^^^ finding the sum of the volumes of the prismoids whose bases are A-^ and J.g, A^ and ^5, A^ and ^7, ••• , A^^ and A^^_^^, The approx- imation is made closer by increasing w. Problem 58. From the prismoidal formula derive the formulas for the volumes of a pyramid and a frustum of a pyramid. Problem 59. The base of the great pyramid of Egypt is a square 764 feet on a side and the altitude is 477.6 feet. Find the volume in cubic yards. Problem 60. The section of a canal is 32 feet wide at the top, 14 feet wide at the bottom, and 8 feet deep. How many cubic yards were excavated in a mile of the canal ? Problem 61. Find the volume of a rectangular wedge whose base is 70 meters by 20 meters, length of edge 110 meters, and altitude 24.8 meters. Problem 62. Find the number of cubic yards of earth excavated from a railway cutting made through ground the original surface of which was an inclined plane run- ning in the same direction as the rails. The length of the cutting is 4 chains 15 links, the breadth at bottom 30 feet, the breadth at top at one end 75 feet, and at the other end 135 feet, and the depths of these ends 20 feet and 46 feet respectively. VOLUMES OF SOLIDS BOUNDED BY PLANES 31 Problem 63. A railway embankment is to be made on ground which slopes 20 feet in a mile in the direction of the rails and the rails themselves slope 1 in 700. The embankment is straight for 2^ miles. The breadth at the top is 29 feet, the slopes of the sides 1 in 1, and the height at the upper end is 3 feet. Find the number of cubic yards of earth required for the embankment. Problem 64. A railway cutting is to be made 30 feet wide at the bottom, the slopes of the sides being 1|- to 1. The depths in feet on one side of the cutting, taken at intervals of 1 chain, are 12, 11, 10, 9, 10, 12, 14, 15, 17, 20, 21, 23, 25, 27, 30, 33, 37, 41, 45, 50, 53 ; the corre- sponding depths on the opposite side, 15, 14, 13, 12, 14, 15, 17, 19, 22, 26, 28, 30, 33, 36, 39, 41, 44, 51, 57, 60, 58. Find the number of cubic yards of earth to be excavated. CHAPTER V COMPUTATION AND USE OP TEIGONOMETEIO PUNOTIONS Art. 17. — On Angles The figure formed by two straight lines proceeding from a point is called an angle. The two straight lines are called the sides of the angle, and their direction is indicated by arrowheads. The point is called the vertex of the angle. One side of the angle is called the initial side, the other the terminal side of the angle. Fig. 10. A circular arrow about the vertex between the two sides with the arrowhead at the terminal side, indicates which side is to be considered the terminal side of the angle. The angle is called positive when this circular arrow is anticlockwise ; negative when the arrow is clockwise. When the terminal side of the angle coincides with the initial side the angle is called a perigon. An angle of one degree is one three hundred and sixtieth of a perigon. 32 USE OF TRIGONOMETRIC FUNCTIONS 33 The ratio of the circuhir arc, described with the vertex of the angle as a center and bounded by the sides of the angle, to the radius of the arc is called the circular meas- ure of the angle. Denoting the circular measure of the angle by 6 and the arc radius of the arc by r, it follows that — = 6 and arc = rd. , Fig. 11. When 6 = 1, arc = r ; that is, the unit angle in circular measure is the angle at the center of a circle which inter- cepts on the circumference an arc equal in length to the radius. The unit angle in circular measure is called the radian. If the angle is four right angles, the arc is the circumfer- ence of the circle ; that is, arc = 2 irr. Hence, the circular measure of an angle of four right angles is 2 tt radians. It follows that 2 TT radians are equivalent to 360°. Hence, 1 radian = and 1° = zr^ radians. Problem 65. Find the circular measure of an angle of 6° correct to four decimal places. 34 COMPUTATION AND MENSURATION Pboblem 66. Find the circular measure of an angle of 14° 21' 2b" correct to four decimal places. Problem 67. Find in degrees, minutes, and seconds the angle whose circular measure is .357. Problem 68. Find in degrees correct to four decimal places the angle 1.258. Art. 18. — Trigonometric Functions Call the vertex of the angle and the indefinite straight line in which the initial side of the angle lies XX' , Call distances measured on XX' from in the direction of the Y Fig. 12. arrowhead on the initial side positive, distances measured in the opposite direction negative. Let OX be the pos- itive side of XX' . Draw a straight line YY' through perpendicular to XX', If OF makes an angle of +90° with OX, call distances measured from towards Y pos- itive and denote this by the arrowhead at Y, Now take any point P in the terminal side of the angle. The distance from to P is called the radial distance of the point P and is denoted by r. The radial distance is always positive. USE OF TRIGONOMETRIC FUNCTIONS 35 The projection of the radial distance r of the point P on the line XX' is the abscissa x of the point P. The projection of the radial distance r of the point P on the line YY' is the ordinate y of the point P. The trigonometric functions of an angle are the follow- ing six ratios : The sine of an angle is the ratio of the ordinate of any point in the terminal side to the radial distance of the same point ; that is, sin = -- The cosine of an angle is the ratio of the abscissa of any point in the terminal side to the radial distance of the same point ; that is, cos = -- The tangent of an angle is the ratio of the ordinate of any point in the terminal side to the abscissa of the same point ; that is, tan 6 = -- The cotangent 6 is the reciprocal of the tangent 0, The secant is the reciprocal of the cosine 0. The cosecant 6 is the reciprocal of the sine 6. Two auxiliary trigonometric functions are the versed- sine and coversed-sine, defined as follows : versed-sine 6 = 1 — cosine 0, coversed-sine = 1 — sine 0. Problem 69. The tangent of an angle is i|. Con- struct the angle and find the values of all the other functions of the angle. Problem 70 The sine of an angle is .375. Compute the cosine and the tangent of this angle. 36 COMPUTATION AND MENSURATION Problem 71. The cosine of an angle is —.45. Com- pute the sine and tangent of this angle. Art. 19. — Computation^ of Trigonometric Functions If an angle is given in degrees or in radians, the angle may be constructed by means of a protractor and the lengths of the radial distance, abscissa, and ordinate of any point in the terminal side measured, and the trigo- nometric functions of. the angle computed. This method does not in general determine the value of the trigonometric functions beyond the first decimal place. If 6 is the circular measure of an angle, sine 6 may be computed to any required degree of approximation by means of the following infinite series, derived in the Differential Calculus, ^^^ 12. 81. 2. 3-4. 5 1.23.4.5.6.7 + ... . Let it be required to compute sine 20° to four decimal places. The circular measure of 20° is (9 = |= .34907. Hence 6 = .34907, f72'73 ^ »QQ707, 1.2.3.4.5 =^ -QQ^Q^. It follows that sine 20° = .3420. The error committed by omitting the terms of the in- finite series after the third may be written ^'*^1.2.3.4.5.6-7L V8-9 8. 9- 10. 11 7 USE OF TRIGONOMETRIC FUNCTIONS 37 '^' ^^^ 1.2.3.t.5.6.7 [-(^ - 8^)-(8.9.lVll 8 -9. 10- 11. 12. 13; J According to (6) the error is negative, and according to (a) the numerical value of the error is less than z — 7^ — ^ — ;i — r — ^ — =. Hence the error committed by omit- ting the terms of the infinite series after the third in com- puting sine 20° is negative and numerically less than .0000002. When sine 6 has been computed, the remaining func- tions of 6 may be computed by means of the fundamental relations, sin2 ^ + cos2 (9 = 1, tan d = ^^^, cot 6 ^ cos 6^ tan 6^ sec = -, cosec d — cos d sin Q The trigonometric functions of any angle wli^tever may be found in terms of the trigonometric functions of an angle not greater than 45°. Therefore if the functions of angles from 0° to 45° are computed, the functions of all angles become known. If Q denotes the circular measure of a small angle and Q" the number of seconds in this angle, approximately sin = tan = 6 = 6'' x circular measure 1" 20t)264.8* The approximation is correct to six places of decimals from 0° to 38' ; to five places of decimals to 1° 20' ; to four places of decimals to 2° 20". 38 COMPUTATION AND MENSURATION Problem 72. Construct the angle 50° and find the trigonometric functions of the angle by measurement. Problem 73. Construct the angle 110° and find the trigonometric functions of the angle by measurement. Problem 74. Compute the sine of 18° to four decimal places. Problem 75. Compute the sine of 115° to four deci- mal places. Problem 76. Compute the sine of 214° to four deci- mal places. Problem 77. Show that the circular measure, sine, and tangent of any angle from 0° to 7° are the same to three decimal places. Problem 78. From the general definitions of the trigonometric functions of an angle, show that in a right triangle, if the hypotenuse is c, the oblique angles A and B and the sides opposite a and b, A CL A b . 4 a smA = -, cos Az=-, tan A = ~. ceo If the side a is so small compared with the sides 5 and c that the angle A does not exceed two or three degrees, r""^ = 206261:8 '*pp'"°"^'""*^^y' A" ^^ = 20 and consequently c = ^2062^4.8 approximately. A. USE OF TRIGONOMETRIC FUNCTIONS 39 Problem 79. Find the area of the equilateral triangle inscribed in a circle whose radius is 15.5 inches. Problem 80. The angle at the vertex of a cone is 25°, the altitude is 12 feet. Find the volume of the cone. Problem 81. A street railway track is 10 feet from the curbstone where the track is straight. In passing a corner where the street is deflected through an angle of 60°, the rail must be 4 feet from the corner. Find the radius of the circular curve. Problem 82. The greatest angle the radius of the earth, 3963.3 miles, subtends by lines drawn from the center of the sun is 8". 8. Find the distance of the sun from the earth. Problem 83. Two sides of a triangle are 76 feet and 125 feet, their included angle is 35° 15'. Find the third side of the triangle. Problem 84. The three sides of a triangle are 17.5 feet, 26.7 feet, and 35.25 feet. Find the angle opposite the last side. Problem 85. Two sides of a parallelogram are 125.75 feet and 153.25 feet, their included angle is 67° 45'. Find the diagonals of the parallelogram. Art. 20. — On Vectors A vector is a quantity which is completely determined by assigning its magnitude and its direction. A vector may therefore be represented by a line segment 40 COMPUTATION AND MENSURATION and the angle which this line segment makes with a fixed line of reference, the X-axis, for example. The direction of a vector is indicated by the arrowhead placed at the terminal point of the line segment repre- senting the vector. Two vectors are equal when they have the same magni- tude and the same direction. The geometric representative of the vector will be used as the vector itself. Y Fig. 18. Let I denote the length of the vector, 6 the angle the vector makes with the X-axis, x the projection of the vector on the X-axis, y the projection of the vector on the P-axis. From the figure (1) I cos 6 = X, (2) Zsin ^ = y. Squaring equations (1) and (2) and adding, l? = x'^-\-y\ Dividing (2) by (1), tan^ = ^. Hence from the projections of a vector on the axes of reference the length and direction of the vector may be computed. USE OF TRIGONOMETRIC FUNCTIONS 41 If several vectors are drawn in such a manner that the initial point of each successive vector is the terminal point of the preceding vector, the vector drawn from the initial point of the first vector to the terminal point of the last vector is called the vector sum of the several vectors. The sum of the projections of the several vectors on any straight line is the projection of the vector sum on the same straight line. Let Zj, ^2, ?3, ?4 denote the lengths of the several vectors, dp ^2' ^3' ^4 ^^^^ angles these vectors make with the -X-axis. Fig. 14. Let I denote the length of the vector sum, the angle this vector makes with the X-axis, x its projection on the X-axis, ^ its projection on the y-axis. From the figure I cos d = X=l-^ cos 6^ + ?2 ^OS ^2 + ^3 ^^S ^3 + h ^^^ ^4' I sin = y = 1^ sin 6^ + l^ sin 0^ + l^ sin 6^ -f- 1^ sin 6^. From these equations the length and direction of the vector sum may be computed. The displacement of a body is determined by its magni- tude and direction. A displacement may therefore be represented by a vector. 42 COMPUTATION AMD MENSUHATIOK If a body has several successive displacements, each displacement may be represented by a vector, and the resulting displacement is represented by the vector sum. The force acting at a point is determined by its magni- tude and direction. A force may therefore be represented by a vector. By experiment it is shown that if several forces acting at a point are represented by vectors, the vector sum rep- resents the resultant of the several forces. Problem 86. Three sides of a quadrilateral taken in order are 11 feet, 1 feet, and 8 feet, and make angles of 18° 18', 74° 50^ and 130° 20' respectively with a fixed line. Find the length and direction of the fourth side of the quadrilateral. Problem 87. A man walks 8, 12, 15, and 20 miles along successive straight lines, making angles respectively of 30°, 70°, 120° 15', and 155° with the E. W. line. Find the dis- tance and bearing of his final position from the starting point. Problem 88. The following are the field notes of a surve3^ Bearing means the direction from one station to the next succeeding station. Stations Bearings Distances A N. 31JW. 10 chains B N. 62f E. 9.25 chains J) S. 451 w. 10.40 chains Plot the survey, find the distance and bearing from C to J), and find the inclosed area. USE OF Trigonometric functions 43 Problem 89. Find the magnitude and direction of the resultant of three forces acting at a point of 11, 7, and 8 pounds and making angles respectively of 18° 18', 74° 50', and 130° 20' with a fixed line. Problem 90. Find the magnitude and direction of the resultant of four forces acting at a point of 8, 12, 15, 20 pounds, making angles respectively of 30°, 180°, 225°, 330° v^^ith a fixed line. CHAPTER VI COMPUTATION AND USE OF LOGAEITHMS Art. 21. — Natuke of Logarithms Examine the tables, I. 100 = 1 11. IQo =1 101 = 10 10-1 = .1 102=100 10-2 =.01 103 = 1000 10-3 = .001 104 = 10000 10-4 = .0001 105 = 100000 10-5 = .00001 Numbers with one integral place lie between 1 and 10, and it is evident that the exponent by which 10 must be affected to give numbers between 1 and 10 must lie be- tween and 1. In like manner it becomes evident that the exponent by which 10 must be affected to give a number with two integral places must lie between 1 and 2 ; to give a num- ber with three integral places the exponent must lie between 2 and 3 ; and in general to give a number with n integral places the exponent must lie between n — 1 and n. Decimal fractions with the first significant figure in the first decimal place lie between 1 and .1, and it is evident 44 COMPUTATION AND USE OF LOGARITHMS 45 from table II that the exponent by which 10 must be affected to give numbers between 1 and .1 must lie be-* tween and — 1. In like manner it becomes evident that the exponent by which 10 must be affected to give a decimal fraction with the first significant figure in the second decimal place must lie between — 1 and — 2 ; to give a decimal fraction with the first significant figure in the third decimal place, the exponent must lie between — 2 and — 3 ; and in gen- eral to give a decimal fraction with the first significant figure in the ^th decimal place, the exponent must lie between —(n— 1) and — n. The exponent by which 10 must be affected to give any number is called the logarithm of that number to base 10. The logarithm to base 10 of a number is called the common logarithm of that number and is denoted by writing log^^ before the number. For example, log^Q 100 = 2,logi„.01=-2. Problem 91. If the base of a system of logarithms is 3, what are the logarithms of 27, 243, 2V? ^^-^? Problem 92. Can 1 be used as the base of a system of logarithms ? Problem 93. Can — 5 be used as the base of a system of logarithms ? Art. 22. — Computation of Common Logarithms (a) Let ?^J, n^, ^3, n^ represent the logarithms of the numbers iVj, JV^, iVg, iV^ to base 10. By definition JV^ = 10"s iV^ = 10«2, iVg = 10«3, JST^ = lO"*. 46 COMPUTATION AND MENSURATION The product of the numbers iV^, JV^, iVg, iV^, by the law of indices in Algebra, is Hence by the definition of the logarithm, log (iV^iiVa^glV^) = ^^ + ^2 + ^3 + n^ = log iV\ + log JSr^ + log JV3 + log iV^, that is, the logarithm of the product equals the sum of the logarithms of the factors. (5) Any number in the decimal notation may be writtea as a common fraction whose numerator is integral and whose denominator is some power of 10. For example, 375.485 = ^llll^ = 375485 x 10-3. Applying the rule just proved to this product, log 875.485 = log 375485 + log 10-3. As a consequence of (5) the direct computatiqn of the logarithms of whole numbers only is necessary. As a consequence of (a) the direct computation of the logarithms of prime numbers only is necessary. The logarithms to base 10 of the prime numbers may be computed to any required degree of approximation by using the infinite series (1) logio(l + ^) = logio^ + 2(0.43429448)^^-ip^ + 1 + L_+ 1 + ...+ 3(2a;+ 1)3^5(22^ + 1)5 7(2a;+l)7 1 1 (2 7i-l)(2 2J + l)2«-i (2n + l)(2 a; +1)2^+1 which is derived in the Differential Calculus. + •■ COMPUTATION AND USE OF LOGARITHMS 47 The application of this formula requires the logi^a:, that is, the logarithm of the number one less than the given prime, to be known. Let it be required to compute the common logarithms of 2, 4, 5 to five decimal places. 1. Place a: = 1 in (1). There results log,„ 2 = log,, 1 + 2(0.43429448)g + _!_ + _!_ + 1 • 7-3' 1 The computations are carried out to six decimal places in order to obtain the correction in the fifth decimal place. The following is a convenient arrangement of the computation. Begin by computing the sum of the series in brackets. ^=0.333333 i = 0.388333 1 = 00.87037 8^ 1.1 = 0.012246 1 = 0.004115 35 1.1 = 0.000823 5 3^ 1 = 0.000457 11 = 0.000065 1 = 0.000051 39 11=0.000006 P = 0.000006 ^^.1_ = 0.000001 1 = 0.000007 0.346574 48 COMPUTATION AND MENSUBATION The sum of all the omitted terms of the bracketed in- finite series is 13 313^15 315^17 317^19 3i9 which is less than L3 313V 32^34 36^ J 13 The series in parenthesis is an infinite decreasing geometric progression, whose sum is found by the for- mula S= -^- to be -. 1 — r 8 Hence the sum of all the omitted terms of the original infinite series is less than ^•p- ^ = 0.00000006, and does not influence the sixth decimal place. It follows that logio2 = logio 1 + 2 (0.43429448)(0.346574) = 0.30103. 2. Iogio4 = 21ogio 2 = 0.60206. 3. logio 5 = logio (1 + 4) = logio 4 + 2 (0.43429448) ri+i.i+i.i+i.i+i.i+...i [9 3 93^5 9^ 7 97^9 99 J Computation of sum of infinite series to six decimal places: COMPUTATION AND USE OF LOGARITHMS 49 ?: =0.111111 9 - = 0.111111 1 = 0.001372 93 1 3 . 13 = 0.000457 1 = 0.000017 95 1 5 .1 = 0.000003 9^ 0.111571 1 = 0.0000002 9' The error committed by omitting all the terms after the third is 1 1.1. 1 + 1. 1+1. J_+... 7' 97^9 99 11 911 13 913^ ' which is less than 1 . Vl + 1 + 1 + 1 + . . .^ = 0.00000C03, 7 97V 92 94 ^ 9« y and does not affect the sixth decimal place. It follows that logio5 = logio4 -h 2(0.43429448)(0.111571) = 0.60206 + 0.9691 = 0.69897. Problem 94. Compute log^^ 3. Problem 95. Compute log^^ 7. Problem 96. Compute logio 11. Problem 97. Compute logjo 13. Problem 98. Compute logio 17. 50 COMPUTATION AND MENSURATION Art. 23. — Arrangement of Tables of Logarithms The common logarithm of a number consists of two parts, an integer and a decimal fraction. From a table of five-place logarithms it is found that logio 97756 = 4.99014, hence lO^-saoi* = 97756. Dividing the last equation by 10 several times in suc- cession, 104.99014 ^ 97750 lience log^^ 97756 = 4.99014 103.99014 _ 9775.6 hence log^^ 9775.6 = 3.99014 102.99014 _ 977.56 hence log^^ 977.56 = 2.99014 101.99014 _ 97.756 hence log^^ 97.756 = 1.99014 100.99014 ^ 9.7756 hence log^^ 9.7756 = 0.99014 101.99014 ^ 0.97756 hence log^^ 0.97756 = 1.99014 102.99014 ^ 0.097756 hence logj^ 0.097756 = 2.99014 103.99014 ^ 0.0097756 hence logj^ 0.0097756 = 8.99014 104.99014 ^ 0.00097756 hence log^^ 0.00097756 = 4.99014 The minus sign placed over the 1, 2, 3, 4 in the last four logarithms denotes that only the integral part of the logarithm is negative. The decimal part of the logarithm is always positive. This manner of writing the logarithm of a number is adopted in order that the decimal part of the logarithm shall depend only on the significant figures of the number and be independent of the position of the decimal point in the number. When the logarithm is w^ritten in this manner, the in- tegral part is called the characteristic; the decimal part, the mantissa. The characteristic is determined by the rule: COMPUTATION AND USE OF LOGARITHMS 51 If the number is greater than 1, the characteristic is positive and less by unity than the number of integral places in the number. If the number is less than 1, the characteristic is neg- ative and equal to the number of the decimal place occupied by the first significant figure of the number. The mantissa of the logarithm of a number is taken from the table of logarithms. If five-place logarithms are used, it is absurd to attempt to compute beyond six significant figures, and even the sixth figure and frequently the fifth figure is unreliable. The truth of this statement will become evident from the computations themselves. Examplp: 1. Let it be required to find the common logarithm of 375.658 from the five-place table. By rule the characteristic is 2. The mantissa corre- sponding to the first four figures of the number is taken directly from the table. The logarithms of the two con- secutive numbers of four significant figures between which the given number lies are log 375.7 = 2.57484 log 375.6 = 2.57473 The difference in the mantissa corresponding to 1 unit of the fourth significant figure of the number is 11 units of the fifth decimal place of the mantissa. Assuming that the mantissa increases uniformly while the number increases by one unit of the fourth significant figure, it follows that the increase of the mantissa due to the fifth figure of the number is .5 x 11 = 5.5 units of the 62 COMPUTATION AND MENSUBATION fifth decimal place, and the increase due to the sixth figure of the number is .08 x 11 = .88 units of the fifth decimal place of the mantissa. Collecting the results, '^^^^f'^^F^N^ log 375.6 =2.57473 ^'VERSITY J increase for .05 = .000055 ^^UFORhU^V^ increase for . 008 = . 0000088 ""^ Ince log 375.658 = 2.57479 Observe that if any figure other than zero stood in the sixth significant place of the number, the logarithm of the number to five decimal places would remain the same. The method of finding the corrections of the logarithm for the fifth and sixth figures of the number is called interpolation. Example 2. Let it be required to find the number whose common logarithm is 3.72564. The characteristic shows that there are four integral places in the number required. The figures of the number corresponding to the mantissa are found from the table of logarithms. Take from the table the logarithm next lower than the given logarithm and subtract it from the given logarithm and from the next higher logarithm. 3.72564 3.72567 = log 5307 3.72558 = log 5306 3.72558 = log 5306 6 9 A difference of 9 units in the fifth decimal place of the mantissa gives a difference of 1 unit in the fourth signifi- . COMPUTATION AND USE OF LOGARITHMS 53 cant figure of the number. Assuming that the rate of change of the mantissa from log 5306 to log 5307 is uni- form, the increase in the number due to the increase of the mantissa by 6 units in the fifth decimal place is 6-i-9 = .67 units of the fourth significant figure of the number. Hence 3. 72564 = log^, 5306. 67. The sixth figure of this number is uncertain. It may have any value from 2 to 9. This makes the uncertainty in the number extend to the fifth significant figure, which may be either 6 or 7 if the number is to be expressed correct to five significant figures. It appears from these examples that the use of five-place logarithmic tables limits the computation to four or five significant figures. Problem 99. From a five-place table find the logarithm of 1476.38. What other numbers of six figures have the same logarithm ? Problem lOO. From a five-place table find the logarithm of 8754.88. What other numbers of six figures have the same logarithm ? Problem loi. From a five-place table find the number whose logarithm is 2.14896 and determine the limits of uncertainty oi the number. Problem 102. From a five-place table find the number whose logarithm is 4.79658 and determine the limits of uncertainty of the number. Problem io3. How does the change in the mantissa corresponding to a change of 1 unit in the fourth signifi- cant figure in the number vary with the number ? 54 COMPUTATION AND MENSURATION Art. 24. — Computation by Means of Logarithms Logarithms are used to shorten the operations of multi- plication, division, involution, and evolution. These operations are performed by applying the follow- ing rules : Let logjQ M= m, logj^ iV= n. By definition M = w\ jsr= io\ The product of ilff aiid iVis Hence by definition logio (MJST) =m-{-n = logj^ M-^ ^ogio^; that is, I. The logarithm of the product equals the sum of the logarithms of the factors. The quotient of il[f and iVis jsr Hence by definition M — = 10"^~". logio 7i? = ^^^ ~ ^ = ^^^10 ^- ^^Sio ^> that is, II. The logarithm of the quotient equals the logarithm of the dividend minus the logarithm of the divisor. Raising both sides of the equation M= IC" to the power jt?. Hence by definition logio MP = p w = p logiQ M 5 that is, COMPUTATION AND USE OF LOGARITHMS 55 III. The logarithm of the power of a number equals the logarithm of the number multiplied by the exponent of the power. Extracting the r root of both sides of the equation 1 m Hence by definition log,, if L !?^ = 1 logio ^; that is, r T IV. The logarithm of the root of a number equals the logarithm of the number divided by the index of the root. In working a numerical problem, first, decide on a method of computation ; second, arrange a scheme of com- putation; third, perform the computation. Example. Let it be required to compute the lateral area and the volume of a cone of revolution, altitude 29.75 feet, radius of base 12.15 feet. Denoting the lateral area by A^ the volume by FJ the radius of the base by i2, and the altitude by IT, (a) A = 'TrB ^E^-\-E% (J) V= 1 ttR^H. Applying logarithms to («), log J. = log TT + log R + 1 log (^ + E^). 56 COMPUTATION AND MENSURATION Scheme of computation : log ^ = 1.47349 i?2^885.08 log H^ = 2. 94698 i^ = 147. 62 log ^2 =2.16916 E^ + ff^ = 10S2.10 log7r = 0.49715 + log 72 = 1.08458 + 1 log (^2 + ji2^ = 1.50698 log v4 =3.08871 ^ = 1226. 63 square feet. Applying logarithms to (5), log r= log TT + 2 log R + log H- log 3. Scheme of computation : log 7r = 0.49715 2 log i2= 2.16916 log ff= 1.47349 4.13980 - log 3 = 0.47712 log F= 3. 66268 r= 4599.2 cubic feet. In these values of A and Fthe fifth and sixth figures are uncertain. Problem i04. The chord of a circular segment of one base is 3.25 inches, the altitude of the segment is 1.15 inches. Find the diameter of the circle. Problem io5. Find the area of the surface of the sphere whose radius is 6.7 inches. . COMPUTATION AND USE OF LOGARITHMS 57 Problem io6. Find the volume of the sphere whose diameter is 15.36 feet. Problem 107. Find the area of the equilateral triangle whose side is 27.16 feet. Problem 1O8. Find the capacity in gallons of a rec- tangular cistern whose dimensions are 10.5, 6.35, and 4:.idb feet. Problem io9. Find the capacity in gallons of a tank in the shape of a circular cylinder, radius of base 8.35 feet, altitude 12.46 feet. Problem 110. Find the volume of a regular pyramid whose base is a square 4.18 feet on a side and whose lateral edge is 11.27 feet. Problem 111. The radius of a circle is 8.57 feet. Find the least chord that can be drawn through a point 3.25 feet from the center. Problem 112. The diameter of a circle is 15.28 inches. Find the length of the tangent to the circle from a point 20.15 inches from the center. Problem 113. In a right triangle the sides about the right angle are 5.75 feet and 8.17 feet. Find the perpen- dicular from the vertex of the right angle to the hypot- enuse. Problem 114. An excavation 1.5 yards deep, rectan- gular at top and bottom, and in the form of a frustum of a pyramid has its upper base 10 yards wide and 16 yards long, and the lower base 7.5 yards wide. How many cubic yards were removed to make the excavation ? 58 COMPUTATION AND MENSUBATION Problem 115. If the atmosphere extends to a height of 45 miles above the earth's surface, what is the ratio of its volume to the volume of the earth, assuming the earth to be a sphere with a diameter of 7912 miles. Problem 116. The circumference of the base of a conic frustum is 49.3 feet, the diameter of the top is 12.5 feet, the altitude is 15.7 feet. Find the surface and volume. Problem 117. On a sphere whose radius is 11.75 inches, what is the area of a zone whose altitude is 3.25 inches ? Problem 118. On a sphere whose radius is 28.5 feet find the area of a spherical triangle whose angles are 120°, 130°, and 140°. Problem 119. On a sphere whose diameter is 25 inches find the area of the spherical polygon whose angles are 110°, 120°, 130°, and 140°. Problem 120. The sides of a triangle are each 12.49 inches. What is the volume of the solid generated by revolving the triangle about one side ? Problem 121. A dome is in the form of a spherical zone of one base. Its height is 30 feet and the diameter of the base 60 feet. Find its area. Problem 122. Find the volume of the frustum of the cone of revolution, the radii of whose bases are 12.5 feet and 7.25 feet and whose slant height is 9.75 feet. Problem 123. The base of a triangle is Q.b feet, the altitude is 15 feet. Find at what distances from the base COMPUTATION AND USE OF LOGARITHMS 59 lines parallel to the base must be drawn to divide the tri- angle into three equal areas. Problem 124. The radius of the base of a cone is 7.2 feet, the altitude is 25 feet. Find at what distances from the base planes parallel to the base must be drawn to divide the cone into three equal volumes. Problem 125. Two parallel planes divide the diameter of a sphere into three equal parts. The radius of the sphere is 12 feet. Find the ratio of the volumes of the solids into which the sphere is divided. Art. 25. — The Compound Interest Formula If a sum of money P is invested at compound interest at R per cent, the amount at the end of the first year is P(l + R). This amount is the principal at the beginning of the second year, hence the amount at the end of the second year is P(l + R)\ In like manner the amount at the end of the third year is found to be P(l + Ry. In general the amount at the end of t years is ^ = P(l + Ry. From this it follows that the present worth of a sum of money A due t years hence, if money is worth R per cent compound interest is P = A(1 + RyK Example 1. What is the present worth of an annual pension of $500 to run for 7 years, the first payment to. 60 COMPUTATION AND MENSUBATION be made one year from date, and money being worth 5 % compound interest ? The present worth of the first payment is 500 x 1.05"^ The present worth of the second payment is 500 x 1.05-2 The present worth of the third payment is 500 x 1.05-^ The present worth of the fourth payment is 500 x 1.05"* The present worth of the fifth payment is 500 x 1.05"^ The present worth of the sixth payment is 500 x 1.05"^ The present worth of the seventh payment is 500 x 1.05"^ The present worth of the seven payments is F = 500[1.05-i + 1.05-2 + 1.05-3 + 1.05-* + 1.05-^ + 1.05-6 + 1.05-7] rnnl.05-8- 1.05-1 rnAl.05-7 - 1 = ^^^ 1.05- 1-1 •^"" 1 - 1.05 ^,,,1-105-7 .05 = 10000(1 - 1.05-7). To compute x = 1.05-7, , pass to logarithms, loga; = -7 log 1.05 = -0.14833 = 1.85167. .'.X = 0.710667. It follows that P= 10000(1 -0.710667) = 12893.33. Example 2. In how many years will an annual pay- ment of $300 meet principal and interest of a debt of f 2000, money being worth 5 % compound interest ? COMPUTATION AND USE OF LOGARITHMS 61 Let t represent the required number of years. At the end of t years the debt amounts to 2000 x 1.05^ If the first annual payment is to be made at the end of the first year, its amount at the end of t years is 300 X 1.05'~i. The amount at the end of t years of the second annual payment is 300 x 1.05'~2, and so on for the successive payments. The last annual payment is to be made at the end of the t years and its value then is #300. Hence the sum of the amounts of the t annual payments is A = 300[1 + 1.05 4- 1.052 + 1.053 + ... + 1.05'-i] = 30oli^^^^ = 6000(1. 05^ -1). At the end of t years the amount of the debt is to equal the amount of the annual payments. Hence 2000 X 1.05' = 6000(1.05' - 1). Solving this equation for 1.05', 1.05' = 1.5. Passing to logarithms, ^log 1.05 = log 1.5. Solving for t, ^^logL5 3 3^ log 1.05 ^ Problem 126. What is the present worth of 82000 due 10 years hence without interest, if money is worth 6 % compound interest ? Problem 127. A man pays a premium of 8104 per annum on a life policy of $3500 for twenty years before 62 COMPUTATION AND MENSURATION death. Money being worth 5 % compound interest, does the insurance company gain or lose, and how much ? Problem 128. A town, whose property has the as- sessed valuation of $7,325,000 and whose annual tax rate is 19 mills on fl, issues bonds to the amount of $225,000 to build a sewerage system, these bonds being at 4| per cent interest and maturing in 15 years. An extra tax is to be laid to meet this interest and to provide a sinking fund to redeem the bonds, the rate of interest in the sinking fund being 3 J per cent compounded annually. How many mills must be added to the tax rate for this purpose ? Problem 129. A town finds it necessary to build a bridge. The cost of a stone bridge is $20,000, the cost of maintenance 150 per year, the life of the bridge 50 years. The cost of a steel bridge is $10,000, the cost of main- tenance $150 per year, the life of the bridge 15 years. The town can borrow money at 4 per cent compound in- terest and realize 4 per cent compound interest on its sinking fund. What is the cost per year to the town of each of the two bridges ? Art. 26. — The Slide Eule If line segments are laid off from one point of a straight line proportional to the mantissas of the logarithms of numbers from 100 to 1000, and the number whose loga- rithm is represented by the line segment is written over the terminal point of the line segment, there will be formed a logarithmic scale. P'rom a logarithmic scale nine inches long constructed COMPUTATION AND USE OF LOGARITHMS 63 in this manner the line segment which represents the loga- rithm of a number of three figures, and conversely the number to three figures whose logarithm is represented by a given line segment, can be read off with considerable accurac}^ To find the product of two numbers, apply to the loga- rithmic scale the sum of the line segments representing the logarithms of the numbers. Over the terminal point of the sum of these line segments stands the number which is the product of the two given numbers. I I I I 1 1 1 1 itiiMiiiiiliiiihiiiliiiiiiiiiMiiiiiiiiiUM^ Fig. 15. To find the quotient of two numbers, apply to the loga- rithmic scale the difference of the line segments repre- senting respectively the dividend and divisor. Over the terminal point of the difference of these line segments stands the number which is the quotient of the two given numbers. The slide rule consists of two logarithmic scales, one of which slides along the other. To find the product of two numbers by means of the slide rule, locate one number on the fixed logarithmic scale and bring the initial point of the sliding scale over this number. Now locate the second number on the slid- ing scale and in line with this number on the fixed scale the product is found. To find the quotient of two numbers by means of the slide rule, locate the dividend on the fixed scale and the 64 COMPUTATION AND MENSURATION divisor on the sliding scale and bring these two num- bers over each other. The quotient is the number on the fixed scale in line with the initial point of the sliding scale. To find the square root of a number by means of the slide rule, bisect the line segment which represents the logarithm of the number and find the number whose loga- rithm is represented by the corresponding line segment. In like manner any root of a number may be found by means of the slide rule. ar^ — a a ar^ r-1 1-r 1-r' a ar^ CHAPTER YII ON LIMITS Art. 27. — The Infinite Decreasing Geometric Progression The formula for the sura of the first n terms of a geo- metric progression whose first term is a and ratio r is whence '- 1-r 1-r In these formulas a and r for a particular progression have fixed values; that is, they are constants ; while n may be any positive integer and the value of aS'^ depends on n; that is, n and S^ are variables. If r is less than unity and n increases without limit, the progression becomes an infinite decreasing geometric progression. The difference between the variable /^„ and the constant — - — is numerically equal to . When r is less than 1-r -^ ^ 1-r one, if any numerical value, however small, be assigned in advance, a value of n can be computed such that for this p 66 66 COMPUTATION AND MENSURATION value of n and all larger values of n the difference between the variable S^ and the constant :; shall be less than 1 — r the value assigned. This is expressed by saying that the limit of the variable S^ when n is infinitely increased is the constant , and it is written 1-r limit aSJ = ^ 1-r In general, if a variable approaches a constant in such a manner that the numerical value of the difference between the variable and the constant becomes and remains less than any value, however small, that can be assigned in advance, the variable is said to approach the constant as its limit. Pkoblem 130. Find the limit of the sum of the first n terms of 1 + |- + -| + 2V + FT + *" when n is infinitely increased. Problem 131. For what values of n does the sum of the first n terms of 1 + J + ^ +2V + "sS: + *'* ^i^^^ from the limit of the sum by less than .00001 ? Akt. 28. — The Length of a Curved Line The approximate length of a curved line is found by inscribing or circumscribing a broken straight line the parts of which fit the corresponding parts of the curved line as closely as may be practicable. The approximation is made closer by diminishing the lengths of the parts ON LIMITS 67 of the broken line. The limits of accuracy of direct measurement are soon reached. The length of a curved line is defined as the limit of the inscribed or circumscribed broken line when the number of parts of the broken line is infinitely increased in such a manner that the length of each part of the broken line is infinitely diminished. By using this definition the length of a curved line may be computed either exactly or to any required degree of approximation. Art. 29. — The Computation of tt By definition circumference of circle semicircumference TT diameter radius Hence if the radius is made unity, tt equals the length of the semicircumference. If (?„ represents the length of the side of the regular polygon of n sides inscribed in the circle, and f„ represents the side of the regular polygon of n sides circumscribed about the circle, the semicircumference always lies between ^nCn and ^nt„ and the semicircumference is the common limit of ^ ncJ^ and | nt^ when n is infinitely increased. It is readily proved that (1) t„ = — % (2) c^n = V2 - -V-i-cl The side of the inscribed regular hexagon is the radius of the circle. Hence Cq= 1. By formula (1) the side 68 COMPUTATIOIf AND MENSUBATION of the circumscribed regular hexagon, ^g, is computed. By formula (2) the side of the inscribed regular polygon of 12 sides, e^^^ is computed. By repeated successive application of formulas (1) and (2) the sides of inscribed and circumscribed regular polygons, the number of whose sides is six multiplied by any power of two, may be computed. The results of twelve successive applications of for- mulas (1) and (2) give the following table of values for the semiperimeters of the polygons; n \nc,. Intn 6 3.0000000 3.4641016 12 3.1058285 3.2153903 24 3.1326286 3.1596599 48 3.1393502 3.1460862 96 3.1410319 3.1427146 192 3.1414524 3.1418730 384 " 3.1415576 3.1416627 768 3.1415836 3.1416101 1536 3.1415904 3.1415970 3072 3.1415921 3.1415937 6144 3.1415925 3.1415929 12288 3.1415926 3.1415927 The semicircumference of the circle, and therefore tt, always lies between the semiperimeters of the inscribed and circumscribed regular polygons of the same number of sides. Hence the value of tt correct to six decimal places is 3.141593. . ON LIMITS 69 By direct measurement the value of tt could not be determined beyond the second or third decimal place. There is no limit to the approximation that may be reached by computation. Art. 30. — An Important Limit The derivation of many of the formulas for the deter- mination, exact or approximate, of lengths, areas, and volumes is based on the limit, when n is infinitely in- creased, of the ratio of the sum of the p powers of the first n natural numbers to the p + 1 power of w, or expressed as a formula, IP + 2^ 4- 3^ + 4^ + 5^ -h ••• + nP limit tP+i If p is any positive integer, and if a > 6, ^ "^ — = aP +aP-^b + aP-^^ + ••• + aJ^-i + h^ a — which may be written (1) CP + i)b''<^^^^^^. a — From (1), by placing h successively equal to 1, 2, 3, 4, 5, ••• , w and a = J + 1, (p + 1) 1^ < 2^+1 - 1^+1, (;?+"l) 2^ < 3^+1 -2^+1, (p 4- 1) 3^ < 4^+1 - 3^+1, {p 4- 1) 4^ < 5^+1 - 4^+1, (j9 4- 1) ^^ < (w + 1)^+1 - ri^+i. 70 COMPUTATION AND MENSURATION Adding these n inequalities, (^p + 1)(1^ + 2^ + 3^ + 4^ + ••• + w^) < (n+ 1)^+1 - 1^+^ Dividing the last inequality by (jt? + l)n^+i, IP + 2^ + 3^ + 4^+ ••• -f nP 1 r A 1 Yl^_l_ I < 1 {l+^J + £C|±_l)+...+iL_±l}, jt? + 1 I n 2!7i2 n^-i J and finally Ip 4- 2^ 4- 3^ + 4^ 4- ••• -^n^ 1 ^P+i JO 4- 1 < jt? + 1 I w 2!?i' n^ p-i If now 9i is infinitely increased, since - becomes in- n finitely small and the number of terms in the bracket is jt? — 1, a finite number, limit 1^ + 2^ + 3^ + 4^4- ••• +71^ ^ _1_ i?+ 1* n=oo ^^+1 Art. 31. — Lexgth and Area of Involute of Circle If a fine inextensible string, coiled around a circle, be uncoiled from the circle and the string be always kept taut, the free extremity of the string will trace the curve called the involute of the circle. Denote the radius of the circle by r and suppose the length of the string uncoiled to be rd. Divide the arc r6 into n equal parts and inscribe in the arc a regular broken ON LIMITS 71 line. Denote by a the length of each part of the broken line. Then a = 2r sin -— . If the string were wound 2n around the broken line and then uncoiled, if S^ represents Fig. 16. the length of the curve traced by the extremity of the string and A^^ the area swept over by the string, A n \ji n n n n J ^„=2 7'2sin2-- -H 1 + ... H . zn\ji n n 7i j When n is infinitely increased, the broken line approaches the circular arc rd as its limit, and the involute of the broken line approaches the involute of the circular arc as its limit. 6 6 Since limit sin—- = limit -— , it follows that 74=00 'zn n=» 2n limit S. = re^ limit p + 2 + 3 + 4 + .■■ +>»1,g,^ limit 4„ = 1 ,-^^3 ii„,it ri^+2^+3^+f+-+>»n ^ i ,.^. n=« n=oo L ^ J 72 COMPUTATION AND MEN8UBATI0N Problem 132. Find the length and area of the involute of the circle radius r extending from ^ = to ^ = 2 tt. Problem 133. Find the length and area of the involute of the circle radius r extending from ^ = 27rto ^ = 4 7r. CHAPTER VIII GEAPHIO ALGEBEA Art. 32. — The Graph of an Equation- If an equation /(a;, y) = is satisfied by the coordinates of every point (x^ y) of a line, and if every point (a:, ?/) whose coordinates satisfy the equation is a point of the line, the line is called the graph of the equation. To construct the graph of an equation /(ic, y) = 0, com- pute the values of y for different values of a;, and locate the points whose coordinates are the pairs of corresponding real values of x and y. The graph of the equation is the smooth curve drawn through the points located. Example. Construct the graph of x^-\-y^ — ^. The corresponding pairs of real values of x and y are a;=_3 -2 -1 +1 +2 +3 y= ±2.24 ±2.83 ±3 ±2.83 ±2.24 0. For a; > + 3 and for x< — 3, ?/ is imaginary. Hence the graph lies between the lines a: = 3 and a: = — 3. The graph also lies between the lines ^ = 3 and «/ = — 3. Locating the points whose coordinates are the corre- sponding pairs of values of x and y and drawing a smooth line through these points, it becomes evident that the graph resembles a circle. 73 74 COMPUTATION AND MENSURATION In fact the form of the equation shows at once that the graph is a circle whose radius is 3 and center the origin. \ r ^ ■^ / \ 1 \ ^ X \ \ J v y Notice that for the circle x^ + y"^- the circle aP'-\-y'^- without the circle Problem 134. Problem 135. Problem ise. Fig. 17. all points (x^ y) on the circumference of -9 = 0; that for all points (a;, y) within - 9 < ; and that for all points (a:, ^) a:2 + /-9>0. Construct the graph of 3a^— 4«/ = 12. Construct the graph of x^-\-y'^ — 16. Construct the graph of — + ^ = 1. Problem 137. Construct the graph of !-■ Problem 138. Construct the graph of ?/2 = 4 x. Problem 139. Construct the graph of :c2 = 4 y. GBAPHIC ALGEBRA lb Problem 140. Construct the graph oi y = Zx— ba^. Problem 141. Construct the graph oiy = b — x-\-Qx^, Art. 33. — Equations of Lines Suppose the relation between the coordinates of any point P (x^ y) to be expressed by the first degree equation (1) Ax^ By -^ C=^. Let PiCa^i, «/i), PgC^ ^2)^ A(% 2/3) be any three points whose coordinates satisfy this equation, that is, Ax^ j^By^-{- 0=0, Ax^ + %2 +0=0, Ax^ + %3 + C = 0. The elimination of A, B^ C gives ^1(^2 - ^3) + ^2(^3 - ^i) + ^3(^1 - ^2) = ^• The left-hand member of this equation is double the area of the triangle whose vertices are P^, Pg' ^3- ^^ follows that the area of the triangle whose vertices are any three points whose coordinates satisfy equation (1) is zero; that is, these three points lie in a straight line. Hence the graph of every first degree equation is a straight line. The equation (2) 0.-2 + ^2 = ^2 expresses the fact that the distance from the origin to the point (x, y) is a. Therefore all points (x^ y\ whose co- ordinates satisfy equation (2) are located on the circle whose radius is a and center at origin. 76 COMPUTATION AND MENSURATION From equation (2) y = Va2 — ic^. Multiplying y by _, that is, multiplying each ordinate of the circle by the same factor -, a a which reduces to The curve formed by the points (x^ y^ whose coordi- nates satisfy equation (3) is called the ellipse. The curve formed by the points (.t, y) whose coordinates satisfy the equation (4) ^ ^-^' = 1 is called the hyperbola. The curve formed by the points {x^ y) whose coordi- nates satisfy the equation (5) y'^ = 1'px is called the parabola. Problem 142. Show that the area of the ellipse ^ + f- = 1 is iTob. Art. 34. — Graphic Solution op Equations Example 1. Solve the equation a^ —1 x + 1 = graphically. Construct the graph oi y = x^ — 1 x -{-1, QttAPHtC ALGEhBA n The abscissas of the points where the graph intersects the JT-axis are the roots of the equation x^ — 1 x + l = 0. These roots are found by measurement to be 1.3, 1.7, — 3.1. Y t 1 __:n:__± ^..., -\-\- ::;::: ::|:: -\-v- :-::o:;:i: mm ■^x > < :;z /\ ^^ ^^ ' ■ 7 V ^^ 7 7 ^ r f 4 X- r / \ A ' ST \/ ^z \ \ ±. Fig. 18. Fig. 19. Example 2. Solve graphically the simultaneous equa- tions ^2 = 10 a; and x^ + ^'■^ = 25. Construct the graphs of these two equations on the same axes of reference. The coordinates of the points of intersection of the graphs are the common solutions of the equations. The points of intersection are (2.1, 4.4) and (2.1,-4.4). Problem 143. Solve graphically x^ — 4:x — 15 = 0, Pkoblem 144. Solve graphically a^-\-7x— 7 = 0. Problem 145. Solve graphically o^ -^ y^ z= 25, ^2= 10 a; — a^. Problem 146. Solve graphically rr^ + ?/2 = 25, z^-if = 4. 78 COMPUTATION AND MENSURATION Art. 35. — Inequalities treated Graphically Example. Show graphically for what pairs of values of X and y the inequalities X - y -\-l>0 2x-\- y -Q>0 -^x-\-2y-\- 10>0 are true ? Construct the graphs of the equations The graph of the equation x — y ■\- 1 == divides the XF'-plane into two parts, such that for the coordinates of every point (x^ ?/) in one part x — y + 1> and for the Fig. 20. coordinates of every point (x^ y) in the other part x — y + 1 <0. Determine by trial for which part x — y -\-l>0 and indicate this part of the plane by the arrowhead placed on the graph. GRAPHIC ALGEBRA 79 Kepeat the same operation with the other two graphs. The shaded triangle in the figure contains all the points (^x, y) whose coordinates satisfy simultaneously the three given inequalities. Problem 147. Solve graphically the simultaneous in- equalities 2— ?/ + 4>0, a:— 2?/ + 5<0, 2; + 5y — 4<0. Problem 148. Solve graphically the simultaneous in- equalities 2;2 -f ^2 _ 25 < 0, 2 2; + y - 10 > 0. CHAPTER IX AEEAS BOUNDED BY OUEVES Art. 36. — Exact Areas If the equation of a curve is known, the ordinate corre- sponding to any abscissa may be computed. If a curve is given whose equation is not known, the ordinate corresponding to any abscissa may be measured. Denote by a and h the abscissas of the end points of a curve and assume h>a. Divide h — a into m equal parts and call each part A, so that mh = b — a. Denote by ^q, y^, i/^, y^, • • • , ?/^ the ordinates of the points of the curve whose distances from a are the multiples of h. ■►X Fig. 21. Construct rectangles on each successive ordinate and the adjacent part oil — a located to the right of the re- spective ordinate. 80 AREAS BOUNDED BY CURVES 81 The limit of the sum of the areas of these rectangles when m is infinitely increased is the area bounded by the curve, the ordinates of the end points of the curve and the X-axis. If' the relation between ordinate and abscissa is ex- pressed by an equation of the form (1) 7/ = A-i-Bx-\-C2^-hI>:i^-{- ••• -\-]Srx\ this limit can be determined exactly. Consider the rectangle constructed on the ordinate 2/1 corresponding to the point whose distance from a is Ih. The area of this rectangle is Substituting for A its value ~ , the expression for m the area bounded by the curve (1), the X-axis, and the ordinates corresponding to x=^a^ x=h becomes l=m-\ Area = limit V A(h-a)--\-B(h-ay I m and finally Area = ^ (/, _ a) + | (6 - a^ + ^(h - a^ N ^7 Nw+i + ...+ (j_«) + When w = 1, equation (1) represents a straight line and the area is given by the formula Area = A(h - a) + f (6 - a)2. 82 COMPUTATION AND MENSURATION When 71 = 2, equation (1) represents a parabola and the area is given by Area = A(h-a) + ~(h + ay-{- ^(h + a^. Problem 149. Find the area bounded by ^ = 3-f-52;, a; = 2, rr = 10, and ?/ = 0. Pkoblem 150. Find the area bounded by ?/ = 2 2: + 6 a:^^ a; = 0, a? = 5, and «/ = 0. Problem 151. Find the area bounded by x^ = 2py, a; = 0, x= a^ and ?/ = 0. Art. 37. — Approximate Equation of a Curve If the equation of a curve is not of the form (1), it may be possible to express approximately the relation between ordinate and abscissa by means of an equation of the form (1) and thus obtain an approximate value of the area. For example, the equation of the circle is x^ + y'^ = r% from which , ^/^ 2 n 1'-^^ 1 Srr* 1 3 5a;6 -] ^ L2r^24r4246r6 J This infinite series is convergent for all values of x numerically less than r, hence for all such values y can be expressed in the form A+ Bx-\- Cx^ + 1>^ + • • • + iVic" to any required degree of approximation, and consequently the area of the circular segment may be computed to any desired degree of approximation by the methods of the previous article. If a curve is given whose equation is not known, and if the abscissas of the end points are x — a and x = h^ divide AREAS BOUNDED BY CURVES 83 h — a into n equal parts and call each part k^ so that h — a = nk. The ordinates corresponding to the points of division may be measured. Denote by y^, y^, y^. Vz^--- ^Vn the ordinates corresponding to the abscissas a, a + ^, a + 2 ^, a + 3 A:, • • • , 0^ + w^. It is possible to determine an equation of the form (1) y^^A^Bx^Cx^^D:^^- ■■• + Nx\ which is satisfied by the coordinates of these n-{-l points of the curve. For the successive substitution of the coordinates of the n-\-l points (a, yo), (a4-/c, y{), (a -{-2k, ^/g), •••, (a + wA:, ?/„) in equation (1) furnislies n-\-l equations of the first degree between the n -{- 1 coefficients of equation (1). These equations determine the coefficients. In general the higher the degree of equation (1), and consequently the greater the number of points the line represented by equation (1) has in common with the given curve, the more closely will the line represented by equation (1) fit the given curve and the closer will be the approximation of the area bounded by equation (1) to the area bounded by the given curve. Art. 38. — Approximate Areas First approximation. When w = 1, equation (1) be- comes y = A + Bx, and Area = A(h — a) + —(b — a^. 84 COMPUTATION AND MENSURATION Denoting the ordinates of the end points of the given curve by ^^ and ^^^ yi = ^^ y^ = A + B(h- a). From these equations A '= y-^^ B = ~ 7 • Substituting these values, Area = -^ (^^ + ^2)- This is the approximation found by replacing the given curve by a straight line joining the end points of the curve. Second approximation. When ?^ = 2, equation (1) be- comes y = A^Bx-{- Qx\ ^^^ Area = vl(5 - «) + f (5 - aj' + ^(h - a^. Denoting the three equidistant ordinates of the given curve by y^, y^, y^, y, = A^^Ch-a}-\-^Cb-ay, y^ = A + B(h-a)+ Q(h - ay. From these equations ^ = Vv h — a (5 - af • AREAS BOUNDED BY CURVES 85 Substituting these values, Area = —^ (t/^ + 4 ^2 + ^s)- This is the approximation found by replacing the given curve by a parabola through three points of the curve with equidistant ordinates. It is known as Simpson's approximation for the area bounded by a curve. Third approximation. When n = S, equation (1) be- comes and Area = AQh - a) +~(h - ay+^^b - ay + ~(h -ay, Z O 4: Denoting the four equidistant ordinates of the given curve by i/^, y^, y^, and y^, y^ = A^ \B(h - «) + \C(h - ay + i^BQy - ay, y^ = A + lB(h - a) + I C(h - ay + ^^j D(h - ay, y^ = A + B(ib-a)-^ C(h - ay + D(h - ay. Solving these equations for A, B, (7, D and substituting, Area =\(h-a) {y^ + 8(^3 + y^) + y^. Other approximations. When ri = 4 or 5, the resulting expression for the approximate area does not take a simple form. When w = 6, if the equidistant ordinates are denoted ^y Vv Vv ^3' ^4^ Vh' y%> Vv ^^^ expression for the area is Area = -^^(h + a) {y^ + ^^ + ^^ + y^ + 5(^2 + i/^) + 6 y^ }. This is Weddle's approximation for the area bounded by a curve. 86 COMPUTATION AND MENSURATION The application of these several formulas for areas will not give close approximations if the equidistant ordinates have a wide range of values. It is therefore important to break up the given curve into parts such that for each part the equidistant ordinates have a small range of values and apply the formulas to each part separately. Problem 152. By using the value of tt compute correct to two decimal places the area of the segment of the circle whose radius is 14 feet bounded by a diameter and a chord parallel to the diameter and distant 7 feet from the diameter. Problem 153. Compute six equidistant ordinates of the circular segment of Problem 149 and find the error of the first approximation. Problem 154. Using the six equidistant ordinates of Problem 150, find the error of Simpson's approximation. Problem 155. Compute seven equidistant ordinates of the circular segment of Problem 149 and find the error of Weddle's approximation. Problem 156. Find, approximately, the area of a plot of ground bounded on one side by a straight line and on the other by an irregular curve. The length of the straight line is 8.97 chains, and the lengths in chains of 31 equi- distant ordinates taken in order, measured from this line to the opposite boundary, are 1.37 1.83 2.31 ' 2.52 2.46 2.73 2.91 3.21 1.98 1.74 1.50 1.59 2.93 3.41 3.92 3.42 1.75 1.50 1.65 2.58 2.47 2.32 2.75 2.63 2.01 1.98 2.37 2.76 2.76 2.42 2.01 CHAPTER X VOLUMES OF SOLIDS Art. 39. — Exact Volumes Suppose two parallel planes to be drawn such that the solid extends from the one plane to the other and is con- tained wholly between the two planes. Divide the distance between these two planes into n equal parts, and through the points of division pass planes parallel to the planes drawn at the start. Starting from one of the original parallel planes, con- FiG. 22. struct cylinders on the successive plane sections of the solid as bases with one of the n equal parts of the distance between the original parallel plahes as altitude. Denoting by Xq, X^, X^, Xg, X^, • 87 X,i the areas of 88 COMPUTATION AND MENSURATION the successive equidistant sections of the solid, and the distance between the parallel planes including the solid hj h — a^ the sum of the volumes of the cylinders is It and the volume of the given solid is the limit of this sum when n is infinitely increased. If the areas of the successive plane sections of the solid can be expressed in terms of the distance x from one of the end planes by an equation of the form X = A + Bx-\-Cx^ + I)3?-\-Ex^+ ••. +iV^% the limit of the sum can be determined exactly. In fact the determination of this limit is identical with the deter- mination of the limit for areas of Article 36. Art. 40. — Approximate Volumes The volume of a solid is given by the expression Volume = iimit(Xo+Xi+ X2+X3+X4+ - +x„_i)^i:^. « = Q0 71 It follows that the approximation to the volume is made closer by increasing the value of n. First approximation. When n=l, denoting by X^ and X2 the areas of the end sections, Volume = ^ (Xj + X^). Second approximation. When ^ = 2, denoting by Xj, X^^ and X3 the areas of the three equidistant sections, Volume = ^ (Xi + 4 X^ + X3). VOLUMES OF SOLIDS 89 This is the prismoidal formula. Third approximation. When w = 3, denoting the areas of the four equidistant sections by X^, X^, X^, and X^, Volume = 1 (5 - a) [X^ + ^CX^ + X^) + XJ. Fourth approximation. When w = 6, denoting the areas of the seven equidistant sections by Xj, Xg, Xg, X^, Xg, Xg, Xy, Volume = Problem! 157. The solid generated by revolving a cir- cular segment of one base about its chord is called a circular spindle. Find the first, second, third, and fourth approximations of the volume of the circular spindle if the radius of the circle is 14 feet and the height of the circular segment is T feet. Problem 158. A vessel laden with a cargo floats at rest in still water and the line of flotation is marked. Upon the removal of the cargo every part of the vessel rises 3 feet, when the line of flotation is again marked. From the known lines of the vessel the areas of the two planes of flotation and of five equidistant sections are calculated and found to be as follows, expressed in square feet: 3918 3794 3661 3517 3361 3191 3004 Find the weight of the cargo removed, supposing 355 cubic feet of water to weigh 10 tons. 90 COMPUTATION AND MENSURATION Art. 41. — Applicability of Prismoidal Formula The prismoidal formula Volume = l(h- a)(Xi + 4 Xg + X3) gives the exact volume of any solid bounded by two par- allel planes provided X=A-\-Bx^ Cx\ where X denotes the area of a section of the solid made by a plane parallel to the end planes, x denotes the dis- tance from this plane section to one of the end planes, and ^ A, B^ Q are constants. a. For prisms and cylinders X=A and the prismoidal formula applies. h. For pyramids and cones X=Cx^ and the prismoidal formula applies. c. For frustums of pyramids and cones X=^A + Bx + Cx^ and the prismoidal formula applies. d. For a sphere X= iri^ax-x^) Fig. 23. and the prismoidal formula applies. VOLUMES OF SOLIDS 91 e. For an ellipsoid of revolution or spheroid or and the prismoidal formula applies. Fig. 24. ^. For any ellipsoid Y X=7r-^C2ax-x^) and tlie prismoidal formula applies. Fig. 25. g. For the paraboloid of revolution Y X= 2pirx >-X and the prismoidal for- mula applies. Fig. 26. 92 COMPUTATION AND MENSURATION Problem 159. The radius of a sphere is 10 feet. Find the volume of the spherical segment whose bases are dis- tant 2 feet and 4 feet from the center of the sphere. Problem leo. Find the volume of the solid generated 2 2 by revolving the ellipse ~-\-^=l about the rc-axis. Problem I6I. Find the volume of the ellipsoid whose axes are 10, 14, 18 inches. Problem 162. Find the volume of the solid generated by revolving about the a;-axis that part of the parabola y2 = 2jt?a; bounded by a: = and x = a. Problem 163. Find the volume of the sphere whose radius is R, Problem 164. Find the volume of the frustum of a cone of revolution whose altitude is A, the radii of the bases M and r. Problem 165. A cask is constructed in the shape of the middle frustum of a spheroid. The inside dimensions of the cask are length 1^ meters, end diameters | meter, bung diameter ^ meter. Find the capacity of the cask in liters and in gallons. THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. ?^^ SEP 28 193^ m%^M flahs i>s --' 4Nov'60EW JCTIT^BO" OCT 11 1938 ^^'■25869