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University of Illinois Library L161— O-1096 ee Cd as ak ‘ Nes i i =, Ae WR (c} - 4) ; Me , oF a Ep at —_ . mn bs H as as Laat } N . al Cenk AD ——— ELEMENTS OF CROMETRY. pe SEAS SST PSE DE SF STORER DORIS PR EASELS EY —— ‘ 1 — | Noe L As ji wi ——— “aN “~A | iy i. s ELEMENTS OF GH Oe Mee Ths THEORETICAL AND PRACTICAL: CONTAINING A FULL EXPLANATION OF THE CONSTRUCTION AND USE OF TABLES, AND A NEW SYSTEM OF SURVEYING. BY REY. GEORGE CLINTON WHITLOCK, M.A, PROFESSOR OF MATHEMATICAL AND EXPERIMENTAL SCIENCE IN THE GENESEE WESLEYAN SEMINARY. New-Dork : PUBLISHED BY PRATT, WOODFORD, & CO. 1848, Entered according to Act of Congress, in the year one thousand eight hundred and forty-eight, by GEORGE CLINTON WHITLOCK, in the Clerk’s Office of the District Court for the Northern District of New-York. SSS SSE PGEE NSS CS SRR TE TH BSP GR a Stereotyped by C. Davison & Co., 33 Gold st., N. Y. PREFACE. WHILE some persons are complaining of constant innovation in text-books, and others finding fault equally with those in use, one scarcely knows whether or not to make any apology in putting forth a new work. One thing however, as it seems to me, is clear: that in view of the importance which is justly attached to elementary instruction, there can be little danger of too greata supply of manuals from which an enlightened community may select. If new books of geography and grammar, of arithmetic and algebra, are not only acceptable to the schools, in their onward march of improvement, but even indispensable for giving them life and vigor, why should objection be made to attempts in adapting the elements of geometry to the wants of the young and to the existing condition of instruction? Why should a blind veneration for antiquity cause the elements of Euclid to continue in one form or other in our schools, when the luminous Grecian himself, if now living, would, we dowbt not, no longer employ them without a material remodeling in conformity with the mathematical methods of the day? If boys are to learn that which they will practise when men, why should tyros be so long restricted to processes which, as mathema- ticians, they will seldom use? Is it essential to the acquisition of competent skill in numbers that our arithmetics should be filled with examples, and to the com- prehension of general principles that our algebras should observe, in the development of forms, an unbroken continuity of progres- sion? why in the elegant science of geometry should there be neither example nor process?’ I am aware that these suggestions are not applicable, in all respects, to certain books on geometry recently published ; but an elementary work of sufficient fullness, yet moderate in magnitude, highly practical, and, consequently, progressive in theory and example, is still, I believe, a desideratum. I have endeavored to prepare such a work ; how well I have suc- ceeded will, of course, be determined by others. Its chief feature 44-0 1o4| 6 PREFACE, will be found to consist, not simply in the acquisition of geomet- rical principles, but in a regular progression of method, whereby it is intended to teach how to geometrize. In pursuing quite out to the end of our geometrical studies, as well as at the beginning, the synthetic and undevelopable methods of the ancients, we acquire little or no power of going alone— we get some geometry, it is true, but still remain almost desti- tute of that education in analysis which is far more important. Why, in investigating the doctrines of forms, should we studiously keep out of sight the general principles of quantity, as if no such principles existed, when even Euclid himself could proceed but a little way without stopping to construct his, to ws, clumsy book of proportion, the best and only algebra at his command ? Why, when so much labor is saved and greater clearness ob- tained, should we refuse to employ an equation like (a+b)? =a?+b*+2ab? Shall we have resources at hand and refuse to use them because Euclid was poor? Is it shorter, more satisfacto- ry, or productive of finer results, to shut up the circumference of a circle between the perimeters of polygons than to avail ourselves of the simple symbol Ba , when employed as the vanishing ratio of the increments of two variables? Has geometry given birth to algebra, and shall she reap no advantage from her offspring? The succinct and methodical Franceeur quoting Lagrange, says, ‘* Tant que l’Algébre et la Géométrie ont été séparées, leurs progrés ont été lents et leurs usages bornés ; mais lorsque ces deux sciences se sont réunies, elles se sont prété des forces mutuelles,et ont marché ensemble d’ un pas rapide vers la perfection.”’* Again the lumin- ous Lagrange, in the first of his ‘‘ Legons ;” ‘* Les fonctions dérivées se présentent naturellement dans la géométrie lorsq’ on considére les aires, les tangentes,’’éc. t In accordance with these views, and in compliance with the re- commendation of Lacroix, avoiding double methods, that we may be ever pressing on in that body of geometrical doctrines that are most useful, I have paid much attention to the classification, endeavor- * So long as algebra and geometry were separated, their progress was slow and their application restricted; but when these two sciences became united, they lent each other mutual aid, and advanced together with rapid pace towards perfection. t Derived functions present themselves naturally in geometry, when we consider areas, tangents, &c. 4 *, a oe t iss 3b PREFACE. 7 ing, consequently, to arrange the subjects in a natural order, so as to fall easily into families and readily develope each other. Thus, by the simple method of superposition, instead of a long, mixed, and circuitous route, the doctrines of parallels are presently arrived at, and, as a consequence, all the elementary theorems relating to angles and independent of the length of lines, are embraced in a first section of moderate length. The comparison of equal figures follows ; then that of propor- tional lines, which prepares the way for the investigation of areas— the doctrines of the circle are not considered till afterwards. Thus, in the first part, the topics are kept distinct, and, it is believed, in their natural order, by which means the progress is rendered more easy and rapid, and the methods of geometrizing are introduced, one after the other, as required by the gradually increasing diffi- culties of the growing subject. For instance, in the first section little or no artifice is employed, and the simplest algebra, amounting to scarcely more than the common symbolical notation, is sparingly introduced in the second, while in the third the algebraic requisitions are somewhat increased, especially in the exercises. The method of incommensurables de- veloped as a part of a system in the introductory book, is employed for the first time in the third section of the second, or first zeomet- rical Book; the correlation of figures and change of algebraic signs find application in the more advanced propositions of the cir- cle in the third Book; and the ratio of vanishing increments draws a tangent to the parabola in the close of the first Part. The ele- mentary properties of the ellipse and parabola, being as simple as those of the circle and as useful in the study of natural philosophy and astronomy, are here introduced. Further, as proportion is generally included in our works on geometry, I have thought it ad- visable to insert an introductory Book, embracing, in a regular series of proportions the first doctrines of algebra, as being convenient for reference to those already acquainted with the science, and indis- pensable to others, who, by taking up these principles as required, may wish to proceed in the same class. ‘The first Part, consisting of a hundred and twenty pages, is designed to embrace, in theory and practice, such an introductory body of elementary geometry— all the more difficult problems relating to perimeters or areas being postponed—as is required, not only to enter successfully upon the study of the higher investigations that follow, but for furnishing, in im PREFACE. some measure, with tangible and useful matter, those who want the disposition, lack the time, or have not the ability to proceed further. The first Book of the second Part consists of an elementary system of functions, depending on a single variable, and presented constantly under the simple notation. It embraces the binomial and logarith- mic theorems, Every teacher must have observed that isolated methods, like those employed by Legendre in cases of incommen- surability, make only such an impression upon the mind as to leave a sort of confusion always hanging about them, while that which forms part of a system readily commends itself to the under- standing, and, consequently, remains ever after a permanent part of our appropriate knowledge. Laplace has well said, ‘+ Préférez, dans |’ enseignement, les méthodes générales; attachez-vous a les présenter de la maniére la plus simple, et vous verrez en méme temps qu’elles sont toujours les plus faciles.”* The method pur- sued in this book has been judged not only the most perfectf in it- self, but, as will frequently happen when connected subjects, instead of being disjoined, are permitted to fall naturally together, at the same time the easiest. But aside from the indispensable matter which it contains, the chief object of this book is to prepare the way for what follows in the arithmetic of signs, the construction of trigonometrical tables, and the mensuration of surfaces and solids. In virtue of the course just alluded to, I have been enabled, in the second Book of the third Part, to make not only a more than usu- ally full development of the trigonometrica] forms with their appli- cation in the practical resolution of triangles, but to embrace also the quadrature of the circle and ellipse. In the next Book I have developed a system of surveying which I regard as peculiarly my own. It is true that the theorem for the computation of polygonal areas, which constitutes its chief feature, may be substantially found in Hutton, yet I have given to it so much of a new form and a demonstration at once generalt{ and of the greatest simplicity, and extended it in so methodical a manner to the laying out and divid- ing of lands, that it becomes altogether another thing. Years of * In instructing, adopt general methods; endeavor to present them in a manner the most simple, and you will see, at the same time, that they are the easiest. t Not all the demonstrations in our algebras are perfect—for instance, the demon- strations of the binomial theorem in some school books, the most widely dissemina- ted, amount to no demonstrations at all. t The demonstration in Hutton is very tedious, and can hardly be said to be general. PREFACE, g instruction have proved, as hundreds of individuals would bear tes- timony, that the theorem here given will save, at the lowest esti- mate, two-thirds of the labor ordinarily incurred by the rectangular method. A further advantage is that, dispensing with a large and faulty table altogether, it is far more accurate—the computations being executed by aid of the common logarithmic numbers, calcu- lated with greater care and usually extending to six or seven deci- mal places, and the operation being so ordered that, without any additional labor beyond what is absolutely essential to an honest confidence in the result, all gross errors, if any exist, whether of the field or the tables, are detected, and if these have no existence, the smaller and unavoidable ones very much reduced. It is hoped that this book, often requested by my pupils, will prove acceptable to the schools generally. Of the third Part, embracing the mensuration of solids, spherical trigonometry, and navigation, time will permit us to say little more than that, by the method pursued, we have been enabled, within moderate limits, to give a fuller development of these subjects than is usually found in our elementary books. The modification and extension of Napier’s Rules demands, how ever, a brief historic notice. I demonstrated and extended these rules by showing: I. When A = 90°, a, = 90° — a, B, = 90° — B, C, = 90° — C, sinb = cosa, cosB, = tanC, tanec, | sinc = cosa, cosC,, = tanB, tand, sina,=cosb cosc = tanB, tanC.,, sinB.= cosC’, cosb = tana, tanc, sinC.= cosB, cosc = tana, tand: II. When a= 90°, B, = 180° — B, A, = A— 90°, &c., sinB,= cosA, cosb, = tanc, tanC, sinC,= cosA, cosc, = tand, tanB,, sinA,=cosB, cosC, = tanb, tanc,, sinb, = cosc, cos Bh, = tan A, tan a sinc, = cosd, cosC, = tanA, tanB,: Il]. When c=a, or the triangle is isosceles, sind, = tanA, tan(+B), sinA, = tana, tan(4d), sin(+b) = cosa, cos(4P),, sin(4B) = cosA, cos(4d). 10 PREFACE, Having shown the above extension to Mr. Dascum Green, then a pupil, he returned soon after, saying that he had not only verified my forms, but had obtained better ones, and presented the modi- fications of Napier’s Rules as I had extended them, substantially as they will be found in the text. This rule, as now extended and modified, possesses a greater simplicity and symmetry, and will enable us, in spherical astrono- my, frequently to dispense with complicated figures. I have added a small set of tables, extending to seven decimal places, calculated to answer the wants of the student while pursuing the work, and to make him more ready in using tabular numbers, by compelling him to interpolate by second differences. After- wards he will find it decidedly to his advantage to possess himself of the tables recently published by Professor Stanley. In conclusion, the advantages which we have endeavored to se- cure are: 1°. A better connected and more progressive method of geome- trizing, calculated to enable the student to go alone. 2°, A fuller, more varied, and available practice, by the intro- duction of more than four hundred exercises, arithmetical, demon- strative, and algebraical, so chosen as to be serviceable rather than amusing, and so arranged as greatly to aid in the acquisition of the theory. 3°. The bringing together of such a body of geometrical know- ledge, theoretical and practical, as every individual, laying any claim to a respectable education or entering into active life, demands, 4°. The furnishing to those who may wish to proceed on in mathematical learning, of a stepping-stone to higher and more ex- tended works. How well we have accomplished our object it is not, of course, for us to say. We have endeavored to render the work as mechan- ically correct as possible, but, residing at a distance from the place of publication, we can hardly expect that it will be entirely free from typographical imperfections. G. C. W. Genesee Wesleyan Seminary, June, 1848, CONTENTS. PART FIRST. ALGEBRAICAL PRINCIPLES—PLANE GEQMETRY DEPENDING ON THE RIGHT LINE, 1. Definitions, Mathematics, quantity, proposition. ........... esses oa oh My daxplanation Gf the Sighs te ya, | 3-g: Ooae soe 63's deh eered sai sihmninie 9 21 Shs A SEOUE BDO COCQUATIOR, a oe Saleikis abu sigergn du'§'on 4: SnaP 405-0 nin boedie fod 23 4. Inversions of additions, subtractions, Gc... ...4 ccs sees cesessce sees OD Go Gorticients aided and subtracted, wots edo css acs neh sme sencas medias 27 G, Palynomiials..<. ., x... «+ dtd de Seti «Pte IENs 6 wks espe ewe se denndl 29 7% Changing the sign of a factor —POWersy... wsie0 os dec aided cdescseonces 30 8. Square of polynomial, binomial, residual —Product of sum and dif- TERROR SOs oe ctnal etn ieee eee ed otis aig dmaiawse se eaen ae ews 6 ccmaee ‘9. Changing the sign ofa -polynomiabiissinn .) eid ved 8. .ee cae ecewsves 32 10. Degree of product, multiplication and division of powers............. 33 il. Multiplying or dividing dividend or divisor... ...... 0% as'osssevccescnce 34 12. A fraction an Expression Of sCiVISIONs woe thos orcs ce sawe's aes 8 sas» os 35 13. Multiplying or dividing numerator or denominator... .........002...- 35 - 14, Multiplying a fraction by its denominator... .... 2... cee cose cece cece 35 PG, BYACtiGis MUS Oe LOPCINET, ono ca costs sense 54 wham Aap sees tncsiac ts 35 16, Involiiign and evolution of fractions... 40. ass 2240 0th ose sans 56 35 i, Divasion Dy @ Traction—Teciprocal J. 2 Fs ences ered rede code a ean a's os 36 Ls. Reduetion® to given dencminetor iss i's nce. os ogls ss syccinia sme cieme’ od ts 36 19, Addition and subtraction Of 1ractions.... . ov. 2.50 coe0 ener osee woesie She 36 20. Kractions cleared of subdenoniimators. «eo. os os nsye scee tees seccwesse 37 91. Schole sasions. common factors, GC... wns none oss cone ues cves sete 37 2a, Equations: Defined ——Mentical ree. 5 os ras es aeay sees so ness coue seas 37 23. Transposition —Simple Equation, how solved,...... casey sas «pe dee 38 SA LAAT ORSG 6a Ov ee Sages be eee bs as phen a4 eb ece'ey + eee 009 sbmiel tage 39 Section II. Exponents—Proportion— Variation. i, Retiprocalepoweisk. ais ox hoigs ch hwnd. clnweelegeies. cds. 40 Root. of 2) powerextensionics . cicsieden gat denenvacccdcmemmieewie. ° Al 3. Fractional exponents......ver sees Meuysiyene cqanlemgeans Mw eeSe Le SE AND ON THE CIRCLE, ELLIPSE, HYPERBOLA, AND PARABOLA. BOOK FIRST. ALGEBRAICAL PRINCIPLES. Section I. Use of the Signs—Fractions—Simple Equations. 12 CONTENTS. Pacs 4, All real exponents subject to the same rules,........ secs ceseceeevese 42 5. Ratio defined—proportion—homologous and analogous terms—antece- Gents, CONSCQUEDES<\..5 ics 6 tks Wels ald Baan einis «ok Site o Wis ale lel 43 6. Inversions—compositions—product of extremes and means—involution and evolution—equal multiples... 0... .ss0ceseccocsvbosssencecn 44 a anverse OY reciprocal Proportion ...-...50.. ines secisens 25% oo eee be) ae ae 8. Continued proportion—geometrical progression .... ..e. cece veces cvden SO) 9. Variation defined—notation.. Sr fe ged... oe shee 10. Inversion — composition “Aiwwatntter-—riniinladt. comparison —com- bined comparison....... ‘ide meet wile a Say eral gs, Jia oiias + f's's Sanaa Section III. Analysis of Equations. 1. A single equation resolvable into several distinct eee Ex- BINPILES Saisie cereale Se SAE ae oo a eae 50 2. Bie taer ee acid for palgiaesira ae product _ PODS: Sakis SP 51 3. Classes of biquadratics solvable as quadratics... ........ 00. Su agi ee 52 4. Rule for putting problems into equation... .... .... eee wees tills dais. vee 5. Discussion of the two values of the unknown quantity........... ... . 5d 6. Equation between constants and variables... .... 2... cece cece cece sees OT i. Goemicients etiitted sia Bilis eet PAW, wea as eee ee a ocick cans ty ae (Pee Bs RePeieea eC Tas aula See Camas Te ee Dey tile oO take a bad Se's cies ewe cteee BOOK SECOND. PLANE GEOMETRY DEPENDING ON THE RIGHT LINE. Section [. Comparison of Angles. ° 1. Definitions. Geometry—solids, surfaces, lines.. .... .... ceee coos woes 63 2. straight line, nafure, origin Of NOMON. Of .scgas etme deceavsenenennes , OD 3. Corollaries—scholium on parallels.......... bo Boge 08 once ae ieee 4, Applications —Straightedge, paraltel nti ane 64 &. Sum of adjacent’angles constant, method... . ..'.'.. 0s 4 as00 00s aves, OO 6. Corollaries—right angles, sum of ni sebht ae lines formite one and the same straight line, vertical angles, sum of angles round a LET OR ie SS SES en, EE ies 5 ph ppeireaténe Papntanele, Guiverore Cree 67 8. Parallels, method, reversion and sneeninae:* bie Prelate sae al : 3 68 9. Corollaries—the converse, conditions determined by the achaltty of alternate angles, secant perpendicular, lines parallel to the same, angles having parallel sides, angles of parallelogram ............. 69 19, Application -—diawih? PAYallelssiys ca dug sir > «eka nev sc ce ees ape hp see 11. Sum of external angles of polygon........ ‘ RPM Se 12. Corollaries—sum of internal angles of pores on, Where son. renteen. quadrilateral, triangle, sum of acute angles of right angled triangle, the external angle formed by producing one side of a triangle. SOMO UID 5:5 oye d staves d Win wagidd d atnionspcisnar w aiden ORME re MATES A em. TO TB; WXCKCISES, 000s owe erage ttdsivanboine nendanacdcivees MAIO’ a J }. Polygons, when equal, how proved, . ..~ sis sined ‘is aoiris wiubatnins omdpecieticd 73 2. Corollaries—equal triangles, parallelogram divided by a diagonal, distance of parallels, diagonals of parallelogram, isosceles triangle bisected, equilateral triangles. ........ er ret et ok eee 3. Relation of angles in a triangle of atieleal sites ee ey Serer 73 4. Corollaries—the converse, a triangle having two equal angles—three. 73 5. Relation of two sides of a triangle to the third... ........ 0.0. cc00 cece 74 6. Consequences—perimeters enveloped and enveloping, the shortest dis- . tance from one point to another, shortest distance from a point to a OR cs oa ns balay ke A an py sca fee ee nS Cae ae 75 7. Triangles having two sides of the one equal to two sides of the other, each to each, but the included angles unequal.... ........ ..02 cece 76 8. Consequences—triangles having their sides severally equal, a quadri- lateral having its oppositesides equal. fos) 0. Fines cies cede wepes oe 76 AUR TE MOL CISES., oo n-a594cass 04t 1448 58 eee ese ated dawn 77 Section III. Proportional Lines. I, Segments of linesintercepted by parallels... .. s.00 <6ss neces osee saves 78 2. Consequences—similar triangles, Sc. 14.2 1... cok cece tee wees sence 80 3. A right angled triangle divided by a perpendicular........ tase cdee ot 81 4, Consequences—relation of the perpendicular to the segments of the hy- pothenuse, &c., square of the hypothenuse, &C........ ..00 sees voce 81 5. Relation between the oblique sides of a triangle, the line drawn from the vertex to the base, and the segments of the base—consequences. 82 6. Distance of foot of perpendicular to middle of base................... 83 7. Line bisecting the vertical angle of a triangle..... 1... eee seeeceeeee 84 SY Exercises. ,.smecen tian $c lemens oan Vues ogee Wediaes slatsvlenswmevies eau (Oo Section IV. Comparison of Plane Figures. 1, Rectangles—consequences, measure, right angled triangle............ 90 2. Trapezoid—consequences, measures, eas triangle, compar- isons, equalities........ acme s = oe 3. Triangles having an pee? of hese one y ede 8 an angle of ‘the eee CONSEQUENCE. «<2, casera sed dase MER UL HK 6 5c» seine nels oe eeits 93 4. Bre geas oh. sais hc OMe ne oe awe Tae kok a8 N aee eee code seins oa HOw BOOK THIRD. PLANE GEOMETRY DEPENDING ON THE CIRCLE, ELLIPSE, HYPERBOLA, AND PARABOLA. Section I. The Circle. 1, Definitions—consequence.. Bing « ave no ccctgnmencn veal. LOL 102 CONTENTS, Section II. Equal Polygons—First Relations of Lines and Angles. 2. Angles at the centre of acnfel iphone oneeqdencks, MeCASUPES. v0 ue « CONTENTS. 3. Inscribed angles—consequences, ..., taNngent....seeececececccsecceee 104 4. Angle embraced by intersecting secants measured... .........002.0-- 105 5. Principle—correlation: of figures... 2... es eek cece Sees cece cece cee te 106 6. Products of the segments of intersecting chords—consequences........ 107 7. Product of the three sides of a triangle, how related to the diameter of fre circumscriping CHClEs. 13 Suse kuee eee eee ets oe eek epee Sear 108 *G. Equation of thecircle and’ consequences esi. 4 We Sees tees 109 Be) SOS ELCRSCR Sete e Tes Sale ow Pin nes bs ro eee e Mae ents seteete toe scale Se Renee 112 Section Il. The Ellipse. 1. Equation of the*ellipse ‘and ‘consequences "1. So Siva. ges ot. ese ese. L1G am Trangent and conseqnemces’*, 71.4". tue i. ve totes vee Sine eRe ete dee bee 118 Ss. Npreral and Corollary sec ves cteeen so VE Sade bees s cu ene stirs see ee RD Se FGMETCISES TSS Sl STAC IT ee ue cacy THUR Fads Howe cede haste aree cst eens 120 Section III. The Hyperbola. t. Equation-of thehyperbola ss. 250 6. dswd ee tote ced searasee WENGE 121 Section IV. The Parabola. By EQUATION ANG CONSEQUENCES 2 v.04 xen dio ons aan tow ausiend os ud nse manent oe 2. Tangent—method, consequences. .... 262. sees cece voce best wee: shad 123 3. Scholia—signification of the symbol ba bial FR ARTS SOP ISUP BOC 124 4. Ellipse, spies and PRR how related.. fa tan mach sassy Loa? i ROC Ses, Loins ass i oe Binet neal Nahe thao Ck aicated aan aaa itene 2 LET PART SECOND. ANALYTICAL FUNCTIONS, PLANE TRIGONOMETRY, AND SURVEYING. BOOK FIRST. ANALYTICAL FUNCTIONS, Secrion I. Primitive and Derivative Functions of the Form xm, 4. Detihinonemnd SyDibols 6.2. -0:.4 segs fyssgecen ee seRe ack eae ks | ee 131 2. Derivativeol y = f& = Go 4 aD aaa A onc ccs cece ce wane uns 133 3. Derivative of the function, y= Ax‘, a being +, —, &e............... 135 4, Derivative of y = fx = Ax* + Bxr>+- Cz +-..., and the converse..... 137 5. Derivatives of equivalent functionsa 4... 4.5: ew exes wes se. o000 cows doce 138 Section If. The Binomial, Logarithmic, Interpolating, and Exponential Theorems. 1, Binomial Theorem, or development of (a+ 2)... cc. cece cence cece 139 a. aozariinms, rules of Gperalon.....-.%. «sis seme ssoc cote atin Gee e 145 CONTENTS, 15 Pace 3. Development of y = fz, when aY = x, computations. ........ ..00000. 146 & Interpolation Airis aadeeon tavc ties eeee Sake CeMeb eee Sater LOG or Exercises... Gta ites eee ena os chee seas sevens enbeertratitee” LOG 6. Exponential theorem, or development of y = a7..... 0... eee sees case. 159 Section III. General Laws relating to the Development of Functions depending on a single Variable. 1. Ratio of the increment of a continuous function to that of its variable 162 2. Derivative of a polynomial and multiple function..............-.... 163 3. Use of intermediate and converse functions ........ 2.06 cee. woes Teves 1O4 4. Derivative ora functional produeh.3 (esses eee bees ens eee LOO oO: Powerdfe function, derivatives 2s. FA Cee eee Pewee ene ess 166 6: Fraction: of fanctions,-defivatives 3. e039. 6.2 voce Oo 8k ceceder ia Lee 7. Substitutes in finding derivatives............. CN Bie naty Be tbe ve 167 8, Expansion of a function, Maclaurin’s theorem.... Sass ere ese tOG er AXETCISOS sd PGi tl. Cate hal ELM Tate atta MET EG beats th LOO BOOK SECOND. PLANE TRIGONOMETRY. Section I. Trigonometrical Analysis. 1. Construction and definitions—complementary arcs, sine, &c.......... 170 2. Sum of the squares of the sine and cosine, the sine an increasing, the cosine. decreasing, fUNctiOn « cin nv eaod “204 bainth* vimsmdccschatenen'g? LAL 3. Tangent, how related to radius, sine, and cosine...................6- 171 4. Tangent and cotangent, how related................. Si ei ts ile 171 Ga Secant and tanzent....+ 2s4+.000% «eee sutvhad s 4s she «meh asc ake 172 6 An Incremental Vanishing ArG..\s i... tiem sthasepenine mas Teer ee 172 TJ, Sine and cosine, derivatives Of .. sc opi s'ws os ss» afk eam ta satenbene 173 8. Sine and cosine developed in terms of arc..... 1... 0.2. wee eer one 174 9. Sine and cosine of the sum and difference of two arcs.. Aapaicaes Lad 10. Corollaries—sin 2x, cos 2x; 1-+-cos 27, 1 — cos2v; 1 ns sin ape 1 — sin 2x; (1+ sin 2x)! + (1— sin 2r)#; sin p+ sin g, sin p—sin gq, cos p++ cos g, cos gy— cos p;(sin p-+-sing);(sin p — sin q),&c.; sin CFE OR ee race thee Vee Beeb arden anes sonnet cantons he's 90% 177 11. Tan‘(@ +0), cotan (a-b4),”.'. tan Za, Cot 2a, KC... 20. cee case seseces 178 12. Denominate equations—radius restored..........se- cece cece cece sees 180 13. Arc developed in terms of tangent, .-. r computed. ..........2--0----. 181 14. Trigonometrical lines computed .... 0... .5.. cc ee cece ceee cece come ees 183 15. Logarithntie simes'and tangents... ici ices cae chee cee'e canis vine sen 188 16. Arc developed in terms of sine, 2. ..:.... cs 0. eens coe cone teen sane cone 190 17. Trigonometrical equations solved............. ee yr Pe Section II. Resolution of Triangles, and Mensuration of Heights and Distances 1, Projection of one line upon another. ........ .c00 cece ce cous sees ceeee 192 DMD OP dD CONTENTS. . Fundamental relation between the sides and angles of a triangle....... 193 . Sum of squares of two sides—corollary..... 1.2. sees cece cecscese sees 194 . Sum transformed into product—object Of... ....ssee cece cece eeceseee 194 . Consequences—sin 3A, cos., tanis, SiN A$. 4%) s.0mp oiey ve oils oss bimmenet LOE . melation of sides and opposite angles 7......4.49s a5 ees 05s eynagabane LOO . Ratiovof sum/‘of.two'sides‘toalifferenceyc. wwii Wess. wes WEE, 195 . Cases in Plane Trigonometry—applications........... és eee 195 Section III. Quadrature of the Circle, the Ellipse, and Parabola. 1.. Circular sector, how measured, ,:.. area Of CITCIE,). « «6:00 be wipe 09 esse den 201 2. Ratio of circumference to diameter; .-. arcs of similar sectors, .°. areas of circles; circles and their like parts, how related............-+.. 202 3. Incremental vanishing arc of continuOUS CUIVE..... 2... 200. eee woeee 203 4. Segmental aroa, derivative of; .... ss cama iacaniils dnible ele Wels lenienieietens 204 5. Area of ellipse—compared, With, CiFcl@.. + uip:sis/} «winise sale ore = "ts 6s kee 5 0 baila canes Sele 205 7. ‘Proximate-area of CONLINGOUS. CULV Os ssc sidan oa sivicdndueas saenteees apa 206 As TORETCISON, i. «de! 40s cp os Grea tae e > he aw iamiea wemneNGslbe ss pas aoee ane BOOK THIRD. SURVEYING, Section I. Description and Use of Instruments. 1, The chain—tlength, division, how used, field notes... .... ..s2 sees see 208 2. The surveyor’s cross—construction and US€.... 2... cee. cane cece eee 210 3. The compass—how graduated and lettered, how used............-..- 210 4 Vernier or nonius......... gecress Ceoe ses cee AEs DNR SSS Eee ee eee 5. Theodolite—how sptsiinted awed Apes aks peue ase cane <5 pc tevatogy ter. 212 Sev aridtin Of NRCS tn ata tle eae yt sae fena as ner ha vinemetens fake ree 214 BAC ELIS eis aati bh waleiwn bee sr GOA Obie Ns aie Fos Op. 34 $4.4 ood aaien oa rate Tae Section II. Plotting. . Graphical problems—perpendiculars, parallels, &c.......--.-....003 216 Problems of construction—-% = @ ++), Ge... 2. wie se ew emecngecs srgmes CLO ~ GtagGation of the circke—-chorgg, . 3.3 od < ceia es arming > do a0 ap ae To plot a fel, ‘Diagonal scale Sector copa tire» cemdoen aa neces canen 222 be ERRORS METAL Oo a a ee he sess clpusimiha bata bigs hin on ghana 224 nat BTEC PVM a at al es as QUA cid, AW DOs A OM dlin a Aachen ote in ne Section III. Computation of Areas. 1. Last side, and diagonals of polygonal fields..... 6... .... cee wees cone 225 2. Corollaries. Similar figures and proportional lines; the Pantograph. 227 Bye LECTCISOS, 5 4s tc k's aie ee AN Sekai AUD. 05:4 bis bccn o>, 9454 4 kip ene 229 4. Area of polygon in terms of the sides and their inclinations,.......... 231 CONTENTS, 17 Pac . Corollaries—triangle, parallelogram, similar polygons, equilateral, 5 regular—equilateral triangle, square, &C..... 0.2... 0. eee coon eee 232 © Form of compuanenayniet ees eRe Ae ee a 234 i. Hercices, pr peasenart series cars eel pee ee tiee Deen eee 236 ©. Dividing-andiwying out lands: cz. cvs. eeaecct ee 12 POAT IN Fe 237 PART THIRD. SOLID GEOMETRY, SPHERICAL GEOMETRY, AND NAVIGATION. BOOK FIRST. SOLID GEOMETRY. . Section I. Planes. 1. Position of plane, how determined, consequences.........++. sees cee: 245 2. Line perpendicular to: plane ,COnseQUen Cesta, ta. «Ni o0sd asian cg vs eb ees 246 @: Parallel planes iniersectetl, Consequences. co. <..ajc.s sae va ee acres gee .. 247 4, Similar figures described by the revolution of a line passing through a fixed point and intersecting parallel planes, cone, pyramid, cylinder, Pisin, Helse dee Se ae Pode eee ees sc OM ET eee vecce beleclascecemss 248 Section II. Surfaces of Solids. in Polyhedron, Surface, CONSEQUENCES, <5. .4< 0 cc eo as 35.4 mente dae ead 250 ®.. Surface of revolution, devivalive, -7..4 550 ese ac 8x «> 0d dae yaw mageetodins 251 ae Spherical zone," comsequenees, cc tess aise ds nape bans.scgetuss nog beean ae cae Be PRERCISER I Bia tad ens eee AO Aa oer de haps nee decks SRR A Mad EN a ee Section III. Volwmes. 1. Rectangular anette re how related, consequences. .......... 253 2c Pyramid, measured: ..... 2.252. sig x's «OOM BEMIS St lore een EBS 3. Frustrum of cone and paraniia, cinseauenee Fas Cbdatteie e 0g vee sy ais 39 255 4. Derivative of solid generated by variable plane...............2. c000. 256 5. Consequences, . phat frustrum, .’. ellipsoid, sphere, ..., Para- poloid.. hi Seis i disk EDR RaW ha we 05 + OT, i eRe ee ee 6. Similar sole; how palaces, oP RIT SRR Re US Ae Fe ESGTCRAE A ede shee wien 6085 6 qe 0 es ETO OU REN oe ede. Sed 259 BOOK SECOND. SPHERICAL GEOMETRY. Section I. Spherical Trigonometry. .. Sphere defined, cOnsequences, i. cies cere cess ewes tevnceeendes sien OL 18¢ CONTENTS. PAGE 2. Spherical triangle measured, CONSEQUENCES. .... sees ceee seve cone sence 262 3. cosa = cosh cosc + sind sinc COSA, CONSEQUENCES. ..+. eee voce cove cers 264 4, Sides, how related to opposite angles, CONSEQUENCES. .... 202 eee vee 267 5. Elimination, first, second, siiciaahaitans kh) were. 268 G> Napier’s analogies 20.26 ¢9<0/<< vo censes+ ives dam sah poe tae eeteeeee See 7. Napier’s Rules ETM hee igh 's bn eke RN ap Apes «54 lee 273 8, Cases in spherical trigonometry..... 6... ceee cece cece cece seen ceceees 275 i EROPCISOG. i aaa cus wee sada s Coes os cea oh pe aes SPte meee 4 cee eevee eee Section II. Projections of the Sphere. 1, Orthographic projection, consequences. .... 2.2. cece eee sees cece cee 280 RAW TRQOG: «dn. tcis nonce, 5 4's he soe Nba y. 59 ok 8 4.8 eels Steet nie a ea Sw OSs 6 0m ie 282 St. AJROMORIO PIOJCCLION.. wv s- »cee's¥os ssh case Lnse Gnne ae chvend yUee tute 283 a, CONIC SECHON.. .5..\ ss «sR RE EGE + SERB REMe 05 54 ee es tips vend aye <> 284 5. Steréographic projection, consequences, .'. 4 20. v= 0c wsse vepe ewes eves 284 ©. Mow made. ....4'0. 5h deh EO eb MEM ae ce cane ngtiayades 286 a. Conical ‘projection, hOW Made. as. aye «04 cmiers os ce oeue s650 meee Aeon 288 SL PACECIBPH, . .. a kno vainh oo ar ere eee eee Ree seas Ones Ac oe She Palys ek ol ee BOOK THIRD. NAVIGATION. Section I. Problem of the Course. L. Pitterence of Fatitude 4 ob saces aac be rng cbc Ae oees DOR ep aels ek & 291 a, Difierence of longitude, consequetices.. oj... <5 5e/swue weSe wees ones a sen 292 Be aPauer saihinig. 2 3S ee hese Wek Si bee ote eae Weep op knee tate 293 SS CHOMM NITE N's 2 SEAN. VST PSUS CER i ok oe Sere ety eee chee eee 294 Jer LERCTCUSCRS 04's St LUGE VLR ee Se Ss oh ee eee ooh eee Ate et Be ee rae 294 Section II. Problem of the Place. De Lavine ‘by meridian AMUUTde. 24 ds vscie ny bagels kde oe é RNP bw oe bu coe 296 an Litae, consequencesuestp7..utiet Ri .sPeGiaa NCE eel 298 seLongitude, by. lunar distance .o.aveseseos «4a000saecaweneah olen 302 Section III. Description and Use of Instruments. 1, Course Mariner's compass, .¥'.9):93h Gn. LG Waka hea o aerg suas yeas 304 Me OE DUG HOG... vsk wo decuck wb Strdd'n's wa eenia pula de a) anne aes 304 3. Zenith distance—sextant, adjustments, use, depression of horizon, re- fraction, marallax, semidiamMeter, ...< ness seen: saanaye v4 ROSS 305 ADDENDA [IME ca es Or + Aggregation, Equality, Inequality, Deduction, Continuation. (), 00) 3€0)23 ae SR ETOMIRT), out outs Thus, that 4 is to be added to 6, is written, 6-+-4, and read, six plus four; that six and four are equal to 12 diminished by two, is written, 6-+-4 = 12 — 2, and read, 6 plus 4 is equal to 12 minus 2; that the sum of 6 and 4 multiplied by 2, is equal to 44 diminished by 4, and the remainder divided by two, is written, (6 +4) x 2 = (44—4) +2, or (6-+- 4) « 2= (44 —4) ;: 2, or 2(6 ++ 4) = (44—4) : 2; for when the omission of a sign, as is done in 2(6-+-4), would not be attended by any ambiguty, it may be dropped. Thus a 5b sig- nifies that a is to be multiplied into 5, Propositions are much shorter in symbolical than in common language, and are, conse- quently, more clearly expressed, as has already been shown, and as will appear in a still stronger light by writing an example or two in corresponding columns, a‘='5, | the a is equal to }, c=b; >same ¢ and ¢ is equal to b; a=. as (therefore a is equal to c. (a+b) (a—-b)| The sum of a and 6 multiplied into the differ- “m _\ence of a and 3}, and this product divided by m, aa—bb gives a quotient which is equal to the quotient Ce a arising from dividing the remainder of the square aa of a diminished by the square of b, by m, which < m again is less than the quotient of the square of a divided by m. Scholium. The student should be accustomed to turn the alge- braical into common and appropriate language ; thus the sign of equality, =, will commonly be read, ‘is equal to,’’ but sometimes, *‘ will be equal to,’ and at others, ‘equalling ;” the symbol of deduction, .., will generally be read, ‘ therefore,” sometimes, ‘hence,’ ‘it follows,” and again, ‘from what goes before, we infer,” &c.; the symbol of continuation, consisting of three points, AXIOMS, 23 wey Will be enunciated, * &c.,”’ and so on,’’ “ continued according to the same Jdw.”’ PROPOSITION I. [axtos.] The whole is equal to the sum of all its parts. (1) This proposition is an axiom, that is, evident of itself; no words about it, therefore, can make it any plainer. Corollary 1. The whole is greater than any of its parts, (1,) or, the whole exceeds any of its parts by those which are except- ed—otherwise the whole would differ from the sum of all its parts. Cor. 2, Quantities which are equal in all their parts, are (15) said to be equal to each other; for the whole is known by its parts --or, quantities which are not evidently identical, can be compared only by a resolution into like or unlike parts. Cor. 3. Quantities which, however resolved, are unequal (1,) in any of their parts, are not equal to each other (1;). Cor. 4, Quantities which are equal to each other, are (1;) equal in all their parts; for, if some of their parts were unequal, they would, by (1,), be themselves unequal; .°. Cor. 5. Unequal quantities are not equal in all their parts. (1,) Cor. 6, Quantities which are equal to the same or equal (1,) quantities, are equal to each other; for they are equal in all their parts, (1), (Is). Cor. 7. Quantities measured by the same or equal quantities, (1,) are equal to each other; for equality of measures implies equality of parts, whether the measuring quantities be of the same kind with those measured or not. Thus, two masses of lead are equal in weight when they both contain the same number of pounds, or when they both contain the same number of cubic inches. Cor. 8. Quantities ate to each other as their measures; .*. (14) Cor. 9. Of quantities having ttnequal measures, that is (1, )) the greater to which the greater measure belongs. Cor. 10. If the same or equal quantities be increased or (1,,) diminished by the same or equal quantities, the resulting quantities will be equal to each other ; since they will be equal in all their parts, (1,), (1). Cor. 11. If the same or equal quantities be multiplied or (1,2) divided by the same ot equal quantities, the resulting quantities 24 AXIOMS. will be equal ; since multiplication is repeated addition, and divi- sion a continued subtraction. Cor. 12. If equal quantities be raised to the same pow- (1,3) ers, or the same roots be taken of them, the resulting quantities will be equal; since a power is formed by continued multiplication, and a root is extracted by the converse operation. Cor. 13. The same or equal quantities, by the same or (1,,) equivalent operations, give the same or equal quantities, Cor. 14. Quantities satisfying the same or equivalent (1,5) conditions, are equal to each other. Cor. 15. Unequal quantities, by the same or equivalent (1,.) operations, will continue to be unequal, and in the same sense— that is, the greater will be the greater still (1,,). See also(1,,), (112), (113). Thus, if a be greater than 4,[a> 6], a increased by c will be greater than 6 increased by c [a+ c,> 6-4 c], a diminished by c will be greater than 6 diminished by c,[a—c > b-—-c], m times a will be greater than m times b,[ma > mb], &c. Cor. 16. If inequalities, taken in the same sense, be (1,7) added, the result will be an inequality also in the same sense (1,4). Thus, if @ be > b and c >d, thena+c>b+d, Cor. 17. When inequalities, whose differencesare the same, (1, s) are added in a contrary sense, the result will be an equality (1,). Thus, if abe as much > db as cis < d, thena+e will =d+d, Cor. 18, When inequalities are added in a contrary sense, (1,4) the sense of the resulting inequality will be that of the greater (1, ,). Thus, ifa > 6 andc < d, then will a +c > 6-+d, provided the differ- ence between a and bd be greater than the difference between c and dfa—b>d—c,..a>d—c+bha+cec>b+d}]. Cor. 19, If inequalities, taken in the same sense, be sub- (leo) tracted, the one from the other, the resulting inequality will be in the same or a contrary sense, according as the minuend is the greater or less inequality. PROPOSITION II. [corotrary From 13.] Magnitudes which may be made to coincide throughout, (2) are equal to each other. The magnitudes are equal in all their parts. Cor. 1, When one magnitude embraces another without (22) being filled by it, the first is greater than the second (1.). INVERSIONS, 25 Cor, 2. Magnitudes measured by the same or equal mag- (23) nitudes, are equal to each other (1,). Cor. 3. Magnitudes are to each other as their measures (1,). (2,4) Cor. 4. Of magnitudes having unequal measures, that (2,) possessing the greater measure is the greater. Scholium I, It is sometimes convenient to make a distinction between equal and equivalent, but the terms will generally be used as synonymous, Scholium IL. It is obvious that all propositions requiring demon-_ stration, must be founded, either directly or indirectly, upon those which do not, or on axioms; and hence our first proposition be- comes the source of a vast amount of knowledge. Def. 4. A coéflicient is a figure employed to show how many times a letter is taken; thus, in 3a, 3 is the coéfficient of a, and 3a=a +a--+a, A letter may be regarded as a coéfficient, as n in na=a+a+a-...[n times], — Def. 5. Operations are said to be relatively free when the result is the same in whatever order they are executed, the one after the other. Thus, O, O,, the two parts of a compound operation, are relative- ly free when O, (O.) =O, (O1). PROPOSITION III. Additions, subtractions, additions and subtractions, are (3) relatively free operations—that is, the terms of a polynomial may be inverted at pleasure. 1°, Additions. That 4+ 3 is = to 3+4 will be evident from counting the units into which the two sums are resolvable ; thus 4+3=(14+14+1+4+1)+(14+1-4)), and 34+4=(1+14+1)+(14+141+4+1); Pe Ee 4+3=344. So a+b=(1+1+4+1+... [a units]) 4+ (1+1+1+... [6 units]) =1+1+4+14..[a@+)] | = (1-414... [6]) + (141 PI+s. [a]) = b-ba, which was to be proved. 2°, Subtractions. The remainder arising from diminishing @ units first by 6 units and then by c units will be found the same as that 26 | INVERSIONS. obtained by diminishing a first by c and then by 5, ora—b—c=ea ~c-—9b; for, let a containr+6-++-c units, or a=r-+b-+-c, which (1°) =r-+-c+b; then (1;,;) a—b=r-+c, subtracting 5 from both sides, and a—b—c=r, subtracting c from the last equation— or a—c=r-+-b, subtracting c from the first, and a—c—b=r, subtracting 0 from the last ; .(1,) a—b—c=r=a—c—bd, Q. E. D* 3°. Additions and subtractions, a+b—c=a—c-+5b; for we obviously have the same number of units whether we diminish the a+ units by taking the c units from 5 or from a, Q. E. D. PROPOSITION IY. Multiplications, divisions, multiplications and divisions, (4) are relatively free operations. 1°. Multiplications. We may know that 3 « 4=4- 3 by resolv- ing the numbers into their component units, and setting these down in an orderly way to count; 4 units. wo thus, by counting, we find 4 units repeated 3 times, : } 1111 the same as 3 units repeated 4 times, or 3 « 4=4 3. a@ units. oe Nati has ia: Dae So, aunits (=1+1+1-+... (a]) repeated btimes, &|11 1... is the same as 5 units (=1-+1+1+.., [b]), repeat- GJ111 ed a times— os or ab = ba; eho ee hence (1,2) a@ times 6 units, repeated c times will be equal to } times a units repeated c times, but 6 times a@ units, repeated c times, by what has just been demonstrated, is = to c units repeated b times a times—and in order to repeat c units } times a times, we may mul- tiply first by 6 and then by a; for, multiplying by 3 instead of ba, is multiplying by a number a times too small, and consequently, the product, being @ times too small, will be corrected by multi- plying again by a; all which may be set down in symbols thus ;— * “Quod erat demonstrandum,” which was to be proved. COEFFICIENTS. 27 ab = ba, Ll x3) : (ab)c = (ba)c = c(ba) = cba, or abc = bac = cba, where it will be observed that a is made to occupy every place in the product, and in like manner the same may be shown of } and c —and the reasoning may be extended by introducing additional fac- tors at pleasure. Q. E. D, for 1°. Cor, 1. The factors of a product may be grouped in multi- (4) plication at pleasure. Cor. 2. Any factor may be regarded as the coéfficient to (4;) the remaining factors of the product (4,). 2°, Division. Assuming any quantity, Q, to be divisible by others, as 6 and c, is obviously the same as assuming Q to be resolvable into factors, two of which are 6 and c; hence, Q being divisible by } and c, denoting the third factor by a, we have Q=abc; *.(1 py) Q : c=ab, dividing both sides by c, and observing that the product abc is divided by c by omitting the multiplier c; .*. dividing the last by 6, there results (Q w¢).: b= a. But (1°) Q = abc = ach; i Qa:Ab = ze: and (Q.2.b) sea; *. (Lz) (Qodeb)uit ci (@ .2.:¢).4,0., .Q- Ey D,. for, 2°. 3°. Multiplication and Division.—Let Q be divisible by 6, or Q=ab; rat Q.: b=a, and, multiplying by c, (Q : b)+c=ac; again Q-c=ab+c=aceb, Ps (Q' 2: ¢)uinbeeges hence (Q :.b)ec=(Qec): b,. QE. D. for3?, and the proposition is proved. PROPOSITION V. Additions and Subtractions in regard to multiplications (5) and divisions, are relatively free. 1°. Quantities otherwise alike are added and subtracted by (5,) adding and subtracting their coéfficients. 28 COEFFICIENTS. 3 + 2 ("I~ ~~ cl Pe hy i | Thus, 3 times 4 and 2 times 4 are obvi- 1 sf Tee ok ously 3-++ 2 or 5 times 4, sn tO IG ati | 3-412.4=(842)-4=5-4, Keane Denies bene ode a + b ee ee oe ae So a times c plus 6 times 1+14+14+..4+14+1+4+1-+.. c is equal to a+b times c, 1+1+4+1+..+1+1+4+1-+... or ac+bc=(a+ djc; 1t+1+14+..+1+141-4.. *, (a+ b)c =ac-+ be, in- f FR Se aes verting the members of the equality, which obviously ° may be done, since it is only asserting hie same proposition in an inverted order ; * (111) (a +5)c — bc = ac, subtracting bc from both sides ; but, as a and b may be any quantities whatever, a+) may be any whatever, and we may substitute m for a+ 6, n for 5, and, conse- quently, m—n for (a+b) —b=a; whence me — ne =(m— nec, (5s) or (m — n)c = mc — ne, Scholium. The last form, resulting from the hypothesis (5,) that m is greater than n [(a-++ 5)>b], must also be adopted when m is less than 2; otherwise general symbols for the representation of quantities would have to be abandoned altogether, as it will fre- quently be impossible, as well as generally inconvenient, to distin- guish between m and n, whether m be greater or less than n. And in order to thus extend the application of the form, or to make it general, we have only to interpret the expressions mc — nc, (m—n)c, both when m is greater and less than n : 1°, when m >n, we have mc > nc and, consequently, mc — nc, m — n, both plus; but 2°, when m < n, we find mc < nc, and mc — nc, m—n, both become minus. It is easy now to extend the operation to any number of terms, whether plus or minus. Thus, azr+br—cer+dzr+...=(a+b)e—cr+dr... =(a+b—c)r+dr... =(a+b—c+d)r-+... =(a+ 6—c+d++...)a. So the first part of the proposition is proved. POLYNOMIALS, 29 2°. Dividing both sides (1,.) of the last equation by xz, we have (az+ bx—cx4+dz+...):xr=a+b-—c+d-... = (ar) 2>@-+ (bz): 2—(cr) :.2-+(dzr): 7+... , and the proposition is proved. Cor. 1. A Polynomial, or algebraical expression consisting (55) of many terms, is multiplied or divided by multiplying or dividing its terms, Cor. 2. One polynomial is multiplied into another by mul- (5,) tiplying all the terms of the one into all the terms of the other. Thus, (a+b+c-+...) (dg +b,+¢,4+...)=(at+b+ce+...) dg +(atd+tet+...)b,.+(a+b+c+...) cg +... =aa, + ba, +cag +...t ab,-+ bb, -+cb,.+...+ ac, + beg +cc, +... (5,). Cor. 3. The number of terms in a polynomial product, is (5;,) equal to the product of the numbers denoting the terms in the con- stituent polynomial factors, Thus, af’ P..’ P;, P:; :.. (nej, denote polynomials of 4,°b,’c,”..: terms, the ae ais being m in number, we have P+ P, =P, 4 polynomial of a times b terms ; Ay Dg) PM egy! ado ed ach EI ghd and generally P,+ P,+ P,«...[m]=Pabc-.., [m]. (53) If wemake the number of terms the same, a for instance,in all the polynomial factors, or puta=b=c=..., there results, Pee, ele ee my Pad... Tiny, or CPA rar, ts Os (5,) Cor. 4. The number of terms in the mth power of a polynomial of a terms, is equal to the mth power of a. Thus, the number of terms in the expansion of the sixth power of the binomial r+ y, [ (c+ y)*®] will be found to be 2° = 64. Def. 6. Operations like those preceding, which may be per- formed upon the whole of a polynomial at once, or upon its parts separately, are denominated linear. Cor. 5. The compounds of the above linear operations, (5,0) are themselves linear. Thus, if we multiply any polynomial, z+y+2+..., by any quantity a, and then divide by 6, we find fa(ct+ytet+...)]:b=(ar): b+ (ay): b+(az):b+... 30 LAW OF THE SIGNS, PROPOSITION VI. A product made up by the multiplication of additive (6) quantities, is itself additive ; and is changed in sign by chang- ing the sign of any one of its factors. This proposition will become evident by comparing (5;) and its extension in (5,) with the first part of (4); for, observing that m — m may represent any quantity either plus or minus, by making m greater or less than n by that quantity, and that m—n, mc — nc are both plus when m is greater than n, [m >n],and both minus when m < n,(m—n) c= mec — ne, becomes + (m—n)e-+e = + (mc —nc), or +++ =-+ when m> n, and —(m—n)++c = — (mc —nc), or — + +=-—, when m=—2 e—TZe —r=-—2', (—x)?*} Be hosts ren ' . * Whatever n may be, 2n is an even number. SQUARE OF POLYNOMIALS, — 31 Cor. 6. An odd root of a minus quantity is minus (6,). (6,) Thus, "°° —a=—r; for —a=(—r)**'=—r*t',—=-, Cor. 7. In products and quotients, like signs[+,+, or—,—] (65) produce plus [+] and unlike signs [-+, —, or —, +,], minus [—]. For products this has already been shown (6,), and, to establish the same for quotients, let g be the quotient arising from dividing any quantity D by any other d. or DD: a= 0, whence (1,2) (D: d) + d=qd, multiplying both sides by d, or (4) (D.d):d=D=qd; = / + D=+q+-+d, whence +:+=4, —D=+9--—d, we tee, + D=—q--—d, we fi-=-, —D=—q-+4, eee —:+=-—. Q. E. D. Cor. 8. In products and quotients the signs +, —, are (64) relatively free (6,). Thus, -+e«—=—+«+4,+:—=—:4. PROPOSITION VII. The square of a polynomial is equal to the sum of the (7) squares of the terms and their double products taken two and two. For in (5,) making a, =a, b, = b, cz =¢,... , and arranging, we have (a+b+c-+...)? =a?+b?+c¢?+...+ 2ab+ 2ac+... +2bc-+... (7) Cor. 1. The square of a binomial is equal to the sum of (8) the squares of its terms increased by their double product. For in (7) making all the letters nothing except a and 6, there results (a+b)? = a?-+b?4- 2ab = a?+2ad+b?=a?+(2a+b)b. (8) Cor. 2. The square of a residual is equal to the sum of the (9) squares of the terms diminished by their double product. For changing the sign of 6 in (8) we have (a — b)? = a? + (— 6)? + 2a (— d), or (63), (6), (a — b)* =a? + b? — 2ab. (9) PROPOSITION VIII. The product of the sum and difference of two quantities (10) is equal to the difference of their squares, and vice versd, the difference of the squares of two quantities is equal to the product of their sum and difference. 32 CHANGING THE SIGN OF POLYNOMIALS. For in (5,) making all the letters nothing except a, b, do, bg, and changing a, into a, b, into — b, we get (a+b) (a—b)=a? —B’, e i BN (ast bla B). (10) Q. What relation does (5,) bear to (7)? (7) to (8)? (8) to (9)? (5,) to (10)? Which is the more general (5,) or (7) ? Scholium. Forms (7), (8), (9) and (10), are important for their applications, and remarkable for illustrating the facility with which general truths are discovered by the employment of algebraical language. PROPOSITION IX. A polynomial may be freed from the minus sign, or on (11) the contrary subjected to its influence, by changing the signs of all its terms. Every polynomial, as 3a — 5b + c+ 2d—e, may be, so far as the signs are concerned, reduced to the form -+ B— C, representing the sum of the plus terms, as 3a-+c+2d, by B, and the sum of the minus terms, as — 5b —e, by C. Therefore all cases of subtraction will be comprised in this general form, A~(+ B-—O), where it is required to take the polynomial (+ B—C) from any quantity A. Now if from A we subtract B, the sum of the plus terms, we have A— B, by which all the terms in B originally + become — ; but, in taking the whole of B from A, we have dimin- ished A by a part of B, namely C, which ought not to have been subtracted, since the true subtrahend, B — C, is only that part of B remaining after the diminution of B by C; therefore, A—R being too small by C, the true remainder becomes A— B+C, on the addition of C—and all the terms embraced in (+ B—C) change their signs in (— B+ C), also A—(+ B—C)=A-—-B+C, or —(+ B—C)=— B+ C, or —BiAC=—(4+B-—C,) QE. D. Remark, The student should be familiar with the resolution of polynomials into factors, not only by the addition and subtraction of coéfficients, but by the employment of the theorems under (7), (8), (9), (10). DEGREE OF PRODUCTS. oo EXERCISES, 1°. What is the sum of fifteen times x and seventeen times zx ? 152+ 172 = how many times x? 2°, axr+br=? 3° 2a+b)4+3(a+b)=2 4° a(c+d) + b(c + d) =? 86°. 1085 —7b=? 6°. 35b—20b=2 7°. ac—be =? 8°. —ad+bd=—(ad—bd)=?2 9°. mz+nr—rzr=? 10°. ac—ad- bd — bc = (a — b) (c —d,), how ? 11°. Resolve a? —b?, a*—b*, a®—d*, a® —B*, at® —H!, a” — b*, into their simplest factors, 12°. Resolve 7? +2241, x?+1-—22, 1—2ry—zr+y)+zrzr + yy, into their prime factors. PROPOSITION X. The degree of a product is equal to the sum of the num- (12) bers indicating severally the degrees of its factors. For, as two factors, ab, multiplied by a third, az, give a product of three factors, aba,, so a monomial of p factors abc... [ p], multi- plied by an additional factor, a,, gives a product of p+1 factors, abc... X @g[p-+1], and by a second factor, abc...X dab, [p+2]; and generally, adc... < Kdgbocq ...[q] = abc... Xdgbec,...[ p+]; abc... [p] X bolo... [Gg] X agbgC3... [7] = ADC. K Aadgly i. X agb3¢3-..[p+q+r] 5 *, in general, abc...[p] X a,b,c, ...[q] X @3b3¢3... [7] X ... = abe woe XK AgD,Cy 00 X Ayb3C3...[pt+g+r+...]. (12) Cor. 1, Powers of the same letter are multiplied together (13) by adding their exponents. For from (12), making 0, Cy... Gado,€q) «++ Aaj03,635-.. all = a, there results, a...[p] X aaa...[¢]K aaa..[7r}] X... = aad... (pay r-..), (13) or Pea ea se are eT ee, Cor. 2. A quantity is itvolved by multiplying its exponent (14) by the index of the power to which it is to be raised. For, making p, q, r,... all= m and m in number (13), becomes Oe Ore os [ ep aati io hid, (14) (a”)*=a™. 3 34 MULTIPLICATION OF DIVIDEND, Cor. 3. Powers of the same letter are divided by diminish- (15) ing the exponent of the dividend by that of the divisor. For, from (13) we have a®ri=—aqdPe a’; se art? s g’=@a?=a?r-7 (15) or{fp+q=2,9g=y], @ :@=a, Cor. 4, When one polynomial is divided by another, the (16) highest power of any letter in the dividend divided by the highest power of the same letter in the divisor, gives the highest power of that letter in the quotient. | For (15), divisor a’ quotient a? = dividend a?**. Scholium. It will be found convenient in division to arrange (3) the polynomials according to the descending powers of a given letter. EXERCISES. 1°. Divide a—b by a—b. 2°. Divide a®?— 5b? by a—d, 3°. Divide a?—b? by a—b. 4°. Divide at —b* by a—b, Operation. a — b) at — bt (a? + a*b + ab? + 53 The division is a* — a*b ; Subtract by here executed Y aeh changing the sign (11). by the powers + a°b — a*b? of a (16), + a?b? + ab? — ab? + ab? — bt + ab? — bt 5°, Divide a> — 6°, a’ — 5S, a’ —B’,..., a*— b”, by a—b. PROPOSITION XI. 1°. Multiplying the dividend while the divisor remains (17) the same, or dividing the divisor while the dividend remains the same, multiplies the quotient ; 2°. Dividing the dividend while the divisor remains the same, or multiplying the divisor while the dividend remains the same, divides the quotient; .*. MULTIPLICATION OF FRACTIONS, 35 3°. The value of the quotient is not altered by multiplying or dividing both divisor and dividend by the same quantity. Def. 7. A Fraction is an expression of division, and arises from an impossibility of performing the operation. Therefore, indica- ting the dividend, now called the numerator, by N, the divisor or denominator by D, and the quotient or value of the fraction by V, we have ‘ N: D= V, or N=DV;; hence (17), Cor. 1. A fraction is multiplied by multiplying its numera- (18) tor, N, or by dividing its denominator, D, [N-R=D-. VR, &c.]. Cor. 2. A fraction is divided by dividing its numerator, or (19) by multiplying its denominator. Cor. 3. A fraction is changed in form, without being (20) changed in value, by multiplying or dividing both numerator and denominator by the same quantity, Cor, 4, If a fraction be multiplied by its denominator, the (21) product will be the numerator, For (18) the fraction 7 multiplied by d= i = (nd): d= d Cor. 5, Fractions are multiplied together by taking the (22) product of their numerators for a new numerator, and that of their denominators for a new denominator, A ee : ‘ N QN For; times any quantity Q = Q times D (4), = D (18) ; N, NN, nes At\s Ne eel substituting D, for geek D times Dt apa ey (18), _ NN, Baris = (21), multiplying both numerator and denominator by D, (20) ; N N,N; _ NN, N; _NN,N, 29 DD, Dp07 6 DD, > Do DDD os Cor. 6. A fraction may be raised to any power by involving (23) its numerator and denominator separately, and consequently, any root of a fraction will be found by an evolution operated upon its terms. For, making N,, Soak = N, and D., D3, ... = D, (22) becomes N N NNN... [m] 7 y"=5 SS D’D DDD... [my ° © noe [0] = DDD...[m] (23) 36 DIVISION OF FRACTIONS, « Cor. 7. To divide by a fraction, invert it, and proceed as (24) in multiplication, Let Q : * be any quantity divided by a fraction ; multiplying both dividend and divisor by es (17), we have a8 (0.2): (5.2)-(a-2) 32 (a.2) ea serge Q. E. D. N Cor. 8. The reciprocal of a fraction is the fraction inverted. (25) For Q : ake Q- =,» becomes 1s a1 oy when Q = 1. Cor. 9. Any quantity, whether whole or fractional, will be (26) reduced to a given denominator, by being multiplied by this de- nominator and set over it. D_ QD F h — e 1 — e— = a or we have Q=Q DD -10 5 «100 79° 662 Th 5 side te eee 6, Pie 3 = ‘662 ; us $= “79 = 79 = $= 09 = Too = 00 ~ 0 7 lo = fi and Bath aor ee whence the rule for reducing vulgar to decimal fractions. Reducing as > ae . to the denominator M, we find MN MN, MN, D PA D, from which it appears that the b TAL oie Mes least common denominator, M, must be divisible by each of the denominators, D, D,, D3, ... of the given fractions, or, must be the least common multiple of all the denominators, if we would have the resulting fractions freed from subdenominators, .-, Cor. 10. To add or subtract fractions, find the least com- (27) mon multiple of the denominators, to which, as a common denom- inator, reduce all the fractions, then add or subtract the numera- tors (5,). This rule will frequently be superseded by the following : SUBDENOMINATORS. 37 Cor. 11. To free a fraction of subdenominators, multiply (28) its terms by the least common multiple of the subdenominators (20), el okey “IES ARTE a, (atop b P d ‘ EUS, - ae gS Aug eM fo wy? where, if we would make the subdenominators 0, d, 2, i, k disappear, M must be divisible by each. Scholium I. The rule for managing the signs has already been given in (6,), but it may be well to observe that, of the three signs pertaining to a fraction—viz., the sign before the fraction and those before its terms—an even number produces plus, an odd number mi- nus, and any two may be changed ata time. Thus, a ark yee ms ers Ae pea eae Ve oe $Ea=th HatWab4+ost:H=tOe5 $oat(-:-)=4)=+42=4+4:37=40=-; ps ONT ONE tat Gy? fy Uy Heeb a —Ta-(b i He-H=5-F=-- =O) =4 a us Scholium II Whenever it may be foreseen that, by performing an operation the same factor would be introduced into the numera- tor and denominator, it should be suppressed. Scholium III, Additions and subtractions will frequently be bet- ter performed in part before reducing to a common denominator. Def. 8. An Equation is an algebraical expression consisting of two members, separated by the sign of equality [=]. Equations are of different kinds, 1°. An identical equation is one in which the members are the same, as ad = d. 2°, An equation of operation has one of its members derived from the other—of these we have already had many examples, such as mc —nc =(m— n)ec. 3°. An equation of condition expresses a determinate relation that must exist between certain quantities, not distinguished as known and unknown, as 4(a+ b) =p—q. 4°, The word equation more commonly indicates a relation be- tween known and unknown quantities, such that the latter may be 38 SIMPLE EQUATIONS, derived from the former, as z+ a= 5, whence, by subtracting a from both members, we have x =b— a. Equations of this kind are denominated simple, quadratic, cubic, &c., or of the first, second, third, ..., degree, according as the un- known quantity is of the first, second, third, ..., power. PROPOSITION XII. Any term may be transposed from one member of an (29) equation to the other, by changing its sign. For, let A+p=B. be any equation, A representing the sum of all the terms in the first member, except p, which, it is to be understood, may be either +or—, and B those of the second member; then, subtracting p from both sides (1,,), there results A=B—p. Q.E. D. PROPOSITION XIII. A Simple Equation will be solved— (30) 1°. By separating the unknown from the known quantities by transposition ; . 2°. By uniting the coéficients of the unknown quantity in one ; 3°. By dividing by the coéficient thus formed ; 4°. By clearing of subdenominators. Scholium. An equation may be cleared of fractions, if desired, by multiplying it by the least common multiple of all the denomina- tors—but the solution will in general be more readily accomplished by the rule just given. The exercises marked I. at the end of the book, may be here introduced. PROPOSITION XIV. From n equations of the first degree to eliminaten—1 unknown quantities. 1°. Whenever the conditions of a problem embrace two un- known quantities of the first degree, it is evident from what has ELIMINATION, 39 just been said that the equations expressing these conditions may be reduced to the form az + by=c, a2+by=c; and .*. Serty=t, a c’ Tae ay whence (+ aaa 5) agen bind b b' b b' an equation involving but a single unknown quantity, which may therefore be solved as above. The method may obviously be ex- tended to three or more unknown quantities, embraced in three or more independent equations—equations not convertible into each other ;—and it will become manifest, by making the elimination, that the number of independent equations must equal the number of unknown quantities to be determined. . From several equations of the first degree to eliminate (31) one or more quanitties : 1°. Make the coéficients of the quantity to be eliminated +1 in each equation ; 2°, Take the differences of the equations thus formed. Example. Given equations (a) to find z, 122 —8y-+ 4z=8 (1) or-+ y— 6z=—13 (2) (a) c+ Ty + 14z = 57 (3) re ay he 32 —Q+27=2 (4) (2): — 6, —4r—tytz=2 (5) {0 (3) : 14, Tye + hy + 2= 44, (6) BO) Ber Wyant Lg (4) — (6), ir — Wy = — 214 (8) ey Pe, —WVerty=r (9) (d) (8):-—23, —hert+y=% (10) (—H-+es)e = ty tt (e) ty 3 He Substituting 1 for z in (9), we find y=trt+ Ht =2; , (4) z=2+2.2—-3+1=3, 40 EXPONENTS, After a little practice it will be found unnecessary to write the equa- tions (5) and (d), it being equally easy to pass at once from (a) to (c) and from (c) to (e). This method will generally produce the greatest amount of cancelation, and, therefore, be the most expedi- tious. The exercises marked II, at the end of the book may be here introduced, SECTION SECOND. Exponents—Proportion—Variation. PROPOSITION I. A factor may be carried from one term to the other of a (32) fraction by changing the sign of its exponent. As in subtracting coéfficients, we must be guided by the same rule, whatever may be their relative values, so, by a like extension in exponents, from (15) we must always have a” : a" = a”~", whether m>or (BP) sata:cte::a—a@:c—c'; pee gee n(d +e from (5°) pare ( ) = pen d ) a= ray c= re Assy c= fos if eee Ca ey gdsertattecyste tal is ose (8°) ~ate:a—c:ira te sa—c, é&c., dc., &c., .*. PROPOSITION IV. If four quantities are proportional : 1°, The sum or difference of the antecedents is to the sum or difference of the consequents, as either antecedent to its consequent ; (39) (40) The sum or difference of the terms of the first couplet is to the sum or difference of those of the second, as antecedent to antecedent or as consequent to consequent ; 3°. The sums or differences, or both sums and differences of the terms, taken in the same order, whether homologous or anal- ogous, are proportional, The principle in (40), 1°, may be extended thus: if a ae ae ee eT @t SothuGe 26 es Pie catnd) BECoy hwy Cot. Bade AE PROPORTION. 45 or if a spo, a= re, a’ =rc', &c., &e. 3 thena+a-+a’+..=r(ete+te’-+...); ” @+at+a+.i:etoeteoe+t.wi:a:e::aic::a'ie':: MOG 7. PROPOSITION V. If any number of couplets have the same ratio: (41) The sum of all the antecedents is to the sum of all the consequents, as any one antecedent to its consequent. We should also have + peewee eee PES ec SES isarse rea 3023 &ee which may be enounced in words, From (1°) a=rc and (2°) reversed re) ="; we have arc =Trca, @iertia se-givesl Farissdeu: PROPOSITION VI. If four quantities are proportional the product of the (42) extremes is equal to the product of the means. Cor. To change an equation into a proportion, make any (43) two factors into which one member may be resolved the extremes and the factors of the other member of the equation the means of the proportion ; or, to read an equation as a proportion, begin and end the reading in the same member. Thus 4.3=2.-6, may be read 4;323::62:3) 2:4::3:6, or 4:6::2 53, eroe1 4: & or 322 5:6 24; e O47 93"! 2, or 3:62 25, sass. 4 t en From (1°) and (2°) we have a” =r"c”, and ata erg" : se ea zich; where m may be any quantity whatever, whole or fractional ; .°, 46 PROPORTION. PROPOSITION VII. If quantities are proportional, their like powers, or roots, (44) or powers of roots, are proportional. Again eel ° me, and ma =r + ma’ (1°) and (2°) Ra =Ts now fo yt ste =r 8 Re: where m and n may be any whatever, whole or fractional; .°. PROPOSITION VIII. A proportion is not destroyed by taking equal multiples (45) or submultiples of homologous or analogous terms. Thus, since 4:1227 8 p24, we have 16 : 24: : 32: 48, [eq. mult, of anal. terms], and 1 : 3:: 1:3, [eq. submult. homol. terms]. From a= TC, we have a (lem) =r. +(l+m) ce, or atma=r (cme); .. PROPOSITION IX. If any two quantities be increased or diminished by (46) equal multiples or submultiples of those quantities, the sums or differences thus formed will be proportional to the quantities themselves. Def. 4. Four quantities are said to be reciprocally proportional when the ratios of the couplets are the reciprocals of each other. Thus if “=e G, Cc; a’, c’, are reciprocally ping ¢ then : and a’=4¢'s proportional. But from the same equations, we have a=Te, é =H 3 Def. 5. Four quantities are inversely proportional when the first is to the second as the fourth to the third. Def. 8. Wath 26:¢c320: ds syateynthehid,.d, £; d, aay Ore said to be in continued proportion, or to constitute a geometrical progression; cis said to be a third proportional to a and 4, and 6 is called a mean proportional between a and c. Since froma: b::6: 0c, &c., we have PROPORTION. AT rains dy rb=c, Tc = d, Teabky multiplying together the Ist and 2d, the Ist, 2d and 3d, &c., there results riG@=<¢, read, ate ee l being the mth term ; .°. PROPOSITION X. In a continued proportion we have the 1st term to the (47) 3d as the square of the 1st to the square of the 2d, or as the square of the second to the square of the 3d, &c. ; and the 1st to the nth » as the (n—1)th power of the first to the (n—1)th power of the 2d, Fc. aus Seve es eee Os Oe ee nt! ne Cor. The last or nth term is equal to the first multiplied (48) by the ratio raised to a power one less than the number of the term [nz — 1]. | | It will be observed that the series is here supposed to be ascend- ing, or thatab>c..., then r will be less than unity—and in all cases By addition we infer that r(a+6+c4+...+h4)=b+c+d+...+1, or r(S—l)=S—a; [S=a+b+c+..47] _tl—a_ a(r*—1) cy ROY hs ee BRE PROPOSITION XI. se ied Asotin. flere (49) terms ratio — one _ (first term) [(ratio)" — 1] .* ratio — one ' 48 VARIATION. Note. Since from (1°) and (2°) we have r a a —=rand —=r7, and .. —=—, y c we may write every proportion as a fractional equation—whereby the many and all the changes that can be rung on proportions, will be reduced to very simple operations on equations. VARIATION, Def. 7. We shall often have occasion to consider quantities, not as proportional simply, but as passing by inapprectable degrees through all magnitudes compatible with certain conditions. Such quantities are denominated VariasLe—and are represented by the last letters of the alphabet, as z, y, z, while the first letters are used to indicate quantities regarded as constant, or such as are independent of the variables. Thus, that y varies as z, a being their constant ratio, is expressed by y = as. In this equation it is to be understood that z, in passing from any one given value to another, is regarded as passing through all intermediate values while a remains unchanged ; and that conse- quently y changes, taking new values depending upon the value of xz, On this account z is called the independent, and y the depend- ent variable. Thus, if the rate of interest, 7, be constant, and the principal, p, be also constant, we shall have a given sum pr, as the interest for one year on the given principal ; then if y be the interest for the time z in years, there will result the variation y = pre, This manner of looking upon quantities, not so much as known and unknown, as constant and variable, is as important as it is simple. PROPOSITION XII. If y = az, e 1 1 7a l= Ae y > 1. €. If y vary as x, x varies as y. (50.) VARIATION. 49 PROPOSITION XIII. Hy ij ¥ = a2, y=r=artxr=(atl)r=(atl)- a os TR, If y vary as x, x+y varies as x or ¥: (51) PROPOSITION XIV. If y= az, ine aft a aT &:, If y vary as x, y” varies as x”. (52) PROPOSITION XV. If y= O27, vi my=a+ mr; i.e., If y vary as x, my varies as mx, (53) PROPOSITION XVI. y = amz, it follows that If y vary as x, y also varies as mx. PROPOSITION XVII. If ¥y = az, and Sab7: og y = abr; i. e¢., If y vary as z, and z vary as x; then y varies as x. (54) PROPOSITION XVIII. If Ba gat and Du be 2 Rs: yoo (ack bles ise, If y and x vary as z, y+ x varies as 2. (55) It will be observed that the above forms embrace, very briefly and simply, not only essentially the whole doctrine of proportion, but a vastly wider field, by reason of the unlimited number of values of which y and z are susceptible. SECTION THIRD. Analysis of Equations. The following is a principle of the first importance in analytical investigations : PROPOSITION I. Certain equations are so constituted that they necessart- (56) ly resolve into, and are consequently equivalent to, several inde- pendent equations. We do not propose, in the present article, to enter into a full de- velopement of the boundless resources which this principle affords, but simply to illustrate it by examples of such particular cases as will be serviceable to us as we proceed. Required two numbers such that if they be diminished severally by a, 5, and the remainders squared, the sum of these squares shall be equal to zero. Denoting the numbers by. x, y, we have (x7—a)*-+ (y—b)? =0; and it follows from (6;) that (x — a)*, (y—b)*, must both be +, whether x2a,* y25; but it is obvious that, since neither of the terms (z — a)’, (y — 5) can be minus, neither can be greater than 0; for if either term, (xr—a)? for instance, have an additive value, the other (y — b)? must possess the same value and be subtractive, in order to satisfy the equation, or that their sum may = 0; whence it is necessary that (x —a)* =0, and (y—b)?=0; x —a=0, and y—b=0, or x =a, and y= b, As a second example, what numbers are those from which if a and b be severally subtracted, the product of the remainders will be = 02 Representing the numbers by z, y, there results (x — a) (y—6) =0; .. dividing by y—b, xz—a=0, or r=a, and dividing by z—a, y—b=0, or y=), * xz a, x greater or Jess than a. a QUADRATIC EQUATIONS, 51 Every equation of the second degree, or, embracing no other un- known quantities than z* and z, may, by transposing, uniting terms, and dividing by the coéfficient of x*, be readily reduced to the form +1e-2?+2ar=5b; (57) understanding that a, b, may be either--or —. In order to find z, we observe that the first member of the equation will become a bi- nomial square (8) by the addition of a? ; of z?+2rat+a?=6b-+ a’, -*» (8), (64), r+a=+ ,/ (b+ a’), and z=—axt,/(b+a’); .., (58) PROPOSITION II. To solve a QuapRaTic Equation : 1°. Reduce it to the form of (57) ; 2°. Add the square of half the coéfficient of x to both sides ; 3°. Take the square root of the members, prefixing the double sign [+] to the second ; ! 4°, Transpose. It will be observed that lf resolves into two independent equa- tions (58), x=—a+./(b+a*), and r=—a—./(b+ 0’), thus illustrating (56). Adding these values of x, we have [—a+./ (b+ a*)]-+[—a— J (6+ a*)] =— 2a, and their product is (?) [—a+ V(b+a*))[—a— (b+0%)]=[— a}? —[ V(b +0%)P =a*—(b+a’?)=— 6, Cor, In a quadratic of the form (57) the sum of the values (59) of x, is equal to the coéfficient of z' taken with the contrary sign, and their product to the second member also taken with the con- trary sign. Queries. What will (57) and (58) become, when a is changed into —a? binto—b? When 6 is minus, what must be its value compared with a’ in order that the value of x in (58) be impossible ? [See (6,.)] Will change of value or sign in a ever render z imagin- ary? Why not? Scholium I. We naturally inquire if the Cubic Equation can be 52 ROOTS OF EQUATIONS. resolved into three independent equations. Every equation of the third degree reduces to the form z?+az?+ br-+c=0. (a) As we must suppose z to have some value—one at least—let r be that value ; then must r satisfy the equation, or r?tar’tbrt+c=0 af, c=—r?—ar’*— bdr, hich value of c substituted above, gives z?+azr’*+ br —r?—ar*—br=0, or (x? —r?)+ a(x? — 7°) + d(x —1r) =0; (5) iz: nei (z+r)+5=0, Hividing by x —r, see examples under (16) ; +(atr)r+r’?+ar+b=0, (c) Phente [(57), (58)] (a) is resolvable into three equations, giving three values for z. Comparing equations (a), (5), (c), and observing that (5) is the same as (a), we learn that, if r be a root of the cubic equation (a), that the equation is divisible by z — r, giving a quadratic (c). The student is requested to prove that the equation zi+pz?+qz’?+rz+s=0, is resolvable into the factors xr—a,x—b,x—c, x—d, or that (x — a) (x —b) (x —c) (uv —d)=2'+ pri +q2r?*+rzr+s=0, d,.D, &,.d,,betng values of 2, or 2 4,9 = 6,7 =r, eo Scholium II. Many Biquadratic Equations may be reduced as Quadratics ; e. 2. 1°. When reducible to the form (2?-+ Ar+ B) (2?+ A’r+ B’) =0, (60) where the conditions are zt+(A+ A')r?+(AA'+ B+B')x? + (AB' + A'B)z + BB =0, ors=BB,qg=AA 4+ BA B,p=At+A,r=AB4S AB. 2°, When reducible to the form (z?+Ar)’?+B(2?+Azr)'=C, (61) where the conditions are zi4+ 2Azr*?+ (A?4+ B)zr*+ ABr=C, or p=2A, r=AB, g=A?+B, 3°. When reducible to the form (61,) a: a reg A em PUTTING IN EQUATION, 53 a for Qere—=24; x a\* a \! (2+) +(2+ =) = b+ 2a. Example. Given zr‘ + 4x° + 62? 4 4x = 15, to find z. 2 = We have A-ha 2, B= a= = and } FRB, (x? + 2x)? + 2x? + 2r)* = 15, (x? + 22x)? + 2x? + 2r)'+1= 16, z?+2r4+1=+4, rtl=+/t4=+4+ 20r=+/—4, ase (2B P/E 4 or S TS A/S 4. Hd many values has r? What values are imaginary? Verify by substituting these values in the first equation. 20] Scholium III, When a problem embraces several unknown quantities, it may be solved by representing these quantities in different ways; but the elegance and facility of the solution will frequently depend upon the notation which we employ. The fol- lowing rule, taken from the Cambridge Mathematics, (application of Algebra to Geometry,) will be serviceable. ‘“< If among the quantities which would, when taken each (62) for the unknown quantity, serve to determine ail the other quan- tities, there are two which would in the same way answer this purpose, and it could be foreseen that each would lead to the same equation (the signs + and — excepted); then we ought to em- ploy neither of these, but take for the unknown quantity one which depends equally upon both; that is, their half sum, or their half difference, or a mean proportional between them, or, &c.’’ Thus, suppose it were required to find two numbers such that their sum should be 4 and the sum of their 4th powers 82. Instead of taking z and y for the numbers, we may make them both depénd equally upon z by putting 2+ z= greater No., 2—xz=less No.; since the sum would be 4 = greater No. + less No., (2+ 2)* =164 3227-4 2427’? + 82? + 2%, and (2—2)* = 16 — 327+ Mr’? —8r?+2'; oT 2r*+2-24r? 42-16= (2+2)*+ (2—7z)* = 82, or zit24r* = 41—16=25; 54 EQUATION SOLVED, xt + 247? + 12? = 25-1 144 = 169, z?+12= +413, 4 4 z= (£13—12)° =+1, or = + (— 28)’ ; 1 ak greater No. = 2+ 2 =3, 1, 2+ (— 25)", or2— (— 25)’, 2: 1 and less No. =2—2=1,3, 2— (— 25)", or 2+ (— 25)”, where it must be observed that if 3 be taken for the value of the greater number, | must be taken for that of the less, and so on for the three remaining corresponding values, As a second example, suppose the equations 4 a ety + (2? +a?) +(y?+a*) =8, and ae ie a given to find + any y. We shall make x and y depend equally upon the same unknown quantity by putting ae: Po SS S25 : ee a whence eid earn which values substituted in the first equation, give a a} 2 z a* 2 t 2h ka Zo a)" + mre = 6, +. 0 =, a or Belek gink Wick le 1 z e: ii b dividing by aand putting 7 Tm to avoid fractions ; (4+) (24+1t=n—2-4 Zz ." Zz’ ah Girt? Lede oth pe n? —2ng — “24 28 1 (Qn 4224 T* _ ns, n? 8 en ecient =— "Fe ee 2 or z*—2mz=—1, putting m= 2 (3 2)? 4(nt1) n+l’ USE OF THE DOUBLE SIGN, 55 oe z?*—Qmz+m? =m? —1=(m+1)(m—1), z=m+[(m+1) (m—1)]*. Scholium IV. From notions derived from the practice of arith- metic, the student, on first being introduced to problems capable of double solution, is frequently inclined to question the propriety of the second value of xz ; and, consequently, to reject or neglect the second or minus sign, in taking the square root of a quadratic equa- tion, Such neglect must zever be allowed; for the double sign (+) is necessary—and, so far from being any thing like an excres- cence or deformity in algebraical language, is a symbol of the great- est utility, as will appear hereafter when we come to apply the fore- going principles to Geometry. The following problem, illustrative of the use of the double sign, will suffice for the present. Required the point in the line joining the centres of the earth and moon where their attractions are equal. Let E be the earth, M the moon and C the point , Brn luh 9 of equal attraction, and let the quantity of matter in © the earth be represented by e and that in the moon by eae m, the radius of the moon’s orbit, or distance from the earth, by r, the distance C M by wu and consequently, C E by r— ux. Now it is a principle of physics that the attraction of a sphere is proportional to the quantity of matter directly and to the square of the distance to its centre inversely, or qt. of mat. to (dist.)2 ; whence Earth’s attraction at C will = eee oe $ uy Moon’s attraction at C will = = : e i Sem = ad ‘ (r—u)? w? 3 AGREE Bue T—Us U of wu /e=tr./m—u(j/m, oa Co) 6.24/21 a= 2-7 ./ m, +r erm m r ee = =, WhO wl Pine qctan AS m 56 USE OF THE DOUBLE SIGN, But the mass of the moon, inferred from her action in raising the tides, is +; of that of the earth ;* oo a5, and a/ <= VB= 86; m mm T 1 1 ~ T8668 °" ~ 966 ~ 766 * or the distance of the point of equal attraction from the moon, is nearly + ;3; of the moon’s radius, or — 38; of moon’s radius, where it remains to interpret the — sign in the second answer. If we sub- tract from uw (that is, from C lay off a line in the direction of M) a quantity less than w, the point of equal attraction will approach M, and the more, the greater the line subtracted, so that, when the line subtracted is = CM, w will = 0, and when > CM, as CC, u will become —, being = MC —CC’=MC—(CM+ MC’) =—MC’. Therefore if u = + 33; + r, gives the point C on the left or this side, — gs «7 corresponds to the point C’ on the right or beyond the moon. Indeed, had we supposed the point of equal attraction at C’ instead of at C, there would have resulted u aM» OT 5 1, é mM Tr a = =e Whence u = : Se boo? ita —14,/£ ' ; m i 1 : 1 or USE aOR) TOG ee St raat The two values of w are therefore perfectly applicable, and, in fact, the problem would not be completely solved without them. The second value has revealed a truth, viz.: that there is a second point of equal attraction beyond the moon—which, though not con- templated in the statement of the problem, is yet a direct consequence of the equation drawn from it ; and which, when once discovered by the aid of algebraical symbols, recommends itself to the under- standing without them, since the attraction of the earth is greater than that of the moon. Suppose, for a simple illustration, the masses of E and M were as 9 and 4, and their distances to C’ as 3 and 2; then their attractions at C’ would be as zs and iS and con- sequently equal, : As a last example in which a single equation is equivalent to two independent equations, we give the following important theorem : * Poisson, Traité de Mécanique, p. 258. CONSTANTS AND VARIABLES. 57 PROPOSITION III. Whenever a problem gives an equality between constant (63) and variable terms, the variables being capable of indefinite dim- inution so as to become less than any assignable quantity ; two independent equations will be formed, one between the constants, the other between the variables. Thus, let AtX=B+Y represent any equation in which A, B, are the sums of ayy constant, those of the variable terms in each member, X and Y, being capa- ble of continually decreasing and becoming less than any assigna- ble quantity, that is less than any assignable part of Aor B. Then will Am Boandak = ¥'; for, if possible, let A be greater or less than B; first suppose A > B by some quantity D, or A=B+D; Ne X= Y-—D, or Y=D+ XX, whence it appears that Y can never be less than D, even should X become actually =0; therefore Y is greater than, or at least as great as, an assignable quantity D, which is contrary to the hypoth- esis, and this contradiction holds as long as D has any value, or as long as Ais > B. Again suppose A< B, or A+ D=B; af, X—D=Y, or A=D-+Y; whence X is always greater than, or at least as great as D, and this is also contrary to the hypothesis, according to which _X, as well as Y, is to be capable of indefinite diminution. Therefore it is absurd to suppose A either greater or less than B, or otherwise than equal to B; * A = B, and consequently X = Y, which was to be proved. The method of proof here employed is called that of ‘ reductio ad absurdum,”’ leading to an absurdity ; where, instead of showing directly that the proposition must be true, we show that it can not be otherwise than true. | As an example, let it be required to find the sum of the repeating decimal fraction 333 &c. ="3-+ 03+ 003-4... [ad infinitum]. 58 LIMITS, Let A = sum of all the terms, X = sum of term after the nth; then A — X will = sum of n first terms, or A— X='3-+ 03+ 003-1... [n terms], _ B3((1)"—1] 3 [P= C1} 43 oy SACL why Woeull Gales ter Wik odode eed But (‘1)" may be made Jess than any assignable quantity by tak- ing n sufficiently great, which may be done since the number of terms is infinite, Thus, if we take 1, 2, 3, &c., terms, or make successively Ss ie ee 4, Bh yive we have (‘1)"= ‘1, ‘O1, ‘001, «0001, ‘00001, ... ; 8 and X becoming < any assignable quantity, we have (63) Ma: it: one ee. So for the circulate ‘23, 23, 23, &e., we have A — X = ‘23 + +0023 + 000023 +... [n], 23 (01)". or A— xX = ‘99° = 99 5 23 23 ~ 499 ~ 99° And generally, if C = any circulate containing c digits, we shall have A— X= C(‘1)'+ C(1)*4+ C(1)*+,,, [2]; € (*1)° C ~T=(lY (10%—1~999.,.cey Why? Rule? We observe that a quantity may increase constantly by an inf- nite number of additions, and yet never exceed a determinate finite quantity or limit; as ‘333, é&c. constantly approaches to 2 which it can never surpass, This gives us our first notion of limits. Def. “When a variable magnitude A —_X can be made to (64) approach another A which is fixed in such a way as to render their difference X lessthan any given magnitude, without, however, their being able ever to become rigorously equal, the second A is called a limit of the first 4 — X,’* A direct consequence of (63) and further illustrating (56), is the following important principle : * Franceur, Mathematiques pures, p. 166. EXERCISES, 59 Cor. If an equation exist between terms affected by whole (65) additive powers of a variable quantity, z, capable of indefinite dim- inution, then the constant coéfficients of the like powers of z, will be respectively equal to each other. Thus, if 4+ A,-2'+A,- 7*?+A,¢ r3+... =a-+a,er'+a,-r*+a,+7°+.,.., where A, egal CO Pg) bo.) Nh | 4 x? + y? = 324900, x?-+-y?=b, a ry =a = ry=a 7°. S$ a2z=5 le Meaceeiag | 9°, (yz=b y : r y+2z=c, adie rent | ee bg x+y =500 r+ y = 2000 =A] 10°. {450-2 22 | 11°.) 1710—-2_17e $ 12.3 eh on 450—y By’ 1040—y 12y'" sagt ED eee pore | chai (10°) 9 LL ytb. 24 y?=514, x+y=b. § ary (a—z) (b—-y)=p 160.) Tiy=min Wa, SS ets (a—zx)* + (b—-y)*=s. w+y +z t+ y= 330. ES 1: } la } : } by ‘ e+ ry + ry? = 126 os gc? —y? = zy. ee ee ee 208 | 2 29) Cal sae Dh cote} mh ed (r+ y) (#* + y*) = 580 r?-+4y?—(x-+y)=42, 90. r+y+ry=a 930, r+y=a z+ y?—(c+y)=b., ides 24°, r+y=a 25°, e+y+r+y=a ry (2? +9’) =b, : shld iadtaak 2 BOOK SECOND. PLANE GEOMETRY DEPENDING ON THE RIGHT LINE. SECTION FIRST. Comparison of Angles, Def. 1. The doctrines of extension constitute the science of Ge- ometry. Def. 2, Solids have three dimensions, length, breadth, and thick- ness, Def. 3. The boundaries of a solid are surfaces, the perimeter of a surface are lines, and the extremities of a line are points. PROPOSITION I. [primary norTion]. A Straight Line is such that it does not change its direc- (66) tion at any potnt in its whole extent. This truth is not produced as a theorem, for it is incapable of de- monstration ; not as a problem, for there is nothing to be done; neither as a corollary, for it is the consequence of nothing ; nor as an axiom, for it is hardly of the unconditional and absolute charac- ter of that enounced in the words, ‘* the whole is equal to the sum of all its parts ;”’ and it is not a definition, for we gain no new idea by the mere terms of the proposition : we only recognize by and in them one of those primary notions which we possess anterior to all instruction, and which, as they are necessary to, lie at the foun- dation of, every logical deduction. It would doubtless be out of place to enter here into any investigation in regard to the origin of our ideas; but I think it will be apparent, that the notion of con- tinuity is codriginal with that of personal identity, and therefore, antecedent to argumentation; and continuity measured out on 64 RIGHT LINE. the one hand, in the lapse of events, as the periodic return of day and night, the revolutions of the celestial sphere, or the index of the chronometer, becomes time, and, on the other, attached to form by a personal passage over the surfaces of bodies, it becomes space ; and, considered in regard to space and restricted by the notion of perfect sameness, continuity gives us the idea of direction, or that of the straight line. Corollary 1. Two straight lines cannot intersect in more (67) points than one; for having crossed once, it is obvious from (66), that, in order to a second intersection, one of the lines, at least, must change its direction, Cor. 2. Straight lines coinciding in two points, coincide (68) throughout, otherwise two straight lines would intersect in more points than one, which is contradictory to (67). Cor. 3. Straight lines coinciding in part, coincide through- (69) out, and form one and the same straight line (68). (Why 2) Cor. 4. Two straight lines cannot include a space (68). (70) (Why 2) Def. 4. A Plane is a surface with which a straight line may be made to coincide in any direction. Def. 5. Two straight lines are said to be parallel when, situated in the same plane, they do not meet how far soever they may be produced. Cor. 5. Through the same point, only one sfraight line can (71) be drawn parallel to a given line ;* for the directions of all the lines save one, drawn through the given point, will evidently (66) be such as to cause them to meet the given line if sufficiently produced. Scholium. The last corollary, though commonly given as an ax- iom, has been thought not sufficiently evident of itself. It is not self evident, doubtless, if regarded as a consequence of any mere definition that can be given of a straight line, but neces- sarily follows from that idea of a straight line, viz., continuity in sameness of direction, which we possess anterior to all definition, As such, it is, I believe, as well established as the primary truths in any department of human knowledge. Application 1. To make a straightedge. Hav- _ ,--_____ ing formed a ruler as straight as possible and drawn (@——=———_y a line with it upon a plane surface, turn the ruler Hiatt ooo * When the wu.d dine is used alone, straight line is to be understood. ADJACENT ANGLES, 65 over upon the opposite side, and, with the same edge, repeat the line, coinciding with, or better, very nearly coinciding with the first; if the lines coincide, or appear equally distant throughout, the edge is sensibly straight. Application 2. To test the parallelism of the edges of the ruler, place it against a second straightedge, and, having drawn a line, turn it end for end and draw a second, (Why 2) Application 3. To test a plane, apply the straightedge to it in different directions. Def. 6. When two lines intersect each other, or would intersect if sufficiently produced, the inelination of the one line to the other is called an Angle. It is obvious that angles are of different and comparable magnitudes, and, therefore, like other geometrical quantities, solids, surfaces, and lines, are capable of addition, sub- traction, multiplication, and division, and, consequently, subject to mathematical investigation. PROPOSITION II. The sum of the adjacent angles formed by one line meet- (‘T2) ing another, is always the same constant quantity. Let the line AO meet the line BOC in O, ma- A king the angle AOB any whatever, and, in like / B c manner, let the line A’O meet the line B’'OC’ in O, making the angle A’OB’ any whatever. Pla- AY N’ cing the first figure upon the second, let the point 3’ — C’ O coincide with the point O and the line OB take Fig. 3. the direction OB’ then will OC take the direction OC' (69) and the line BOC will coincide with BOC, also OA will take a certain di- rection OA", which we are at liberty to suppose between OA’ and OC’; whence the angle AOB will be equal to the angle A”OB,, the two becoming identical, and AOC to A”OC’ ; but the angle A” OB’ is, by hypothesis, the sum of the angles A’OB’ and A’'OA”, and therefore greater than A’OB' by A’OA" (1,), while the angle A” OC’ is less than A’OC’ by the same quantity ; therefore the sum of the angles A’OB, A”OC’ is equal to the sum of the angles A’OB, A‘OC' (1,,), whence the sum of the angles AOB, AOC, is equal to the sum of the angles A’OB’, A’OC’ (1,) which was to be proved. The same in the language of symbols ; Let BOC coincide with B‘OC’,, and OA take the direction OA" ; 5) 66 RIGHT ANGLES, Z AOB= A'OB = A'OB' + A'OA" and Z AOC= A"0OC' = A'0C — A'OA" ; Z AOB+ AOC = A‘OB'+A'0OC.. Q. E. D.* The method employed in the preceding demonstration is obvi- ously that of superposition, and the principles upon which it is grounded are the nature of the straight line, the whole is equal to the sum of its parts, and, equals added to equals, the sums are equal, Def. 7. If the angle AOB= AOC, then AOB and AOC are called Right Angles ; hence, AOB or AOC ri being the half of AOB-+ AOC = a constant quantity (72), is also a constant quantity, or, Cor. 1. All right angles are equal to each other. (73) Cor. 2. The sum of the adjacent angles formed by one (74) line meeting another is equivalent to two right angles. Cor. 3. Conversely ; two lines met by a third, so as to (75) make the sum of the adjacent angles equal to two right angles, form one and the same straight line. For if not, let BOX be a straight line, while Z AOB+ AOC =+—,,t [by hypothesis; ] then (74) AOB+AOX=-; Beettaey .. subtracting AOX — AOC = 0, Fig. 33. therefore BOC does not differ from the straight line BOX, Cor. 4. The vertical angles formed by intersecting lines (76) are equal. If a@ and 6 be vertical angles, while c is adjacent to 7 hi aa oil (97). ¥ Fig. 144. Cor. 5. An Isosceles triangle is bisected by the line (100) which bisects the angle embraced by the equal sides. Cc For let CA = CB and 7 ACD =BCD; Q . then (96) will A ACD = BCD; .-. ee Fig. 145, Cor. 6. In an isosceles triangle, the angles opposite the (101) equal sides are themselves equal; as 7 A = B, and Cor. 7. The line bisecting the vertical angle of an isos- (102) celes triangle bisects the base, to which it is also at right angles or perpendicular, Cor. 8. Equilateral triangles are equiangular (100). (103) PROPOSITION II. In any triangle, that angle is the greater which is op- (104) posite the greater side. Let CB be > CA; then will 7 CAB > CBA. } For, taking CD, a part of CB, equal to CA and join- ing AD, we have ys Be ZCAB>CAD=CDA=ABD-+BAD>ABD. “* ve Q. E. D. [Give the reasons, Consult (?), (101). (94), (?).] Fig. 15. Cor. 1. Conversely, the side opposite the greater angle, (105) is the greater, For if the side opposite the greater angle be not the greater, it must be equal or less; it cannot be equal, for then would the angles be equal (?), neither can it be less, for then would the opposite angle be the less (104), which is contrary to the hypothesis. 74 SHORTEST DISTANCE. Cor. 2. A triangle having two equal angles, is isosceles (106) (105). Cor. 3. An equiangular triangle is equilateral (106). (107) What is an Isosceles triangle? What an Equilateral triangle ? PROPOSITION III. \ Any two sides of a triangle are together greater than (108) the third. Let ABC be any triangle; then will the sum of /D any two sides, as AC-+ CB, be > the third AB, a Produce AC to D, making CD = CB, and join ae DB. ’ A> then (101) ZCDB = CBD < ABD; B “ (105) ABK< AD=AC+CD =AC-+CB. Fig. 16. Q. E. D. . Cor. 1. The sum of the lines joining any point within a (109) triangle and the extremities of one of its sides, is less than the sum of the other two sides. C Let D be any point in the triangle ABC ; produce AD -to méeet'CBin Fothen. ©) 3 ' “© Weenie AE AD+DB< AD+ DE+ EB< AC+ CE-+ EB. : (Why?) Q. E. D. +. B Fig. 16. Cor. 2. Any side of a polygon is less than the sum of all (110) the other sides. [Letter and show how. ] Fig. 163. . Cor. 3. Of two polygons, standing upon the same (111) base and one embracing the other, the perimeter of the enveloping is greater than that of the enveloped. [See the last figure and explain. ] Cor. 4. The straight line is the shortest that can be (112) drawn from one point to another, First. Let APB be any line, whether curved or broken, but concave toward the straight line AB, then APB will be greater than AB. For, taking any point P, in the line APB, and revoly- Fig. 164. SHORTEST DISTANCE. 75 ing the branches AP, BP, about A and B till the point P coincides with the straight line AB, first in Q, then in R, the straight lines AP, BP, in the positions AQ, BR, will overlap each other by a certain line RQ, since the st. line AP + st. 1. BP.> AB—and there- fore, the branches AP, BP, falling on the same side of AB, since, by hypothesis, the path APB is concave toward AB, will intersect each other in some point E; whence the path APB, being equal to the sum of the paths AE+ EQ and RE-+ EB, will be longer than the path AEB by the sum of the parts RE and QE. The same may be shown of any other path that is concave toward AB except AB. Second. Let the path AcdefghB be any what- eee ever, having flexures either continuous or abrupt Zz, on in c, d, e, f, g, h; then drawing the straight lines A B Ac, cd, de, ef, fg, zh, hB, Fig. 16s. we have (110) AB < polygonal perimeter (Ac + cd-+de+ef +fe+geh+hB) < the path (AcdefghB), as shown aboye, since all the parts of AcdefghB are concave to- ward the several sides of the polygon. Q. E. D. Cor. 5. ‘* Hence also, we may infer, that of any two (113) paths, ACB, ADB, leading from A to B, and every- c where concave toward the straight line AB, that “1-2 — which is enveloped by the other, as ADB, is the 4 B shorter. For of all the paths not lying between Fig. 166. ADB and the straight line AB, there is none, ADB excepted, than which a shorter may not be found. And this is the case whether the paths ACB, ADB, be both of them curvilineal, or one of them, (ACB or ADB), rectilineal.”’ * Cor. 6. The perpendicular is the shortest distance from (114) a point to a straight line; and of oblique lines, that is the shorter which is nearer the perpendicular—and those equally distant are equal. For Jet PA be perpendicular to CBAB, and AB’ P = AB < AC, and produce PA in Q, making AQ = AP; eri joining QB, QB, QC, we have Cc BA B PC+QC> PB+QB> PA+QA_ (2) b .. dividing by 2 PC>PB> PA, (2) Fig. 167. also PR= PH. (2) * Library of Useful Knowledge, Geometry. - 716 ANGLES AND SIDES. Cor. 7. All points equally distant from the extremities (115) of a straight line are situated in the same perpendicular passing through the middle of that line. The points C, B, A, B’, are equally distant from P and Q. PROPOSITION IV. If two triangles have two sides of the one equal to two (116) sides of the other, each to each, but the included angles unequal, then the third sides will be unequal, and the greater side will be that opposite the greater angle. Let two of the equal sides be made common in AB, and the other two, AC, AD lie on oppo- site sides of AB; draw AI bisecting the angle CAD and terminating in I, which will be a point in CB on the hypothesis that CAB is the great- 17. er angle, since then we have 7 CAB> CAI= DAL S DAB; — finally, join ID. We have (96), (108), BC=BI+I1IC=BI+I1D>BD. Q@. E. D. Cor. 1. Conversely, if two triangles have two sides of (117) the one equal to two sides of the other, each to each, but the third sides unequal, the opposite angles will be unequal and that will be the greater which is opposite the greater side. For it can be neither equal nor less. (Why 2) q Cor. 2. If two triangles have the sides of the one seve- (118) rally equal to the sides of the other, the angles opposite the equal sides will be severally equal. For, if AB= A'B, BC=BC, AC=AC'; C old *- ZC cannot be 2C, for then would AB2A'B ; -- ZO=C, and, for a like reason, 7 A =A’, <3 4G’ B=B. Fig. 172. Cor. 3. When the opposite sides of a quadrilateral are (119) equal, each to each, the figure will be a parallelogram [ fig. 14,]. EXERCISES. v iy j EXERCISES, 1°, It is a principle in Optics that a ray of light, AP, impinging upon a reflecting surface, MM’, makes the angle of reflection BPP’, equal to the angle of incidence APP’, PP’ being per- pendicular to MM’ at P. Given in position the luminous point A, the eye, B, and the mirror, Fig. 18. MM’, to find the point P from which the light is reflected to B, and to prove that APB is the shortest path from A to B by way of the mirror, or that AP-+ PB < AQ+ QB. 2°. Having given in position the elastic planes A PQ, QR, and the points A, B; itis required to find the track of an elastic ball projected from A upon er R PQ and rebounding from QR so as to hit a pin DAL Fig. 182 td 3°. Given in position the straight line MN and the jf \ points A, B; required the point O in aN such that AO shall = BO. WY 4°. Prove that the perpendiculars drawn through mS MIDS the middle points of the three sides of a triangle, will LOX intersect in the same point. 5°. Having given in position the lines AB, CD, ; and the point P,it is required to draw through Pa i line mn, which shall make equal angles with AB, “2—— x B OM Fig. 21 6°. Let AOB be a right angle, COB an equilateral - c triangle, and OP perpendicular to CB. Prove that the angles AOC, COP, POB are all equal, How e could you trisect a right angle? K \. Fig. 22. 7°. Let A, B, two angles of an equilateral triangle, be bisected by the lines AO, BO, and from O draw OP, OQ, parallel to the sides CA, CB, and terminating in the base in P and Q. Prove that AP = PQ=QB. [The student will construct the figure. ] 78 PROPORTIONAL LINES, 8°. Prove that any side of a triangle is greater than the differ- ence of the other two. 9°, Prove that the angle included between two lines drawn from the vertex of any triangle, the one bisecting the vertical angle and the other perpendicular to the base, is half the difference of the basic angles. 10°, Prove that if a line be drawn from one of the equal angles of an isosceles triangle upon the opposite side and equal to it, the angle embraced by this line and the production of the base, or un- equal side, will be three times one of the equal angles of the triangle. 11°. Draw three lines from the acute angles of a right angled triangle—two bisecting these angles and the third a perpendicular to one of the bisecting lines—and prove that the triangle embraced by these lines will be isosceles. SECTION THIRD. Proportional Lines. PROPOSITION I. The segments of lines intercepted by parallels are pro- (120) portional. In the first place, let. the secant line AC be divided into commensu- rable parts by the parallels AA’, BB’, CC, for instance, such that AB containing three measures, BC shall contain two of the same; it may be shown that any other secant line, A'B'C' will be divided in the same ratio. For, dividing AB, BC, into 3 and 2 parts in a, 6, c, and drawing aa’, bb’, cc’, through the points of division parallel to AA’ and through A’, a’, b’, &e. (points of A'C’), drawing A’a”, a'b", &c., parallel to AC and terminating severally in the parallels aa’, bb’, &c., we have (?) (2), PROPORTIONAL LINES. 79 A‘a’ =A =eb3rb", and /a’A’a =6'ab, Za'aA' =b'd a; «. A’a’ =a’b’, and for a like reason, a’b' = b'B’ = Bc’ = c'C’. * AB: BO: : 3:3, and A'B’: B@i: 33% 2. ow AB?BC:.: AB #aBC’. The same reasoning will be applicable, it is obvious, whatever may be the number of parts into which AB, BC, may be divided, or whenever AB, BC are commensurable. Next, let the segments a, b, be incommensurable. Produce 6 so that 6+ 2 shall be commensurable witn @, and through the extremity of z draw a parallel increasing the corresponding segment b’ by 3 xz; then will 6'’+ 2’ be commensurable with qa’; Fig. 232. and, by what has already been demonstrated, we shall have the proportion b ba - Bt 2 (ee HT oh Thun ath ake ek al but z can be made less than any assignable quantity, since it is less than the measure of a, which may be diminished at pleasure; Ul whence it follows that a? qi are variable quantities capable of indefinite diminution, and, consequently (63), that vale. past ha a s37@:8. QE. D. mee Cor. 1. When three lines, two of which are parallel, are (121) cut by two others so as to make the segments of the secant lines pro- portional, the third line is parallel to the other two. my For, if the line CC’ be not parallel to AA’, let CC” be so; then there will result (120) AB:BO: 2 A'R:? BC’; , rs Sree Ws c” BC els vat? Fig. 233. but, by hyp., AB: BC: : A’B’: BC, or BC = soe be B : “. BC’ =BC, which is absurd; therefore CC’ is not otherwise than parallel to AA’. Cor, 2. The sides of mutually equiangular triangles are (122) proportional. For, let any two of the equal angles be made vertical, then will 80 SIMILAR TRIANGLES, the bases m, n, be parallel, and, if through thie common vertex a line be drawn parallel to m, n, dividing their intercepted perpendicular into the parts a’, 5”, we Ay find (120) a:b: ih eae. DS ie te, Remark. The pcb sans sides are opposite equal angles. Cor. 3. If two triangles have an angle of the one equal to (123) an angle of the other and the sides about the equal angles propor- tional, the triangles will be mutually equiangular and consequently similar. Draw a line through the extremity of 5 [ fig. 23,], see (122), and imitate the reasoning under (121), Def. Figures which have their angles respectively equal when taken in the same order, and their sides proportional, also taken in the same order, are called similar, Cor. 4. The altitudes of similar triangles are to each oth- (124) er as their bases. For we have (122) tt oa b 2 des 64° mark Cor. 5. If lines be drawn from the vertex of a triangle to (125) points in the base, all lines parallel to the base will be divided into parts proportional to the segments of the base. For the segments ab, bc, cd, have to AB, BC, CD, the constant ratio of the perpendiculars Op, > Pie: 23 OP. Cor. 6. Triangles having their sides severally parallel are 12% similar, la Fig. 236). Cor, 7. Triangles which have their sides respectively per- pendicular, are similar. Thus, a, d, c, are to each i other as a’, 5’, c’, where it is obvious that the homol- b ogous sides are those that are perpendicular to each other. PROPOSITION II. Fig. 23 . A right angled triangle, by a perpendicular let fall from (128) the right angle, is divided into two partial triangles, similar to itself, and consequently to each other. HYPOTHENUSE, 81 Let a, b, denote the sides about the right angle, \ h the side opposite, or the hypothenuse, and m, n, b the segments of A made by the perpendicular, p. ns The angle (a, h) is common to the triangles (a, b, Fig. 24. h), (a, m, p), which also have a right angle in each, and are there- fore equiangular, (7 (d, h) being = Z (p, a),) and consequently sim- ilar. So it may be shown that A (d, n, p) is similar to A (A, b, a), and the proposition is demonstrated. Cor. 1. The perpendicular is a mean proportional be- (129) tween the segments of the hypothenuse. For we have ey ee a Cor, 2. The sides about the right angle are mean pro- (130) portionals between the hypothenuse and the adjacent segments. For we have (122) h AMP a — =—, and —=—. a m b n “Cor. 3. The segments of the hypothenuse are to each (131) other as the squares of the sides opposite them ; for, from the above we have hm = a’, and hn = 6°; : .*. by div., mwa 5.5%, Cor. 4, The square of the hypothenuse is equal to the (132) sum of the squares of the other two sides. For we have ae (m+ n)h = a? + B?, or hh = h? = a? + 32, Scholium. This proposition, the celebrated 47th of Euclid, is obviously a direct consequence of (122), and with that theorem, which embraces the characteristic property of the triangle, consti- tutes the working rules of all geometry. Cor. 5. Conversely, if the square of one side of a tri- (133) angle is equal to the sum of the squares of the other two, the trian- gle is right angled. For suppose the triangle (h, a, b) to be such that h?=a?-+ b?; then if on @ we construct the triangle (h', a, b') making b'= 6 and the Z (a, b') = “, there will result A’? =a?+ b2=@?4+ B=h’, .. h' =h, and ya e . Z(a, b)=(a, b)=+. Fig. 242. 6 82 . BASE DIVIDED. Cor. 6. The diagonal of a square is incommensurable (184) with its side, the former being to the latter as ./2 to 1, For we have xz? =a? + a*® = 2a? ie w= da/d, or 2: a= fe J. Hae. The student might very naturally suppose that a common mea- sure to any two lines could be found by taking the measuring unit sufficiently small; but, from this corollary, it is shown that the con- trary may be true, and we are therefore taught the necessity of extending our demonstrations to the case of incommensurability, PROPOSITION III. To find the relation between the oblique sides, a, b, of a triangle, the line c drawn from the vertex to the base, and the segments, m, n, of the base made by this line. Drop the perpendicular p, intercepting the seg- ment z between the foot of p and that of c; we have (132), (8), (9), a? = p? + (m +2)" =p? + m* + Ime + 2%, b? = p? + (n — zr)? = p? +n? —2Qnz4+ 22, c? =p? +2; .. subtracting the 3d from the Ist, we have a? —c* = m*+ 2mz, from the 2d b?’—c?=n?—2nz; ... dividing one by m and the other by n, =m + 22, b? — ¢? =n— 27, a®*—c*? b?—¢? m n If in any triangle, a line be drawn from one of the angles, terminating in the opposite side, and the squares of -the sides embracing the divided angle be severally diminished by the square of the dividing line, then the sum of the quotients arising from dividing these remainders by the corresponding subjacent segments, will be equal to the divided side. =M-+N; 1, Gy (135) QUADRILATERALS. 83 If we take the particular case in which m and n are equal, there will result a*—c*? b*%—¢? m whence a? + b? = 2c? + 2m?, or, Cor. 1. If a line be drawn from the vertex ofa triangle (136) to the middle of the base, the sum of the squares of the oblique . sides will be double the sum of the squares of the middle line and the half base. If the triangle be isosceles, there will result =m+m, gece? g® ~ ¢2 peat or a? —c? =™mn. =m+n, Cor. 2. In an isosceles triangle, the square of the oblique (137) side diminished by the square of the middle line, is equal to the product of the segments of the base. Let ABCD be any quadrilateral, M, N, the middle points of its diagonals, AC, BD; join AN, NC; then (136) AB? -++ AD? = 2AN? + 2BN?, and CB? -+ CD? = 2CN? + 2BN?; Fig. 252. AB? + BC? + CD? + DA? = 4BN? + 2(AN? + CN?) = 4BN? + 2(2AM? + 2MN?) = (2AM)? + (2BN)? + 4MN? =AC? + BD? -+ 4MN?, .. Cor. 3. The sum of the squares of the sides of a (138) quadrilateral, exceeds the sum of the squares of its diagonals by four times the square of the line joining the middle points of the diagonals, Cor. 4. The sum of the squares of the sides of a parallelo- , (139) gram, is equal to the sum of the squares of its diagonals. : For (99) the diagonals bisect each other. y PROPOSITION IV. To find the distance of the foot of the perpendicular of a trian- gle from the middle of its base. Let a, b, be the oblique sides of a triangle, and p its perpen- dicular, dividing the base 2m into two parts, m-+7, m—2Z; 84 PERPENDICULAR. where, consequently, xz is the distance of the foot of the perpendicular to the middle point of the base. We have (132) a* = p?+(m-+ 2)? and b? = p?+(m—z)?; i a® —b?=4mz, or (10), (a+ 5b) (a— b) =2m « 2z; s*. (43) 4 2m:a+b=a—b: 2x, The base of any triangle is to the sum of the oblique (140) sides as their difference to the double distance of the foot of the perpendicular from the middle point of the base. This theorem is convenient for finding the perpendicular let fall upon any side of a triangle; for we have p=[a?—(m-+2)"]' =[(a+m+2)(a—m—2)j* (141) PROPOSITION V. If aline be drawn bisecting the vertical angle of a (142) triangle, the segments of the base thus formed, will be to each other as the sides opposite them. In figure 16, draw CE parallelto DB, E beinga point [fig. 16] in AB; then (81), (80), Z ACE= CDB = CBD = BCE, and (120) ABei EBs AC: CD=CB. Q. E. D. EXERCISES, 1°, Wishing to ascertain the dis- tance AB of an in- accessible object B, I measure a line AC at right angles 1 to AB, equal to 12 i Fig. 26. chains, then I take CD back and at right angles to AC, equal to 5 chains ; and find that a line sighted from D to B intersects AC at O, distant from C 325 chains. What is the distance from A to B? EXERCISES. 85 We have AB: AO=CD:CO, o 5 5 «8°75 or 12 305 ~ 305 °° = 305° = = 13°46. How can the same thing be done by (129)? 2°, What must be the length of a ladder to turn between two buildings, one 20, the other 30 feet high, and 40 feet apart; and what must be the distance of its foot from the first-mentioned building ? Let the length of the ladder be denoted by y, and the distances of its foot from the buildings by 20+ z aN and 20—z; then ‘ (20+ ayn + 20? — yf = ae (20 te! x)? cs 30°, ie meas or —- A’. 2- Wr+ 2? + 2? = 2’ —2- Wer+2’ Fri + 30°; 4 + 202 = 30? — 20°’ = (30 ++ 20) (30 — 20), 50-10 50_ and the distance required is 20 + x = 2625 ; length of ladder = y = (207+ 26252)" =? 3°. Wishing to ascertain the distance from A to D (fig. 63), rendered inaccessible by the intervention of a ledge of rocks, I fetch a compass, ABB'CC’D, viz., AB=11 chs., BB’ =8, BC=8, CC’ =4, and C’D=5. Whatis the distance from A to D? Ans. 15 chs. 4°. Given the perpendiculars let fall from A and B_ [jfg. 18] upon the plane MM’, the one 7 the other 5 feet, and the distance of these perpendiculars = 10 feet; to find the distance of P from the greater perpendicular. Ans. 5 feet 10 in. 5°. Given AB = a and the distances of A and B from [ fig. 19] MN, =p, p'; to find AO=zc. 6°. Given the hypothenuse of a right angled triangle, h, and the sum of the sides about the right angle, = 2m ; to find these sides. Denote them by m+ 2, m— 72; aa * AY 4 a \ then (132) 2=+ (4h? — m?)” ; Fig. 28, ae m+ xr2=m = (th? —m’)’, and m—xr=m= (4h—m ah ;—where it is ob- J PRE vious that both values given by the double sign + Fig. 282. equally satisfy the conditions of the problem. 86 EXERCISES. 7°. Given the hypothenuse of a right angled triangle and the dif- ference of the sides about the right angle, to determine the triangle. If the hypothenuse be denoted by 2a and the difference of the legs by 2m, the answers will be a: 1 zt m = + (2a? — m’)* + m, x — m = + (2a? — m?)* —m. 8°. Given the base of a right angled triangle, b, and the sum of the hypothenuse and perpendicular 2m, to find these sides. 9°. Given the base of a right angled triangle and the difference of the hypothenuse and perpendicular, to determine the triangle. 10°. A liberty-pole 100 feet long was broken, and, resting upon the stump, its top fell at the distance of 40 feet. Required the length of a ladder, planted 10 feet distant, that shall reach the break. 11°. A greyhound is 10 rods distant from a hare; the hare starts off at right angles to the line joining them and the hound pursues in the hypothenuse of a right angled triangle to intercept her. Now the velocity of the dog is to that of the hare as 3 to2. How far does each run? 12°. The same conditions as in the preceding only that the hare runs obliquely from the hound at half a right angle. Ans, The hound runs 60 (,/2 + 4/205), the hare 40 (,/2-+ ./2°05), rods. 13°. Given the perimeter 4a, of a rectangle (right angled par- allelogram) and the diagonal, d, to determine the figure. 14° Given the perimeter, na, of a right angled triangle, and the perpendicular, a, let fall from the right angle on the hypothenuse, to determine the triangle. Let the segments of the Reraiiennae be re- 7 3- Ne presented by x and y; then shall we have wee Be, ef vs r+y+ (2? +a)" +(y?+ @)* =na, Fig. 29. and = a * which ratio may be put = z; \ a then z will = az and y= >? 1 2 2 a whence az +5 + (az? + a’) + ( S + a’) i na, or zt— * 4 (2 41P4 <(2+t=a; EXERCISES. 87 $ ee ie ie Je +1) il ih etal and eA Y Dg Vp opin ced ie Goal oP atest ae i ea ee z 2 or (n+ 2) 2-4 Fa “iene n’ ez=-—f], m+ 2 n n‘ n' ss. (58),, 2* On 2 °Z+ (4n 4) au ap 16(n +1)? —1 _ m—16 (n+ 17 on, 10nd) a Balt Ie ey aft Cente (n+l) : 4(n+1) Tat) se ee (Ce ae ‘ = 4(n+1) and y 2G or eet apa sO ot ayienis, a(n +1) 15°. Given the two lines, m, n, drawn from the acute angles of a right angled triangle to the middle points of the opposite sides, to determine the triangle, 16°. In a triangle, given the segments a, b, of the base formed by the perpendicular let fall from the vertex, and the ratio, m : n, of the oblique sides, to king the triangle. (a+ b) (a—b) (a + b) nay rH | eae: (m—n) ee" of 17°. In a triangle, given the base, b, the eRe p, and the difference, 2m, of the oblique sides, z + m, x — m, to determine these sides. Ans, r+m=? zr—m=? 18°. In a triangle, given the segments, a, b, of the base made , ees Re sides. 88 EXERCISES, by the line bisecting the vertical angle, and the sum, 2m, of the obligue sides, to determine the triangle. Zbm 19°. The same as in the preceding, ie the difference of the oblique sides is given instead of their sum. 20°. Given the three sides of a triangle, 12, 18, and 20 chains, to find the perpendicular let fall on the last-named side from the op- posite angle. Ans. 1066 chains, 21°. Given the base, b, and altitude, A, of a triangle, to determine the inscribed rectangle, the base of which shall be to the altitude as m to n. Whe mbh nbh Ans. The sides of the rectangle are Rea an 22°. The same as the preceding, except that the sum of the sides containing the rectangle is given, 23°. The same, only the difference is given. 24°, Being on the bank of a river and wishing to ascertain its breadth, for the want of instruments I set up a stick at A and then take 25 measures with a rod happening to be at hand, from A ina straight line to B, where I set up a second stick; I then take 5 measures of the same rod directly back from A in line with X, an object on the opposite bank, to C; also from B backward in line with X, 5 measures to D; and finally I find BC to be = 25 mea- sures and a remainder, also AD =26-+-a remainder. Now to de- termine these remainders, I proceed as follows : For the fractional part of BC, I apply to the mea- —; ae Ks! suring rod, m, a second rod, 7, = the excess of BC over 25m, and find r contained in m once, with a Fad Fig. 30. remainder, 7 ; therefore — are | ree Again, applying 7’ to r, I find it contained once with a remainder, 7’; .°, 4 =] Le . Ap- plying r” to vr’, I find it contained 4 times with a esate P's Radic wat , 7’ . } Finally, I find r” contained in r” just twice ; EXERCISES, 89 Whence ee ee "14> 14— 14— { 4 ees rs af Za 1 a 1 7 1 1 oe 1+ _ 1+ = seis i Pe, er 1 a on * jamie 7" Be ge 1 u 1 Ve ee m i. F sies Olde: m 1+ 1 i¥ 1 par 1 2 11 1+—; rt ple Bory *. BC =25'55; by a like process I find AD = 26+ a 26‘65. 1+ i 1+ i I+— We have then (135) putting AX =z, BX = y, 25'55? — 25? -y? — 25? eininatate S mate li Bes t@52 __ 2 2 re rn ae 25 nite - md aint i. r= y= : 25°, Given the hypothenuse of a right angled triangle = 35 rods, and the side of the inscribed square = 12 rods, to determine the triangle. Ans. base = 28, or = 21; perpendicular = 21, or = 28. SECTION FOURTH. Comparison of Plane Figures: PROPOSITION I. Rectangles are to each other as the products of their di- (143) MeENSIONS. We will take the general case at once, and sup- pose the containing sides, a, a’, of the rectangle A are incommensurable with 0}, b’,those of B. Let b be increased by zx, and 0’ by z’, soas to make 6+2z commensurable with a, and b'+ 2’ with a’. Suppose, for instance. that while a is divided into m parts, that b-+-z contains n of the same parts, or that (2,) SU, ben” and that, while a’ is divided into m’ equal parts, which may be dif- ferent from those of a, b'+ x’ contains »’ of the same parts, or that a ™ Digest 0.7. Through the points of division draw lines parallel to the sides of the rectangles and denote the part added to Bby X. Then will the partial rectangles be all equal (95) ; and the number constructed on the base a being = to m, and the number of tiers corresponding to the divisions in a, = n’, we have mm’ for the total number of partial rectangles in A; so the rectangle B+ X contains nn’ of the same. Hence (2,) A: B+ X=mm': nn; but mm':nn'::aa’:(b+2)(b' +2’), since a, a’; b+ 2, b'+ 2’, are as their measures m, m'; n, n’'; here iB teen Pa) (P +2") _ 66 + br tb arte 6 A aa aa BX bh be’ +24 e2' | He Demo. ag BB bb . (63) be Q. E. D RECTANGLE AND TRAPEZOID, 91 For it is obvious that X depends for its value upon z, 2’, decreas- ing as these decrease so as to become nothing if these were to be- come = 0, and that z, 2’, being less than the parts into which a, a’ are divided, may be made less than any assignable quantity, by suf- ficiently increasing the points of division. Therefore xX a+ bz’ ional ex’ A aa r! may be made less than any assignable quantity. Cor. 1. The rectangle is measured by the product of its (144) dimensions, For suppose A to become a measuring square, or that while A = 1 is taken as the unit of surface, a= a’ = 1 | i ; I becomes the linear unit, we have Fig. 31s. Cor. 2. The right angled triangle is measured by half (145) the product of its base into its altitude. Thus T = 40h. “ Fig. 313. PROPOSITION II. The Trapezoid is measured by half the sum of its par- (146) allel sides multiplied into their perpendicular distance. Let a, 6, be the parallel sides and from their ex- b y tremities drop the perpendiculars h, h, on a, b, pro- ; “sh i duced, forming the rectangle Nines X4Trp+Y, : : aaa Fig. 32. composed of the trapezoid Trp and the right angled triangles X, Y, having the bases 2, y. We have (144), (145), X+ Trp+ Y= (r#+a)+h=4+2(c4+a) eh =7(t+a+b+y) eh, and X + Y=4trh+4yh; Trp=t(a+b)eh. Q.E.D. Cor. 1. The parallelogram is measured by the product (147) of its base into its altitude. For, when b=a, the , « trapezoid becomes a trapezium or parallelogram, BpeNOESP and Trp =+t (a+b) eh, Fig. 320. becomes, P=4t(a+a)+h=ah. 92 PARALLELOGRAM AND TRIANGLE, Cor. 2. The triangle is measured by half the predees of (148) its base and altitude. For Trp =4(a+b)-h, hi becomes T = tah. hid Fig. 323. Cor. 3. Parallelograms are to each other, and triangles (149) are to each other, as the product of their bases and altitudes. Cor. 4. Parallelograms of the same or equal altitudes are (150) to each other as their bases. Cor. 5. Triangles of the same or equal altitudes are to .(151) each other as their bases, f Fig. 324. Cor. 6. Parallelograms of the same or equal bases are to (152) each other as their altitudes. Cor. 7. Triangles of equal bases are to each other as (153) their altitudes. Cor. 8. Parallelograms of equal bases and altitudes, are (154) equivalent, WT Fig. 325. Cor. 9. Triangles of equal bases and altitudes, are equiv- (155) alent, Ww Fig. 326. Cor. 10. Parallelograms, or triangles, between the same (156) parallels, and on the same or equal bases, are equivalent. Cor. 11. Conversely, if parallelograms or triangles, stand- (157) ing on equal bases in the same straight line, and having their ver- tices turned in the same direction, be equivalent, the line of ver- tices will be parallel to the line of bases, For if the triangles, having the bases a, a, and V- Vv" the perpendiculars p, p’, be equivalent, there results |, /\ JN w sap = tap, me ee ig. 327 P=P; whence the line of vertices VV’ is parallel to the 1 of bases BB’. Cor. 12. The parallelogram is double the triangle of the (158) same base and altitude. [Compare (147) with (148).] Vs a@ Fig. 328. PARALLELOGRAM AND TRIANGLE. 93 PROPOSITION III. Two triangles, having an angle of the one equal to an (159) angle of the other, are to each other as the products of the sides about the equal angles. Let the equal angles of the triangles A, B, be made vertical, and join the extremities of the sides a, 6, forming the triangle X; then (151) Ul wo I S|/& ofa ole and | ~ Q. E. D. ~ e byl bm byl be be Cor. 1. Equiangular parallelograms are to each other (159,) as the products of their dimensions (158). Cor. 2, Similar triangles are to each other as the squares (160) of their homologous sides. For, then we have ZI (122) a’ a. a Cee a a Oe OE May dias Robbe Dele | ioe eet Cor. 3. If figures, resolvable into the same number of (161) similar triangles, be constructed upon the three sides of aright angled triangle, that on the hypothenuse will be equal to the sum of those described upon the other two sides. For we have Davey CoP BEC EE a Fie or and ay © oes: RE wer re =4 ae ai ee eer Scholium. It is obvious that (132) is but a particular case of this more general theorem. EXERCISES. 1°, Demonstrate (132) by turning two squares into one. . SLY Fig. 34. 2°. On the oblique sides of any triangle describe parallelo- 94 EXERCISES, grams, any whatever; andon the base, construct a par- allelogram having its sides parallel to the line drawn tL» through the vertex and the intersection formed by pro- ducing the sides of the parallelograms described on the oblique sides of the triangle, and terminating in these same sides. Prove that the parallelogram on the base of the triangle will be equal to the sum of the other two; and that, if the vertical angle become right, and the parallelograms squares, (132) will be proved. 3°. It is required to straighten the line ABC, sepa- rating the estates of two gentlemen, by aid of the af 3 cross alone. Draw BD parallel to CA, then DC will be the re- quired line. (Why 2) 4°. Straighten the line in figure 36,. MA 5°. It is required to change the direction of the line AB, so that the extremity A shall be at A’, still retaining r Fig. 363. Fig. 35. Fig. “36. the same amount of land on each side, 6°. To turn a quadrilateral into a triangle, af bem either by the cross in the field, or by the right- peas angle straightedge on paper. Fig. 364. 7°. To turn a pentagon into a triangle. Mears Fig. 365 8°, To turn a triangle into a rectangle, JANN Fig. 366 9°. To turn a rectangle, A, into a square, B. 10°, Draw any polygon on paper, and find a square that shall be equal to it in area, 11°. There is a well at P, in the side of a triangular Pp field; it is required to draw a line from P so as to Fat divide the field into two equal parts. Fig. 368. 12°. What is the area of a parallelogram, 20 chains in length and 15 in breadth ? Ans, 30 acres. 13°, The longer side of a rectangle is r times the shorter, and EXERCISES, — 95 the area is = m*, or to a square having m for its side, What are its dimensions ? We have TreL= wv, f _ m _me/r < as SPRY aie ec and rz=m./T. 14°. A surveyor would lay out a rectangular field of 20 acres, such that the length may be three times the breadth. Required, the dimensions, [Solve and prove. ] 15°, The area of a parallelogram being represented by m?, and the base by b, how shall we find the height, ?—given A, how shall we find 6? m? 2 Ans. h= = 7° How expressed in words ? 16°. A man has 10 acres to put in a’parallelogram, having one side 12 chains. How wide must it be? Ans. 8 chs. 334 links. 17°. If the area of a triangle be m?, and its base b, how shall we find its height, h? 2 Ans. h= — . [Enounce in words, ] 18°. A surveyor would lay out a triangle of one hundred acres on a base of 25 chains. Required, the altitude. 19°, A joiner has a board 10 feet long, 2 feet. wide at one end, and 2 feet 6 inches at the other. How many square feet in the board? (146) : Ans. 224 feet. 20°. Given the sides of a triangle, 15, 16,17 chs, Required, the area. Taking 16 for base, we have (140) 16: 17+15::17—15 : 2z, or Sx se cil s Sp. ods . “., (141) p = [(17+ 8+ 2) (17 —-8—2)]* =[9 f8 S27) 8 Y/OT Area = 8p = 24 VJ 21 sq. chs. = 11 — 54, acres, nearly. 21°, What is the area of a triangle, the sides of which are 13, 19, 34 chs, 2 What difficulty arises in attempting to solve this problem? and why ? 22°. Wishing to ascertain the area of the quadrilateral field ABCD ( fig. 6,), I avail myself of the measures in the third exer- cise of the preceding section, and find it to be nine acres and one- fifth, 23°. In order to obtain the area of a pentagon ABCDE, I mea- 96 EXERCISES. sure the sides BC, CD, DE, and find them BC =6, CD =4, DE = 10 chains; and, by aid of the cross, determine the perpendicu- lars AP = 3, AP, =8, AP, = 1158, let fall upon CB, DC, DE, pro- duced, The area will be found to be eight acres and two-fifths. 24°, Given the sides of a triangle ABC, AB =30, AC = 40, BC = 50 chs., to cut off a triangle ABD (D being a point in AC) equal to 45 acres. We find the area of the triangle ABC = 60 acres ;_ therefore (151) AD = 30 chs., the line required. 25°. The triangle being the same as in the above, it is required to cut off the 45 acres by a line B’C’ parallel to BC—B' being a point in AB, and C’ a point in AC, We find (160) AB = 25:98, AC’ = 34°64 chs. 26°. The same conditions as in the preceding, except that 30 acres are to be cut off by a line PQ, P being a point in AB, 20 chs. distant from A. Required AQ, measured off from A towards C. Join PC, then 30 chs. : 20 chs. : : 60 acres : APC, .. = 40 acres; and 40 acres : 30 acres :: 40 chs. : AQ, .. =30 chs. 27°. To draw a line through a given point in the plane of a given angle so as to include a given area. Let the given point P be embraced by the sides of the given angle CAB. Suppose the problem solved, and that YPX is the required line: we are to find z= AX, a portion of AB. As the position 4@—=x of the point P, in regard to the lines AB, AC, is Pigiet given, we may suppose the perpendicular p, let fall from P upon 7 and a, the distance of P from the line AC, measured parallel to AB, to be known. Further, let the area of the triangle AXY be denoted by m’, that is, by a square whose side is m. We have, y being a perpendicular let fall from Y upon a, ry + p) = 2m’, and x a fe L—a or ps = ¥ EXERCISES, 97 ee eee pi hag uny tae Ty oy +p)=2( ee +p) = (BS + px = 2m?, wo == @& t—G or par + px? — par = 2m*r — 2m*a, 2m? 2m*a zr? ezrt=-— 9 P P a m*\?> m* 2m?a. m r?—Be—eer (~) = — — —_ = — (m*? — 2pa p) pp pi pa) 2 o— T= = (m* — &pa) ; 2 1 and =¢= - +7 (m* —2pa)* = 5 [m + (m? —2pa)?}. 4 Since (m*? — 2pa)* is < m, the two values of z, z= — [m+ (m* — 2pa)*}, and z= = [m — (mn? — 2pa)*], are both plus; and, therefore, equally applicable to the problem, as shown in fig. 37,, where we have : t= AX, zr = AX’. P We observe here since the tr. AXY = m?= AX’Y’, that subtracting the quadrilateral AX’PY from both, there results the triangle XPX’= YPY’. But if we take the difference of the values of X, we have XX says 7 + 2 (m? — 2pa)*, Fig. 372. which is the solution of the following problem : 27°,. Through a given point, P, in the side of a given triangle, AXY, to draw a line X’Y’; terminating in AX, AY, produced, if necessary, so that the areas PXX’, PYY’, shall be equal. We remark that m? must be at least as great as 2pa; for, were m?* < 2pa, (m? pa)" would be imaginary (6;), and the values of xz, © = AX, x = AX’ unreal, or the problem impossible. That is, the area to be cut off cannot be greater than the double inscribed rectangle standing upon a. 98 EXERCISES, We have here an example of a quantity capable of a minimum, or least value ; and, in order to find this minimum, it is obvious that we have only to solve the quadratic and put the part under the radical equal to zero, Thus m? — 2pa = 0, gives m? = 2pa, for the minimum of m?, If we inquire what is the greatest value of m? in this problem, we shall find it infinite, or that m? has no maximum. ax [See the value of z and fig. 37;.] ee x —> If now we subtract from a by insensible degrees, Fig. 373. or, as it is commonly expressed, diminish a according to the Law oF ConTINUvITY, a will at length become = 0, m mS when te 5 [m + (m? — 2pa)*], 2 gives BD ee ee mdr oe Fig. 374. where the first value only is applicable. Continuing the same motion, a will evidently become minus, and the point P will at the same time take up its position on the left of the line AC, or without the angle, and we shall have . ry <3 A z= = [mt (m? + 2pa)*] : rs Fig. 375. where both values of x are applicable, only that the second, being minus, requires the area m? to be laid off on the left. If, in like manner, we make p pass through the value 0 it will change its sign, and P will take up a position be- low the line AB, os 1 whence Rae ss [m + (m?— 2pa)*], of 1 becomes Bae et ee m 1 or t=—~ [— m = (m* + 2pa)?], where both values are applicable, as indicated in the figure. If, at the same time, we make a and p both minus, there will result B r= — [m+ (m?—2+—p+—a)4), EXERCISES, 99 or r=— $ [m = (m? — Qa)? ], where we have the same values as in the original solution, only changing + 2 into —2z, as we evidently ought to do, since z is measured from A in an opposite direction. This is obvious also from the comparison of the figures 37, and 37, where all the parts of the one correspond to all the parts of the other. The figure 37, will have the same double construction that 37 has in 37,. What would be the result of making p=0? what, when p= 0, a=0? . The problem we have just been discussing is admirably adapted to show the correlation of algebraical signs and geometrical figures. This problem will also enable us to divide any polygon into any required parts. 28°. The parallel sides of a trapezoid are a, 6, and the altitude, h, it is required to cut off an area m? adjacent to b by a line paral- lel thereto. What will be its breadth? bh 2hm? bh \?|% pine e=— + [57+ (3) J 29°. Given the sides of a triangle, ABC, viz., AB=10 chains, AC =8 chains 43 links, BC = 470; also the position of a point, P, viz., distant from AB by 1‘80, and from AC by one chain, Re- quired to draw a line through the point P, that shall divide the tri- angle into two equal parts. : The student will solve and verify. 30°, What is the greatest rectangle with a given perimeter ? We may denote the sides by a+ x and a —z, then will the peri- meter be = 4a, and we shall find z= (a? — m?)?, m* denoting the area; whence it is obvious that m? cannot exceed a*; and therefore that z= 0, when the area m? is a maximum, or that among rectangles of the same perimeter, the square is the maximum. 31°. What is the maximum triangle that can be constructed ona given base, and of a given perimeter ? Ans. The triangle must be isosceles, 32°. Given the area and diagonal of a rectangle, to determine its sides, 33°. Let ABCD be a quadrilateral field, of which the sides AB, q 100 EXERCISES, BC, CD, DA, are, respectively, 16, 34, 30, 29 rods in length ; also, let the diagonal BD = 374 rods. Itis required to divide the field into two equal parts, by a line cutting the opposite sides, AB, CD, so that the ratio of the segments of the one shall be equal to that of the corresponding segments of the other. Ans, AX=? DY =? [Produce AB, CD, until they meet, and consult (140), (141), (132), (159). ] BOOK THIRD. PLANE GEOMETRY DEPENDING ON THE CIRCLE, ELLIPSE, HYPERBOLA, AND PARABOLA. SECTION FIRST. The Circle. Definition 1. The circle is a plane figure described by the revo- lution of a straight line of invariable length about one of its extrem- ities as a fixed point. Def. 2. The describing line is called the Radius [rod] of the circle, the fixed point the Centre, and the curve line that bounds it its Circumference. : Cor, All radii, or lines drawn from the centre to the cir- (162) cumference, are equal to each other. PROPOSITION I. Angles at the centre of the same or equal circles, are to (163) each other as their subtending arcs ; and the corresponding sec- tors have a like ratio. In the first place, let the angles be commensurable. For exam- ple, suppose the angle AOB to contain the angle BOC twice ; then it is manifest that in applying the angle BOC twice to the angle AOB, the point C will fall on C’, the middle point of the are AB, since x (162) OC’=OC. Therefore the are AB contains the arc BC twice; and, in the same way, the sector AOB is double the sector BOC. “. ZAOB: ZBOC:: are AB: arc BC: : sec. AOB: sec. BOC, each being as 2 to 1. | Fig. 38. 102 SECTORS. So in general, if a, 6, be any commensurable an- gles, or such that when a is divided into any num- ber, m, of equal parts, b shall also be exactly divisible into some number, n, of the same equal parts ; then it will follow, by superposing one of these equal Fig. 389. angles m times upon a and nv times upon 3, that the corresponding arcs A and B will be divided into m and n equal arcs. The same of the sectors K, L. a:b6:: A: B:: K: L, being asm to n. Finally, let us take the most general case, or that of incommensurability, where a and b having no common measure, are incapable of being divided into the same equal parts. Let. 0d be increased by the angle z, so that b-++ x shall be commensurable with a, then will the corresponding are B+ X be com- mensurable with Aj; and, from what has just been proved, there results the proportion & p Fig. 383. b+2 Bx i bitten Ae ou! Ie re eon «See bs naa nrriats taste bus B iv A , (63) aaa TB Bere oti A: B, and the same is equally true of the sectors. Q. E. D. Cor. 1. In the same or equal circles, arcs may be taken (164) as the measures of their angles at the centre. Thus, if 6 be taken for the unit of angles and B for that of arcs, the proportion Se al : Fig. 384. b «cB becomes ban = 1 1 or , Gite JA § Scholium. As the right angle seems the most suitable for com- paring angles, so its measure, the Quadrant, or quarter circumfer- ence, would appear to be the appropriate unit of arcs, and for this purpose the French have sometimes employed it, dividing the quad- rant into a hundred equal parts, which they called degrees: but custom has established a different unit, the jth part of the quad- SECTORS, 108 rant, which is denominated a degree, and written 1°. The degree is divided into 60 minutes, marked 1’ and the minute into 60 sec- onds, written 1”, 2”, 3”, .... Cor. 2. A quarter circumference is the measure of aright (165) angle, a semicircumference of two right angles, and a circumfer- ence of four right angles. How many degrees in half a quadrant? in one third of a quad- rant? in4d? 427424247247 4? Cor. 3. In the same or equal circles, the greater arc (166) subtends the greater angle at the centre, and, conversely, the greater angle is subtended by the greater arc. Cor. 4. In the same or equal circles, equal arcs sub- (167) tend equal angles at the centre, and the converse. Cor. 5. In the same or equal circles, the greater arc sub- (168) tends the greater chord, and conversely, the greater chord is sub- tended by the greater arc. For, let the arc B> A, .. (166) 7 b>a; «. (116) chord d>c. Conversely, let chordd>c, 7 * Zb>a;., arciBiu. ; é Fig. 385. Cor. 6. In the same or equal circles, equa] arcs subtend (169) equal chords, and the converse. Cor. 7. The Diameter, or chord passing through the (170) centre, is the greatest straight line that can be drawn in a circle. Why? Fig. 386. Cor. 8. The diameter bisects the circle, and its circum- (171) ference. PROPOSITION II. An angle inscribed in a circle, is measured by half tts (172) subtending are, Let AOB be any diameter, and C any point of the circumfer- ence: join CA, CO, then (94), (162), (101), 104 INSCRIBED ANGLES. Z COB = OAC + OCA =2 # OAC; .. (164), measure of 7 OAC=meas, of + 7 COB =4arc CB; so meas, of 7 DAB=#4 arc DB, and meas, of 7 EAB=4 arc EB; .., by addition and substraction, : meas, of / CAD = 4 are CD, [Centre where ?] Fig. 39. and meas. of 7 DAE =¢t are DE. [2] Q.E. D. Cor. 1. In the same or equal circles, angles at the cir- (178) cumference, subtended by the same or equal arcs, are equal. ne " ; Fig. 392. Cor, 2, Of angles inscribed in the same or equal circles: (174) 1°. An angle subtended by an arc less than a semi- (fig. 39,.) circumference is less than a right angle ; 2°. An angle subtended by an arc greater than a (fig. 39,.) semicircumference is greater than a right angle ; . 3°. An angle subtended by a semicircumference is a right angle. Fig. 393. Cor. 3. The sum of the opposite angles of a quadrilate- (175) ral inscribed in a circle is equal to two right angles. (fig. 39.) eee igh. Cor. 4. A Secant line is always oblique to the diameter (176) drawn through either point, in which it cuts the circumference ; and, conversely, a line drawn through the extremity of a diameter, and oblique to it, is a secant line, or cuts the circle. For, let SS’ be any secant line, cutting the cir- “ cumference in any points A, X, and draw the diame- ter AB; then the 7 BAX is < |, being measured by the are BX < semicircumference. Conversely, let the line SAS’ be oblique to the diameter AB, and suppose that BAS’ is the acute angle, then will SS’ cut the circle.. For, take the point X in the . Fig. 394. SECANTS AND TANGENTS, 105 semicircumference on the same side of the diameter AB with the angle BAS’, so that the angle BAX shall be equal to the angle BAS’, which may always be done, since, by taking the point X nearer and nearer to A, the angle BAX may be made any what- ever less than a right angle; then will the line AX coincide with the line AS’, and the point X will be a point of the line AS’, Therefore AS’ will cut the circumference a second time in X, and be secant to the circle. Cor. 5. A Tangent, that isa line which touches a circle (177) without cutting it, is at right angles to the diameter drawn through the point of tangency, and the converse. Let TAT’ be tangent to the circle at A; then will TT’ be. perpendicular to the diameter AB; for, if TAT’ were oblique to AB, it would be a secant line (176). Again, let TAT’ be perpendicular to AB, then is TT’ tangent to the circle; for, if AT were a secant, then would 7 BAT < +, which is contrary to the hypothesis, Illustration. Imagine a line to revolve about a point in the circumference of a circle, there will be but one position in which it will be a tangent, or in which it will not cut the circle. Fig. 39«. Cor. 6. The angle formed by a tangent and secant is (178) measured by half the included are. For the Z BAT,= +, is measured by 4 arc. BXA, (fig. 39;.) and Z BAX is eta by 4 arc BX; ‘ [—] Z XAT is measured by 4 arc XA. PROPOSITION III. The angle formed by the intersection of two secant (179), lines is measured by the half sum or the half difference of the included arcs, according as the point of intersection is within or without the circle. First, let the lines AY, BZ, intersect in X, a point within the cir- 106 CORRELATION OF FIGURES, cleABYZY’; join BY, then the measure of Z BXA =the meas, of (BYX-+ YBX ) = the meas. of BY X-+ meas. of YBX, =4 arc AB+¢ are ZY, =+ (are AB-+ are ZY). Now let the point X glide along the line BX ZX’ until it take up a position X’ without the circle ; as a consequence Fig. 40. of this motion of X, the arc ZY will diminish, vanish, and finally re- appear measured in the opposite direction from Z, as ZY’. But this diminution of the arc ZY may be regarded as produced by subtracting from ZY, a quantity greater than itself; therefore, the remainder ZY’ must be a minus quantity, ZY being plus: Thus ZY +BY VY, w-ZY' = VY ZY, or — ZY'=ZY—YY'. Whence the measure of 7 BX’A = 4 (arc AB —arce ZY’). Q. E. D. Remark. The second part of the demonstration may be made independently by joining AZ; for the measure of 7 BX’A = meas. of (BZA — X'AZ) = 4 (AB — ZY’); but it is important to arrive at a principle by the aid of which we may be enabled, as above, to comprehend the mutual dependencies of all the particular cases of a general proposition. PRINCIPLE.—Whenever a Geometrical Magnitude and its Algebrai- (180) cal Representation can be made, by CONTINUOUS Diminution, to pass through the value nothing, and, by the same “Law of Continuity,” to re-appear—then will the Geometrical Magnitude be opposite in position, and its Algebraical Re- presentation be affected by the contrary sign. For it is evident that the magnitude, whether line, surface, solid or angle, on passing through zero, will change direction, and it is equally obvious, from the theory of algebra, that any algebraical expression, as ad — b, by which it may be represented, will at the same time change from plus to minus or vice versd. Scholium. Hence magnitudes, as two lines measured in oppo- site directions, are usually affected by contrary signs, but not al- ways. Thus, two radii of the same circle, though measured from the same point, the centre, in opposite directions, constituting a diameter, are not to be regarded the one as + and the other as — ; since the one cannot be derived from the other by passing through 0. INTERSECTING CHORDS. 107 Cor. Parallel lines intercept equal arcs; and, conversely, (181) lines intercepting equal arcs are parallel. For the lines AY’ and’ BZ being parallel, the angle (fig. 40.) X'=0, .*. meas. of X’=0, or 4 (arc AB—are 2 B “ZY )=0, « are AB= ZY. . Again, if AB = ZY' then 7 X' = 43 (are AB — are x r, ZY’) =0, and .. AY’ is parallel to BZ. This corollary might have been placed under the i Fig. 402. ceding proposition. How? Join AZ. PROPOSITION IV. When two intersecting lines cut a circle, the product of (182) the segments of the one, included between the point of intersec- tion and the circumference, is equal to the product of the cor- responding segments of the other. Let the chords AY, BZ, intersect in X; join BY, AZ, (fig. 40.) then the triangles AXZ, BXY, are equiangular and similar, whence the proportion “ AX BX XZ XY? 5 AX @ XY = BX e XZ, and the theorem is proved when the point X of intersection falls within the circle. Now let the point X glide along the line BXZX’ till it takes up a position X’ without the circle; then will the segments XY, XZ, decrease, pass the value nothing, and re- appear as X'Y’, X'Z, measured in the contrary direction ; (180), AX « XY = BX « XZ, becomes AX’. — X'Y’= BX’. — XZ, or AX’. X'Y'=BX’.X’Z. Q.E. D. Cor. 1. Through three points not in the same straight (183) line, a circumference of a circle may be made to pass, and but one, Let A, B, Y, be three points not in the same straight (fig. 40.) line; join BY, YA, then draw BX’ at pleasure, and make the angle YAZ = YBZ;; it follows, from what has been proved, that Z will be a point of the circumference. In the same way any other point may be found; therefore the position of the three points A, B, Y, determines the positions of all the points of the circumference. Cor. 2. A straight line cannot cut the circumference of a (184) 108 CIRCLE AND TRIANGLE INSCRIBED. circle in more points than two; since through three points of a straight line no circumference can be made to pass. Cor. 3. If from any point without a circle a tangent and (185) secant line be drawn, the portion of the tangent intercepted between that point and the point of tangency will be a mean proportional between the segments of the secant line. For, let the points A,Y’ approach each other so as to (fig. 40.) become united in ,T,,,; then will the chord AY’ vanish, and X’Y’A become a tangent X’ ,Ty.; <4 x! further X'Y’ will = X'A=X’ ,T,.; ee “. XB +s X'Z= XA. XY’ =X', Ty, + X iT, de Corea D te Py, fs rk Fig. 403. Cor. 4. From the same point two equal tangents can be (186) drawn to a circle. | [Unite B, Z.] Cor. 5. If a line be drawn bisecting the vertical angle (187) (a, 6) of a triangle, and terminating in the base, the product of the oblique sides a, 6, will be equal to the product of the segments m, n, of the base, increased by the square of the bisecting line c. Suppose a circle described about the triangle, pro- S duce the bisecting line c, so as to form the chord LBS c+ x, and join the extremities of c+ xz and bd; then, by similar triangles, we have a” c+e ore otra ab=c?+cxr=mn-+ c?, Fig. 404. PROPOSITION V. The continued product'of the three sides of any triangle (188) ts equal to its double area multiplied into the diameter of the cir- cumscribing circle, Let ABC be any triangle; suppose a circle cir- cumscribed, and draw the diameter BD, also CP perpendicular to AB; then, by similar triangles, CP CB AC -CB AGA Bei Ss oApas AB- AC. CB A =aith e@ =: ZAABC=CP.- AB BD : or AB-+BC-CA=24 ABCXBD. Q.E.D. EQUATION OF THE CIRCLE. 109 PROPOSITION VI. To find the Equation of the Circle referred to Rectangular Coérdinates. Let OX, OY, be two lines, intersecting each other at right angles in O; also let 0, situated at the distance 56 from OX and a from OY, be the centre of any circle; further, let P be any point of the circumference, distant from OX and OY by the variable lines y and z; then will the radius, 7, be the hypothenuse of a right angled triangle, of which the base will be z —a, and the perpendicular y — b, whence The Equation of the Circle is (y— 5b)? + (4@— a)? =r%, (189) Scholium. The variables zx, y, are called Codrdinates of the point, P, or of the circumference, when spoken of together ; when one is to be distinguished from the other, y is denominated the ordinate and zx the abscissa; OX, OY, are called the axes of co- ordinates, OX is the axis of z, OY that of y—O is their origin. If the centre o of the circle be on OX, the axis of zx, b = 0, and the equation (189) becomes y?+(x—a)?=r’. (190) to. ‘ Fig. 422. If, in addition tob=0, we make a=Q, the centre, 0, will be transported to the origin, O, and the equation becomes y? + 7? = 72, (191) Fig. 42. Resolving (191) in reference to y, we find two equal values of y for every value of z, y =: (r? — 2°)8, i.e. y=+(r? — x)?, or y = — (r? — 2)?, iN Fig. 423. Now, if we diminish the arc AD, its ordinate AB =y diminishes also, and becomes=0 when AD=0; finally AD reappearing as A’D measured in the opposite direction, its ordinate A’B reap- pears, likewise measured in the opposite direction from B, and 110 INTERSECTING CIRCLES. is therefore = — y, according to the principle laid down in (180). Whence, Cor. 1. The circle is perfect:y symmetrical, any diameter, (192) as DD’, bisecting allthe chords to which it is perpendicular[ fig, 42, ]. Cor. 2. The perpendicular which bisects a chord passes (193) through the centre of the circle and bisects the arc. Cor. 3. The greater chord is less distant from the centre, (194) since the semichord y increases as x diminishes; and the diameter is, therefore, the greatest straight line that can be drawn in acircle. By comparing equations (189), (190), (191), with the figures (42), (42,), (42;), which illustrate them, it is obvious that the co- ordinates y, x, represent, in general, totally different quantities in these different forms, so that one cannot be combined with another, as in ordinary algebraical operations. But if the circles intersect, as in figure 42,, then the point of intersection, I, being the same for both circumferences, the codrdinates of this point, y;, 2, will satisfy the equations of both circles, and from (191) and (190) de- noting the different radii by 7,72, we have ¥ +2P=7’, y?+(%—a) =r; whence 2azr;,— a* = r* — 72. Resolving the last equation in reference to a, we find @=2,4(2°+ 7,7 — ry? whence, for any compatible values of z;, r., r, we find two values for the distance of the centres, a; or, for any given radii and a giv- en chord, the circles can be made to intersect in two ways. Ifr,, the values of a will be, cc other minus ; and Fig. 427. if : Qt 22) Fe we take — 7; instead of +2, @ = 27,4 (277+7.2—r?)?, 4 + becomes a = — x, + (22+r,2—r?)*, or —a=2, (22 +7 —1?)? ; whence we have the same constructions over again, only on the op- posite side of the origin. 2 t ps2. pe, 2 If we substitute the value of x = oar in y?+27%=r?, aot 2 2\ 2 t a + pile (dibs ids ] we find y; . ( a whence it follows that the circumferences intersect in two points, I, I', equally distant from the middle, B, of their common chord ; and .*, that, as long as y, has a real value, the triangle OI, must be possible. From what has been demonstrated, we have the follow- ing corollaries: Cor. 4. Two circles of given radii will have for a com- (195) mon chord four positions of intersection. Fig. 42s. Cor. 5. The line joining the centres of intersecting cir- (196) cles, is perpendicular to, and bisects their common chords, Fig. 429. Cor. 6. In order that two circles intersect, of the three (197) quantities, their radii and the distance of their centres, any two must be greater than the third. Cor. 7. The distance of the centres of two tangent cir- (198) 112 EXERCISES. cles, is equal to the sum or difference of their radii, according as they touch externally or internally. Fig. 4210. Cor. 8. The line joining the centres of tangent circles, (199) passes through the point of tangency, and the converse, [ fig. 42,..] EXERCISES. 1°. What is the greatest triangle that can be inscribed in a semi- circle? Ans. An isosceles triangle, standing upon the diameter. 2°. What is the greatest rectangle that can be inscribed in a given circle ? Ans, A square, 3°, What is the maximum triangle that can be inscribed in a given circle and standing on its radius? Ans. A triangle right angled at the centre, 4°, How many boards 15 inches wide and an inch thick can be cut from a log 20 inches in diameter ? Ans. 5 ./7, 5°. What must be the diameter of a water wheel to fit an apron 2a feet across and b feet deep? 6°. How much must a plank be cut out to makea felly 1 ft. 6 in. long to a wheel 6 ft. in diameter, the measures being taken on the inside ? 7°. What is the radius of the largest circle that can be cut from a triangular plate of silver, measuring 23, 3, 3‘5 inches on its sides ? 8°. Three brothers, residing at the several distances of 10, 11, 12 chains from each other, are to dig a well which shall be equally distant from them all. What must be that distance ? 9°, The diameters of the fore and hind wheels of a carriage are 4 and 5 feet, and the distance of their centres 6 feet. At what point will a line joining these centres intersect the ground, sup- posed to be a plane? 10°, There are two wheels situated in the same vertical plane, and their centres in the same vertical line ; the largest, the centre of which is 10 feet below the floor, is 8 feet in diameter, and the smaller, the centre of which is 6 feet above the floor, is one foot in diameter. Where must we cut through the floor for the passage of the strap that is to embrace the wheels ? 11°. The same as in the above, except that the strap is to cross between the wheels. EXERCISES, 113 12°, The same conditions as in 10°, except that the centre of the lower wheel is 4 feet from the plumb line dropped from the centre of the wheel above the floor. 13°. It is required to construct three equal friction wheels to run tangent to each other and to an axle two inches in diameter. What must be their common radius, and what the radius of the circular bed cut for them in the centre of a wheel? 14°, If the top-masts of two ships, having elevations of 90 and 100 feet above the level of the sea, are seen from each other at the distance of 25‘7 miles, what is the diameter of the earth ? 15°. How far can the Peak of Teneriffe be seen at sea? 16°. How far will a water level fall away from a horizontal line, sighted at one end in a distance of one mile, the diameter of the earth being estimated at 7,960 miles ? 17°.* If AC, one of the sides of an equilateral] triangle ABC, be produced to E, so that CE shall be equal to AC; andif EB be drawn and produced till it meets in D, a line drawn from A at right angles to AC; then DB will be equal to the radius of the circle described about the triangle. 18°. If an angle B of any triangle ABC, be bisected by the straight line BD, which also cuts the side AC in D, and if from the centre A with the radius AD, a circle be described, cutting BD or BD produced in E; then BE : BD :: AD: CD. 19°. Let ABC be a triangle right angled at B; from A draw AD parallel to BC, and meeting in D, a line drawn from B at right angles to AC; about the triangle ADC describe a circle, and let E be the point in which its circumference cuts the line AB or AB pro- duced; then AD, AB, BC, AE, are in continued proportion. 20°. Let ABC be a circle, whose diameter is AB; and from D any point in AB produced, draw DC touching the cirele in C, and DEF any line cutting it in E and F; again, draw from C a perpen- dicular to AB, cutting EF in H; then, ED*: CD? :: EH: FH. 21°. Let ABC bea circle, and from D, a point without it, let three straight lines be drawn in the following manner: DA touch- ing the circle in A, DBC cutting it in B and C, and DEF cutting it in E and F; bisect the chord BC in H, draw AH, and produce it till it meets the circumference in K; draw also KE and KF cut- ting BC in Gand L. The lines HG and HL are equal. * “ Prize Problems,” Yale College, 1840 8 SECTION SECOND. The Ellipse. Def. 1. An ellipse is a plane curve described by the intersection of two radii, varying in such manner as to preserve in sum the same constant quantity, while they revolve about two fixed points as centres, PROPOSITION I. To find the Equation of the Ellipse. Let P be any point of the curve, formed by the intersection of the vari- able radii a+ u, a — u, which are equal ' in sum to a constant quantity, 2a, and which revolve about two fixed points, Fig. 43. F, F’, distant from each other by the line 2c. Take the middle point O between F, F’, for the origin of rectangular codrdinates, _ the line passing through F, F’, being the axis of X. There results, (a+ u)? =y? + (¢+2)? and (a—u)? =? + (c—2)*5 [+h @+ur=ytor+z%, and, [—], au=cz, c or t= te Ee a 2 a+ — (Pat w=yr +e? + 2, or a?a? + c*z? = a*®y*? + a?c? + az? ; ri a*y? + (a? —c*)xz? = a’(a’ — c*), (200) the oanaten of the ellipse. Now, it is obvious that, if the curve cuts the axis of 2, y for that point will be reduced to nothing ; therefore, if we make y = 0, and denote by z,_, what x becomes for this value of y, we have (200) a . 0? + (a? — Oe a (a? — c’) ‘ Ty) = =a; hence AXES, 115 Cor. 1. The ellipse cuts the axis of abscissas at equal (201) distances on the right and left of the origin, which x°' x i é =a. en SAS distane a Fig. 430. When z =0, the curve cuts the axis of y, but this condition gives (200) ays _g + (a — c%) +» OF = aa? — ) 5 y, p= (C—&)*; hence, Cor. 2. The ellipse cuts the axis of ordinates at equal (202) distances above and below the origin, which distance, de- jy noted by Yreso b, = (a? —c*)z. Def. 2. The line MON = 2a, is denominated the Major Avzis or the Conjugate Diameter, and passes through the Foci,* F, F’; the line POQ = 20, is the Minor Axis or the Trans- verse Diameter, being perpendicular to the for- mer, [6 can never > than a.] If in (200) we substitute 5? for (a? — cc”), the equation of the el- lipse becomes Q Fig. 434. a’y? + b’2* = a’b*, or == |y (203) Ye r a where the constants are the semimajor ae iid axes, Cor. 3. The ellipse is symmetrical in reference to both (204) axes ; since, For every value of z, whether + or — % ae , 2 we have y = + (b? — ans 2?)2, two equal values Mh il Fig. 435. of y, one +, the other —; and, for every value of y, whether + y or — y, a? 1 we have tr =+ (a — oe y’)*, two equal values of xz, one +, the other —; hence, Fig. 436. Cor. 4. The major axis bisects all chords parallel to the (205) minor, and the minor axis bisects all chords parallel to the major. Cor. 5. The origin bisects all chords drawn through it, (206) + Foci, plural of focus, fire-place. 116 DIAMETERS, and is, consequently, the Centre of the Ellipse ; therefore these chords are Diameters. sa) Fig. 437. Cor, 6. Diameters, equally inclined to the major axis, (207) are equal; and the converse. Cor. 7. Any ordinate of the ellipse is to the correspond- (208) ing ordinate of the circle, described on the major axis, as the semi- minor axis is to the semimajor. For let y, Y, be corresponding ordinates of the ellipse and circle; we have (203) b2 y= (a — 2%) and (2) Y?=a*—~z’; a es ae ee ee Fig. 43. Cor. 8. The circle, described on the major axis, circum- (209) seribes the ellipse. Hence, Cor. 9. The angle embraced by chords, drawn from any (210) point of the ellipse to the extremities of the major axis, is obtuse. EN [How 2] Fig. 439. Cor. 10. Any abscissa of the ellipse is tothe correspond- (211) ing abscissa of the circle described on the minor axis, as_ the semi- major axis to the semiminor. For, we have 2 r= as —y*), and X? = b?—y?; : Tis um ee Fig. 4310. Cor. 11. The circle described on the minor axis is in- (212) scribed in the ellipse. Cor. 12. The angle embraced by chords drawn fromany (213) point of the ellipse to the extremities of the minor axis, is acute. Def. 3. The double ordinate drawn through the focus, is denom- inated the Parameter of the major axis, and sometimes the Latus Rectum. To find the parameter we have only to make x =c in (200) ; whence there results PARAMETER AND ECCENTRICITY. 117 a*y? __. + (a? — c*)c? = a*(a? — c?), a*—c* .2b% 4b? (2b)? oo a 2a) hte or 2a : 26 :: 2b: Parameter ; i. e. Cor. 13. The Parameter is a third proportional to the (214) major and minor axes. Putting the parameter = p, and substituting in (203) we get Parameter = 2y,.,= 2 y? = =e (a? — x2), (215) for the equation of the ellipse, in terms of the parameter and semi- major axis. We may transform (200) into a*y? = (a* — c*) (a? — x), 2 or y? = (1 = 5) (a* — 7”), or y* = (1 — e*) (a? — 2”), (216) putting e= < ‘ (217) Def. 4. We call e the Eccentricity because it expresses the ratio of the distance, c, of the focus from the centre to the semi- major axis, and thus determines the form of the ellipse, as round or flat. When the eccentricity is = 0, the ellipse becomes a circle. Equation (216) is that of the ellipse, referred to its eccentricity and semimajor axis, and is convenient in astronomy. It is sometimes desirable to have the equation of the ellipse, when the left hand extremity of the major axis is made the origin of abscissas. In order to this, we have only to substitute z—a instead of - zx in (203), as the new z exceeds the old by a, y remaining the same ; which done, there results, he ie 2 y= 2 (2ax — x”), (218) The origin might be transported to any other point, either in the curve or elsewhere, by changing the value of y as well as that of 2. 3 Scholium. It is easy to show that any equation referred to rect- angular coérdinates, and of the form py + 92? = 7; is the equation of an ellipse; for we have 118 TANGENT TO ELLIPSE. y? zr? a 1 eo ery ee ae which will agree with (203), by putting b=, a? = oy Od q er r r buneniting b—./ (=) eceal a PROPOSITION II. A tangent to the Ellipse makes equal angles with the (219) lines joining the foci and the point of tangency. Let P be any point of the ellipse, through which the line P’PR is drawn so as to make the angles ? FPR, F'PP’, equal; then will P’PR be tangent at the point P. For, producing F’P to Q, and making #* Fig. 44 PQ =PF, we have FP'+P'Q> FP+PQ=FP+PF. Now, if P’PR be not a tangent, let the second point, in which it cuts the curve, be P’, which we are at liberty to suppose, since P’ may be any point of P’PR; then the definition of the ellipse gives F’'P’ + PF = F’'P+ PF, which is less than F’P’+ P’Q; “ PF QPR=F'PP, which is contrary to the hypothesis ; hence, so long as the 7 FPR = FPP’, the line P’PR cannot cut the ellipse, and is tangent to it. Q. E. D. Scholium. It is on this account that F, F’, are called the Foci of the ellipse ;_ since, from the principle of light and heat, that the angle of reflection is equal to the angle of incidence, if the curve were a polished metallic hoop, and a flame placed at F, the rays, reflected from all points of the ellipse, would pass through F’. Cor. 1. The tangents drawn through opposite extremities (220) of any diameter, are parallel. L-> [See (206), (99). ] IN <—Z Fig. 442. NORMAL TO ELLIPSE. "Spe Cor. 2. A parallelogram circumscribing an ellipse will be (221) formed by drawing tangents through the opposite extremities of any two diameters. Cor. 3. The tangents drawn through the extremities of (222) the axes are at right angles to them, and the circum- scribing figure becomes a rectangle. Cl, Fig. 443. Cor. 4. The Normal, or line drawn through the point (223) of tangency perpendicular to the tangent, bisects the an- gle embraced by the lines drawn from the point of tangency to the foci. Fig. 444. Cor.5, The axes are normal to the tangents drawn through (224) their extremities, PROPOSITION III. To find where the normal intersects the axis of abscissas. Let TX, =7 be the normal, intersecting the, axis T of X in X,; from (142) we have the proportion dt EF’ xX, F = " TF od C—f, bt , putting OX, = 2, Fig. 45. Sy PE es ty Ge Qu Cc =—_,.. Z,=—-«4u, in which, substituting the oo ea. a : f 6 value of u, = Sa x, found under Prop. I., we have a c? F Xe, ha ake the point required. (225) e We observe that when c becomes = 0, z, [= iat x | a becomes = 0, and .*. the normal passes through the centre, O, as it ought to do, since the ellipse becomes then a circle described with the radius a. We have Subnormal = XX, = OX — OX, =2r—-7, (226) c? a* — c? b? T gah oie Lema 2 es z t » es We have (129), Fie. 459. ° 120 EXERCISES, Subtangent[=XX,] Ordinate [TX] _, Ordinate [TX] — Subnormal [XX,] ’ (Ordinate)? TX? Subtangent = Subnormal XX, Se 2 y? 5 (ae a? — 72 We observe here that the subtangent is independent of the value of 6b; therefore, Cor. If upon the same major axis, any number of ellipses (228) of different breadths, and also a circle be described, then their tangents, drawn to the same abscissa will intersect the axis of z in the same point. Fig. 453. EXERCISES. 1°. Prove that if two points of a straight line glide along two other lines intersecting at right angles, any third point of the first line will describe an ellipse. 2°. The equation of an ellipse referred to rectangular codér- dinates is 9y? + 42? = 36. Required the distance from the origin to the point in which the normal cuts the axis of z, the abscissa of the point of tangency being z,=1. 3°. Where does the normal in 2° cut the axis of y ? 4°, Required the subtangent in 2°. 5°. Required the length of tangent in 2° intercepted by the axes of codérdinates. 6°. How far distant from the centre are the foci in 2°? 7°. What is the eccentricity in 2° ? 8°. Required the parameter in 2°, 9°. It is required in 2° to transfer the origin to a point in the ellipse, the abscissa of which shall be z = 1°2, the new axes being parallel to the old. This will be done by substituting for y and z, y+ y and z+ 2, = x-+ 12, and observing that 9y,? + 42,? = 36, or 9y,? + 4 + 12? = 36. 10°. Given 9y? — 907+ 42? + 56x + 385 = 0, the equation of a HYPERBOLA, 121 curve referred to rectangular codrdinates ; it is required to ascertain whether the curve be an ellipse. Substitute for y, y+ y, and for 7, 7+ 2z,; find the coéfficients of the first power of y and the first power of x in the new equation, and put these coéfficients separately = 0, from which deduce the values of y,, 2, The resulting equation will be found to be 9y? + 4x? = 36. 11°. According to Sir John F. W. Herschel, the equatorial di- ameter of the earth is 7925‘648 miles, and its polar diameter 7899170 miles. The situation of a place in north latitude is such that a perpendicular dropped upon the earth’s axis will intersect it at the distance of 2456 miles from the centre. Required the point in which the direction of a plumb line suspended at the place, will cut the axis of the earth; the meridian being regarded as an ellipse and the plumb as perpendicular to the surface of still water. SECTION THIRD. The Hyperbola. Definition. A Hyperbola is a plane curve described by the inter- section of two radii varying in such manner as to preserve in dif- ference the same constant quantity, while they revolve about two fixed points as centres. Ordering all things as for the equation of the el- x. lipse, except that the radii are to be denoted by w+ a, +5 ae u—d, we find Fig. 46 a*y? — b?x*? = — a*b’, (229) for the equation of the hyperbola, The properties of the hyperbola are obviously analogous to those of the ellipse. The student will exercise himself in ascertaining them. It has been remarked that the equation of the ellipse, a*y? ++ b?z? = a*b*, becomes that of the circle, y? + 2? = a®, by making b =a; so the equation of the hyperbola becomes y? — z* = — a’, by put- ting b =a, an equation much resembling that of the circle ; hence, the curve which it represents, the Equilateral Hyperbola, pos- sesses properties analogous to those of the circle. SECTION FOURTH. The Parabola. Definition, The Parabola is a plane curve, such that any one of its points is equally distant from a fixed point and a line given in position, which line is denominated the Directriz. PROPOSITION I. To find the Equation of the Parabola. Take the directrix, D, for the axis of y, and for P the axis of x the perpendicular to D, drawn “i through the fixed point, F, whose distance from D yy we will indicate by p. Then, by the definition, Fig. 47. P representing any point of the curve, we have PAE ey es we y? = 2px — p’, (230) is an equation of the parabola. If in this equation we make y=0 for the purpose of finding where the curve cuts the axis of z, there results 4 ar \GPE oe gat Fig. 479. Cor. 1. The Parabola cuts the axis of x midway be- (231) tween the fixed point F and the directrix, In order to transport the origin to this point, we have only to substitute in (230) for 2, 7-+49; doing which, there results Fig. 473. | y? = 2pe, (232) the equation of the parabola. Here we observe that as z increases y increases, and that with- out limit, and for every value of z there results two equal values of y; also, z does not admit of any minus value, since in that case y, [ =(—2)2], would be imaginary ; hence, TANGENT TO PARABOLA. 1% Cor. 2. The parabola opens indefinitely to the right in (233) two symmetrical branches, but, unlike the Hyperbola, has no branch on the left of the origin. PROPOSITION II. To draw a tangent to the Parabola. Let PP, be any curve, cut by the FP line P,PX,, intersecting the axis of xz in X,; then, denoting the codrdi- nates of the point P by y, z, and of P, by y+hk, c+h, we have Xs Xe My x Fig. 48. y ae subsecant XX, h’ et) Now it is obvious, that, if the point P, be made to approach the point P until the two coincide, the line P,PX, will cease to be a secant, and consequently become a tangent at the point = To effect this we have ee to make h diminish till h = 0,and .. k=0; indicating what the ratio © be- comes under this condition, by in- cluding = in brackets, we have y c- ay A subtangent X,X, _ & : (254) where the value fe is to be drawn from the equation of the curve. Applying this process to the parabola, we have y' = 2px — p’, and (ytky= ee +h) — (y+) —y° = 2p(z + h) — 2pz, or Qyk +k? = Wh; ae sales which substituted in (234,) tg es subtangent y’ hs 124 SUBTANGENT AND SUBNORMAL, 2 subtangent = : (235) But we have in all cases [ fig. 482] subnormal >. y y ~ subtangent ” (208) for the parabola we find 5, JRE subnormal X,X, = awe # =p (237) P Whence we have Y FX, = X,X,—X%,F=p—(p—2)=2; 2 Ey but FP =7z, .°. FX,= FP, 7 Exe = FPR: xX F x, therefore, if we draw a line PX, parallel Fig. 483. to the axis of x, the angle X,PX,= X,PF. It follows that all rays of light or heat, or of sound, parallel to the axis of the parabola, will be collected in F.. Hence F is denominated the Focus of the parabola, Cor. 1. The points where the normal intersects the para- (238) bola and its axis, are equally distant from the focus, and the nor- mal is consequently equally inclined to the axis and to the radius vector, terminating in the same point of the curve ; Cor. 2. The tangent makes equal angles with the axis (239) and the line joining the point of tangency and the focus. In equation (232) the abscissa of the focus is 4p ; Parameter = 2y,_ 4, = 2p. (240) Scholium I. The method here employed for drawing a tangent to the parabola is obviously applicable to all curves ; and it is recom- mended to the student to make himself familiar with it by drawing tangents to the circle, ellipse and hyperbola, and to verify his re- sults by the properties already demonstrated in regard to these curves. Scholium LI, We must not pass unnoticed the remarkable symbol, by which we have readily arrived at important relations, Since in ea we have reduced both h and & to zero, it is natural to regard this expression as equivalent to 0° and this in itself, abstractly con- sidered, has no meaning at all, for to it we cannot attach any idea SCHOLIUM, 125 independent of its origin. But to regard the symbol H as des- titute of signification, or not indeed as possessing an important one, would be to attribute to it an altogether erroneous interpreta- tion. In truth, it not only indicates a quantity, but that quantity as evolved, by a peculiar operation, from specific conditions, The symbol [z] signifies, 1°. There are two quantities which are regarded as variable, y and xz. 2°. y is regarded as depending upon z. 3°. Increments [increases], k and h, are given to these vari- ables, y, z. 4°, The ratio, ie of the increment of y to that of x is found. 5°. That particular value, sel of this ratio is taken, which is obtained by diminishing h, and consequently k, to zero. It is also to be observed that the ratio, Ea will generally it- self be a variable quantity. Indeed, in this particular case of a tangent to the parabola, we have <1 pe _ , which may vary from _P _ — infinity to By Rab! Va = 0. In the next book we shall y= y = infinity give the symbol, [F<] , a name, and a further investigation. PROPOSITION III. If a curve be such that the distance of any point of (241) it toa point fixed in space shall bear a constant ratio, e, to the distance of the same point of the curve toa given line or Direc- trix,then will the curve be either the Ellipse, Hyperbola or Par- abola, according as the ratio, e, is less than, greater than, or equal to unity, [e $1]. 126 GENERAL EQUATION, - Let the fixed line or directrix be the axis of y y, and the perpendicular to it drawn through the fixed point F, the axis of z, and let the distance a \z of F from the origin be denoted by d; there re- +—~—*—~.—> sults, Fig. 49. y? + (d— zx)? = 2? = ex’, since i =e, by hypothesis ; y? + (1 — e?)x? — 2dz + d* =0, which becomes at once y” = 2dzr — d’, the equation of the parabola, when e = 1, or (1—e?)=0. In order to make the term 2dz, affected by the first power of x, disappear, we will transport the origin to the right a distance = m, so that we shall have r-+m instead of z, *. y? + (l—e?) (c«#+ m)? — 2d(r + m) 4+ d?=0, or y?+(1l—e’)z?+[(1 —e?) « 2m—2dJzr+4+(1 Fig. 499. — e*)m® — 2dm + d? =0, from which, attributing such a value to m as to make the term af- fected by the first power of x disappear, we have (1 — e?) «2m — 2d=90, and y? + (1—e?) x? + (1 —e?) m?—2dm+ d?=0; whence, eliminating m, there results e2 y’ + (L—e*) =, y" x e?d? a eed? I—e? (1—e*)? or 2 2 which is (1°) the equation of an ellipse _ + - =1, Ey had eo according as 1—e? is +, or —, thatis,e <1, ore > 1. Q.E. D. Scholium. It is to be observed that the Ellipse, Circle, Hyper- bola, and Parabola may be represented by the same general equa- tion, and are therefore to be regarded as nothing more than species of the same curve. 2 or (2°) the equation of a hyperbola, eet EXERCISES. 127 EXERCISES, 1°. From the top of a tower 48 feet high, a cannon ball is fired in a horizontal direction with a velocity of 1000 feet per second. Required the distance from the foot of the tower where the ball will strike the horizontal plane on which it stands ; no allowance being made for atmospheric resistance, and the vertical descent being in the times 1, 2, 3, é&c., seconds, 1? + (16-4), 2? « (1674), 3? + (1674), &c., feet. 2°. To transport the origin to a point of the Parabola, the new axes being parallel to the old. . Let b, a, be the codrdinates of the new origin re- ferred to the old axes ; we have | (y +5)? = 2p(z + a); but b? = 2pa; y? + Why = pz, is the equation required. Fis: 50. 3°, To Spee whether a board cut from a log next the roots without having been squared, may be regarded as an inverted par- abola. Let the middle line of the board be taken for the Y axis of z and the broader end for the axis of y; the preceding problem gives us (c — 9)? + 20(c — y) = pz, where there are three constants, b, c, p, to be de- ~ termined, one of which, c= half the width of the broader end, may be supposed known. We must therefore deter- mine the values of 5 and p from values of z and y taken in two dif- ferent places, and then see if 5 and p remain the same, or nearly the same, for measures taken throughout the length of the beard. 4°, The length of a board, of the form given in 3°, is 8 feet, the ends are 4 and 2¢4 ft. broad, and the breadth of the middle is 3 ft. Required the equation of its edge, the axes of coérdinates being as in the last, Ans, (2—y)?-+ ‘7(2—y) = ‘152, or y? — 4°7y = ‘1dr — 544. 5°. It is required to form a gauge by which to turn a parabolic mirror 18 inches in diameter, and having a focal distance of 10 inches, measuring from the diameter. Required the depth of the mirror and its equation. Ans. Depth = 1‘7268. Equation, y? = 469072 « z. Fig. 51. 128 EXERCISES. 6°. Fold the lower corner of the left hand page, so as to make the area of the folded part constantly equal to a given square (a*) the side of which is a; and find the locus of the corner, or the curve in which it is constantly situated. If we take the position of the corner before folding for the origin of codrdinates and the two edges of the leaf for the axes, we shall find (y? + 2%)? =2 (2a)*yz, for the equation sought; from which it appears that the curve is symmetrical in reference to the axes or edges of the leaf, and that it begins and ends at the origin, since z =O gives y =O. oe sa wiht Yi) Wy ay me 3, |e alls hl Be ANALYTICAL FUNCTIONS, PLANE TRIGONOMETRY, AND SURVEYING. mes ee ie ‘Sesh mae commen as tho? \s ery: | enna te x a ie £ igh Ebadi dl Ste a ghvem wigdaie F ae | ie bir kalig eh pi ap ‘2 fitted ia dy ey ‘he, ial ae. : Bao, sharacgeh bi % oun estab Ct Vat ‘ 2. oe Flee ae fr onesona a serve sive | bapen Sion K san. 0. hes can, hia at Bin | sve he where, HE, caver bing jesiadan 1 dct skh ie enh aa se ey Pop. ih ai . . ie we The PST bes Foam os tlie oA Bay att erica ete 1 aR anoose THAT | at Muon avets anrron’ saemretarh FA eine suena awa... rk, a | | semen A cerns ceeatiaats Ne nee sey cere a Bar Te eT eel sgn BOOK FIRST. ANALYTICAL FUNCTIONS. SECTION FIRST. Primitive and Derivative Functions of the Form a Definition 1. When one quantity depends upon another, so that the first varies constantly by a continuous change of value in the second, the first is said to be a Function of the second. The first is also called the function and the second the independent variable. Thus, in the equation of the parabola, y? =2pz, y is a function of x, or y is the function and z the variable. Def. 2. An increasing function is one which increases when the variable increases, and a decreasing function is such that it de- creases when an increment is given to the independent variable, Thus, in the equation y*? =2pz, y is an increasing function of z, . , ‘ bY. 2 but in the equation of the ellipse, y? = a (a? —x*), y is a decreas- ing function. Def. 3. An explicit function is one in which the value of the function is developed or expressed; an implicit function is one in which the value of the function is implied. Thus y = (2p)? . 2, is an explicit function; and y is an implicit function of z, or x an implicit function of y, in the equation a*y* + 6?2? = ab’, As it is useful to denote quantities and their combinations by let- ters and signs, so it will be found advantageous to indicate differ- ent functions by appropriate symbols. The letters f, F; fi, Fi, &c.; f', F’, &c., as well as others derived from the Greek alpha- bet, such as ¢ (phi), ¥ (psi), are employed for this purpose. Thus 132 DERIVATIVE. fz signifies, not that f is multiplied into z, but that some operation is to be performed upon z, such as squaring it, taking its square root, &c., in which case we have fx = 2’, or fr = ./z, &e.; fx will signify a different function or operation from f, z, and the distinc- tion in reading will be, the f function of z, the f sub one function of x. Further, the small letters may be used to express known re- lations, while the capitals denote unknown functions. The letter h will always be employed to denote the Increment or increase of the independent variable z, while & will as constantly indicate the consequent increment of the function y. Def. 4. The Derivative of a function, or the Derived Function, or simply the DerivarTive, is the function obtained by taking that particular value of the ratio of the increment of the dependent variable to that of the independent when the latter reduces to zero, Thus, if pee foes, then* y+k=(e-+h)'=2"+2ch- +H wae k=22rh+h?; k ae =2r+h 9 and .°. be | = 2z, the value of the ratio a whenh=Oand ..k=9, « indicated by the brackets [ ], is the derivative of the function 2°. And, as y is put for the function 2’, so it is natural to indicate the derived function 2z by y’, which will therefore be read the deriva- tive of y —, and we shall write y-[E]-m The letter f’ will be employed for the same purpose, and f’z will be read the derived function of z, As a further illustration, let us take the function y = fr =azr>+b, then* y+ =f(2-+h) = a(a+h)? + b = a(x? + 3827h + Brh?+h')+. k=f(x+h) —fe = a(32*h + 3rh?-+ A’), k rth)—fxr and - uiticid we = a(32z*°+ 32h -+ h?) ; * Increasing x by 4, and, consequently, y by &, or changing x into x-++-A and y into y +k, or, y depending on x by a constant law, y+ is the same function of x-++-h that y is of x. DERIVATIVE, 133 yay Ea = pao =ae Sz’. It is obvious that y’, f’z, Ea : ey , are but different expressions for the same thing. Observe further that y’, being equal to a « 32’, is also a function of x and its derivative may be found ; so that from y =f = an dz", we have Yah 2A aeZe We, which is the derivative of y’, or the derivative of the derivative of y, or simply the second derivative of y. In like manner the third derivative of y would be indicated by y'” or fx, the fourth by y” or f*, &c. PROPOSITION I. To find the derivative of any function capable of being devel- oped in integral additive powers of the variable. Let the function be denoted by y=fr=aqtazr'+a zr? +az7?4+..ta2°+... (242) Increasing z by h and the function by k, we have ytk=f(x +h) =a,4+a,(c+h)!+a,(r+h)? +... + a,(c+h)"+ o05 . k=f(e+h)—fe=af(et+h)'—2z']+af[(x+h)?—27]+... +a.[(z7+h)*—2z"]+..., a. Peal oo ae Ae ae a ee q h dif AER irk ay (f+ h)isaeh oh. but. aaa OT eg ee erat ag tnt b Seuvn-at 2p 2 3 ——- bast slp B32? + (3r-Lhyh, ta de Siar will be found=4z*-+ PA, where P depends upon z and h; and we are led to infer that Ge = ng Xh, 134 DERIVATIVE. where X is such a function of xz and h that Xh shall disappear when h is diminished to zero. But to prove this let us put xrth=a};x=b, and ..h=a—); Ra have (e+h)'—2'*a—b' (cx—h)’?—2z’ | a’—5? oe Lem a—b’ h a—b’ (x+h)?— x8 Pahari" ee (x +h)" ~ 2" _ a" —~ be h Oe Bi h a—b’ so that we fall upon the examples under (16), and we have only to solve the problem whether a" — 0” is divisible by a—b. In order to this we execute the multiplication, (¢— b) (a"-' + a? b+ a" b2 + ar 53 degree £ ab 5s), and we find the product to be a” +a" b+ a” b? + a®$b3+ a? bt +... + a?b"? + ab —a"*b—a?* b? a 63 — a" Bt — .., —a*b"* — ab" — 3’, which is equal to a* — 8"; a” — b” air es = qr + ah fe a’ $2 +...+a7b"?+ab""+6"", (243) Hence, the difference of the same powers of any two quantities is divisible by the difference of the quantities themselves, and the quotient is homogeneous and one degree lower than the power, the leading quantity descending one degree each term and the follow- ing ascending by the same law. If in (243) we now replace a and b by their values z +h and 2g, we find } 2 A man nie ar al 2 <= (2 hy (wth) te + (ah) ea? +... 2; whence the fourth equation becomes k z+h) — FEE «a, 4a, (Qe +h) +05 (80?-+8zh-+ M) +o +a, [(a + hy + (x + hye + (2 + hye? 4... + 2]; therefore, making h = 0 and observing that there are n terms in the expression (x + h)""'+..., which all reduce to x”, we get y =f'x = ka = Peto] 2 at a8 Qa-ba, © Bx® (244) +a,+ 47+ ,..+a,+nz"'; which is the derivative sought. We perceive that (244) is derived from (242) by multiplying by the exponents of z in the several terms and decreasing them by DERIVATIVE OF Az’. 135 unity ; and hence that the term a, independent of z, or, which is the same thing, the term a, + 2°, disappears in (244) ; so that, if we were to pass from (244) back to (242), it would be necessary to introduce a term independent of z, or to add a constant quanti- ty, which may be represented by C. PROPOSITION II. To find the Derivative of any real power of a variable. Let y = fr = Ar’ be the function, a being any whatever, whole or fractional, plus or minus, but not imaginary, not of the form a=-/—c. Suppose, in the first place, that a is fractional and additive, ™m or Ce ee then y= fr= Aa" = A(x") . Now, if we substitute z for 27, we get y =f.z = Az”, and z=fyr =a". Suppose then that, while z is increased by h, z receives the in- crement 7, and, as a consequence, that y is augmented by k; we have y+k= A(z-+ 7)”, and z-+i=(c+h), or (z+7)"=zr+h; w k=A(z+i)™— Az”, and (z+ 7)"— 27=h; k.. (z-+%)"— 2 i z x 1 B) gem [ ase aaaP ES NG leneeaact (z+ 7)" = 2"’ a haley RM, ” Lacie mdi 1 ; ; hi ‘4 i (z+i)"—2"’ | t k k al wey he Ee eee . (E)- EEE] earnest ghee Be eA (Se) we rz", n n ™m =i Or Yynse= Yong 29 Miape= Ave @ * y’y-fe, read, the derivative of y, y being a function of z. 136 DERIVATIVE OF Az’. Here it is worthy of remark that we have found the derivative of y a function of x, by taking the product of the derivative of y a function of z, and that of z a function of x We observe the same rule, then, for the derivative, whether the exponent be frac- tional or integral, we multiply by the exponent, and diminish it by unity. Let a be now taken subtractive, and either integral or fractional, or let a=—T, » where 7 is either a whole number or a fraction ; we have y= Ar*= Ar; * Ytk= A(zr+h), and k= A(x+h)"— Az”; k as (z+ h)7 — Ghat er (c+h)7—a27 (e¢#+hyr’ h h h "(et hyn ace (2- hye" — a2 -+- hy iy x2’ —(x#+hy (2 +h)'2’h fe (c+ h)a2"h (cx +h)— 2” h A EE k r—1 Bi Sh | =| =_— A e ae =A e —rz¢ 7, the derivative sought. Hence, RULE. To find the Derivative of a variable, affected by any (245) exponent whatever that is not imaginary; multiply the variable by its exponent, and decrease that exponent by unity. Thus, the derivative of z? is 2x”, of x* is 32°, of x is 4z°; of —1 —4+ Be Pe 2 hd 3 3 ° 4 zi isfo. = f2,°=F+—, of g istsa” =fex", of a is x3 1 : 210g Tee Os of Az* is Aar*"; of ris] egiIt=1-. 7° =i Le 1: uf Ape ae) Lie As De 4 Oe & DEVELOPED FUNCTIONS. 137 PROPOSITION III. The Derivative of any function, capable of being devel- (246) oped in real powers of the variable, will be found by multiply- ing in each term by the exponent, and decreasing it one. ConverRsELy: if a derived function be developed in powers of the variable, we shall return to the original function, by increas- ing the exponents by unity, and dividing by the exponents thus augmented, taking care to add a constant. Thus, let y be such a function of z, or depend upon x in such way, that we have y =fe = Ar*+ Ba’+ Czr°+..., (247) where A, a; B,b; C,c; &c., may be any real quantities, + or —, whole or fractional, we have (245) y =f'r= [+] = Adar + Bo 4Coa +... (248) We observe that if a= 0, or there be a constant term, Az’ =A, in (247), it will disappear in (248), since Aaz*" then becomes 1 A+0+—=0; therefore, in passing back from (248) to (247) we must add aconstant for that which may have disappeared. We shall see how this constant will be determined in any particular case by the nature of the problem, To illustrate, suppose we have found the derivative y = 3x — f05 +7, returning to the function (246) we obtain y =8r? —Z- bys + 7x + constant. Whenever, then, we can find the derived function developed in powers of the variable, there will be no difficulty in ascertaining the primitive function. PROPOSITION IV. The Derivatives of equal functions, depending upon (249) the same variable, are themselves equal. This is manifest from the nature of the operation; thus, if we have any two functions of zx, such that Ex = $2; 138 © SCHOLIUM. whatever may be the value of z, then are the functions equal when zx is increased by h, or when z becomes x -+-h, and we have F(x+h)=f(x+h), from which subtracting the first equation, there results F(z +h) — Fr =f(z +h) —fa, which, divided by h, gives F(x+h)— Fr _f(x+h)—fe. h r h ; and this equality, having been obtained independently of any sup- position in regard to the magnitude of A, is true for all values of h, and therefore true when A becomes less than any assignable quan- tity, or when / is diminished to an equality with zero. Therefore, Relea aed mes de oa = [ Age ==) : h h or Se MS Be Thus, if A +A,e z'+A,-27?+..=aq+a,¢zr'+a,¢ r?+..., then shall we find A, +A, ¢2r7+...=a,+a,¢ 2r+.... We observe that if the last equation be multiplied by x, and sub- tracted from the preceding, we obtain A,— A, + 2? — A, « 27? — ... = d)— Gye ©? — Og e AT? —..., from which the first power of zx is eliminated ; and, in like manner, we might eliminate in succession the terms affected by x’, 7°, &c., to the last, if the series were finite, and to a term less than any as- signable quantity, if infinite and convergent, that is, if z were less than unity; hence we should obtain A, =a,, and, pursuing the same process, A, = a,, A, = a, and so on, which is in accordance with (65). Scholium I, The student will not embarrass himself with unne- cessary difficulty in reference to the symbol ’ [1 or its equivalent Sea ’ by supposing it to signify nothing more than is indicated by re in itself considered, It is true that the numerator as well as the de- ° ° k ° e . nominator in loge is =Q; but we are not to infer from this, either that the quotient, [=| ,1s equal to unity, or that one 0 can SCHOLIUM. 139 have a different value from another 0, in order to make the quotient, 0 @ ’ represent different quantities ; but that [+] is simply asym- BoL of a determinate operation, whereby we have derived one quantity from another. We are not, therefore to.separate the & from the A as if they were determinate quantities, and as such could be managed separately, nor even to remove the brackets ; the whole expression, [+]. let it be remembered, is one indivi- sible symbol, the sign of a process, it does not so much signify what the result is as how it is obtained. In any function, de- pending upon z, we give to x an increment h, we subtract the un- augmented function away, we divide by the increment h, and, in the quotient, eliminate 4 by reducing h to 0. Nor is this a singu- lar case; all algebra is but a system of symbols, a mathematical language, telling what is to be done, and showing how results are obtained. Thus, ab, a: b, a’, /a, are not simply the re- presentatives of quantities, but indicate the operations, multiplica- tion, division, involution, evolution, whereby certain quantities are obtained. . | Scholium II. It is obvious that the derived functions will differ from each other according as the primitive functions, from which they are derived, are different. SECTION SECOND. The Binomial, Logarithmic, Interpolating, and Exponential Theorems. PROPOSITION I. It is required to develope (a+ x)" in powers of x, n being any real quantity, plus or minus, integral or fractional. First, suppose n to be plus and integral, that is, any of the num- bers, 1, 2, 3,4, 5,.... It is manifest that the development of (a-+ x)" can contain no other than plus integral powers of x, and must contain these, since the multiplication of whole positive 140 BINOMIAL THEOREM. powers by whole positive powers gives whole positive powers ; thus, (a+ 2)? =(a+2)(a+27)=@?+2ar4+ 2’, (a+ 2)? =(a+ 2) (a?+42ar+ 2’), &e. ; hence, (a+ x)" must = A,+ A, + z'+ A, x*+ Azo 7+... +A,*2"+..., (a) - arranging x in ascending powers, and denoting any term by A, « 2”. In order to determine the coéfficients A,, A,, As, ..., A, of the several powers of z, it will be necessary to take the derivative of (a); therefore, to find the derivative of the first member (a+ 2)’, put y=(a+2)"=2", and .. z=at+z; ytk=(2+72), and zt+i=a+e2t+h; _@+ipne a t and =1; +. agri a on Yy=fe = tf |= E 4, [= | ant anata, whence the same rule holds for the derivative of (a-+ x)" as for 2’, substituting a+ 2 for z. The derivative of equation (a) gives us then (249) the equality n(a+ar)""= A,+ A, + 2r+ A, + 32?+ Aye 47? +... +A, * mz”""+ ...3 (6) Taking the derivative of (b), we have n(n —1)(a+ar)’?= A, + 2+14A, +36 27+ A, eo 40 Bz? +..+ A, + m(m —1)r"?-+...3 (c) the third derivative gives n(n —1)(n— 2) (a+ 2r)"%=A,+-3-2e-14+A, 2463 Bz +..+A, © m(m —1) (m—2)z"*+...5 (d) the fourth derivative will be found n(n —1) (n — 2) (n— 3) (@4+2)"%= A, + 4-3-2-1+4,,. + A,, « m(m — 1) (m— 2) (m—3)2""* +... 5 (e) therefore, observing the law of derivation in (a), (b), (c), (d), (e) -., and proceeding to the mth derivative, we obtain _ n(n — 1)(n—2) (n—8) (n—4) ... (n—m+1)(a+-2)"™ = A,, + m(m — 1) (m— 2) (m—3)...¢3eBelt+..., (1) BINOMIAL THEOREM. 141 From the last derivative we observe that, if 7 be a whole addi- tive number, m can never exceed n; for if m were =n-+1, 2—m-—+1 would be =0, and the (n-+1)th derivative, and con- sequently all after it, would disappear. his is as it should be, since it is obvious that the nth power of (a+ 2) cannot give any term in the development of a higher degree than n. As the coéfficients Ay, A,, Ag .., Any +, are independent of z, we may make z =0 in (a), (0), (c), (d), (e), -.., (2), ..., and we ob- tain a" = A,, na" = A,, n(n — 1)a** = A, oe, n(n —1)(n—2)a"* =A, + 3-21, n(n —1) (n — 2) (n—3)"4=A, + 4-3-2 1,... n(n — 1) (x — 2) (n — 38) (n —4) ... (2 —m- 1)a"™ = A ® m{m —1)(m — 2)... 3 e¢2el, Substituting the values of Ay, A,, A..., A,, drawn from these €quations in (a), we obtain Newrton’s Binom1aL THEOREM. {atzr)*=a"+na™" + x+n(n—I1)a"* ss (250) + n(n —1)(n—2)a"*e aoe aby + n(n—1) (n—2) (n—3) ... © (n—m-+1) a « sat : Since the same rules of addition, subtraction, multiplication, di- vision, and derivation are applicable, whether the exponents be plus or minus, integra] or fractional, we might infer that (250) would hold in all cases, whatever might be the character of n. But to prove it observe that if m be any real quantity [not imagin- ary |, there can be no other than real powers of z in the develop- é ment/such a mo, ily A, DiS Be egos vole Vike Uy, Blea. Therefore the form (a+2)"= Ayer’ pAyeat tif Ae oF tif A, ea Fi + A_,¢ + AL, oot. must at least be sufficiently general, Suppose now that the minus powers are arranged in a descending order, that is, that r>s, s>t..; .. multiplying by z*, we have 142 BINOMIAL THEOREM. (a-+eyea=(Ab Apert. t Ae eae.) a" Seaside Me of BBY hp eo where it is obvious that all the powers of z are plus; .*. making z = 0, there results 0=0+4 A_,+0+..., .. A_L.=0; and in the same way it may be shown that A_,=0, A_,=0,...; that is, the expansion contains no minus exponents, Rejecting the minus powers and taking the mth derivative, we have n(n —1)(n—2)...¢(n—m+1) (a+ 2)""= A, + m(m—1) (m—2)...e3e2e1+A,.,°(m+1)m(m—1)... 27+... $A S(4-1) (4-2) ..+(S—m4ijor 4... The order of this derivative may be any whatever, if nm be minus or fractional, since in this case the coéfficient n — m-+ 1 can never become = 0, m being a whole additive number; we can therefore take m so great that the exponent Ar m shall be minus, (+ —m can never = 0, m being = 1, 2, 3,...), But n(n —1)...° (w»—m-+1) (a+2)" cannot have a minus exponent in its development, for the same reason that (a+ 2)" has none. It follows that the development of (a+ x)” will not be affected by any fractional power of x, what- ever n may be; (250) is therefore true for all the real values of n, since for all such values the coéfficients of the expansion have been determined. It appears, however, that if n be either minus or frac- tional, the series will never terminate, As the form of (250) is independent of the value of a or x, we may change z into — x, doing which and observing that the terms affected by the odd powers of z will be minus, we obtain (a—z)"=a"—na""-r+n(n—1) Z n(n —1)(n=2) v3 eon We aa ne (251) If in (250) and (251) we make a=1 and x=1, and remember that all the powers of 1 are 1, we get Feng n(n—1) , n(x—1) (n—2) =ltat 5) iit e ae Daytime n(n—1) n(n—1)(n—2) and Valea en a cet ee ee BINOMIAL EXPANSIONS, 143 Thus it appears (1°) that the sum of the coéfficients of any bi- nomial power is equal to 2 raised to the same power, and (2°) that the sum of the coéfficients of any binomial power taken alternately plus and minus is equal to zero. This furnishes a convenient method of verifying an involution. For example, (a+z)'=1lea+lezr, 14+1=2=2', and+1—1=0; (a+z2)*= a? + 2ar+ 27, 14+241=4=2', and+1—2+41=0; (a+ 7)? =a?+3a*%r+ 3axr? + 2%, 14+-34341=8=2', and 1—3+3—1=0. . Expand (a+ z)*, (a+ 2)°, (a+ 2)§, (a+ 2)’, by aid of (250), and (a—z)', (a—z)5, (a—z)*, (a—z)’, by aid of (251), and verify. If in (250) and (251), we make n = MS , there results ox thawte cesar arith in altage 1h. gel) 2° 2Qr—1 38r—L P.O. RF. Zare rT? «23a * og 0 Pee 4 wl 1) (2r 1) Ors 1) Hb ft af Bile (252) 1 1 and (a—2r)7=ar—-—_{ — 5,5, -—- >a —s (2888) rar TeeQar Te W3e4ar Making r = 2, 3, in (252), (253), we get a a 2 3 (a-+ajtaah ++ —_* 4 Qa” 2? « Qa*, BeBe Ba? 3s 524 pe ein ey tg, (254) 2.2.3 4a’ a pit 2 3 (o— x) =< gag nae tah teh es (255) Qa* 22a" 2.2. Ba? 5 Gee Qr 2. 52° (a+2)? =a + —-— 4 a> 32. Qa® 3? «2+ 3a° 2.5 8rt nO seamen Xda Pe (256) 31-23. 4a° 144 BINOMIAL EXPANSIONS. ee we ar | (257) 3a*° 3?» 2a° In (254) "RB a=c*, we find ee 323 Q 7 ee ee (c +2)" =e+ 5 Bo 2. 2e8 + 33.29.30 3-524 ag Sct sn oat (258) from which, substituting z? for x, we obtain 4 OS Zz 326 2 2) 4. — — ——__—. ++ —_____ (oT 2) =O as oF. ac8 T BF. Bee 3° 52° TH. 2.8 cas ears ow) changing z into — z in (258), ct— zr ew ard ieee 80 Ee a ( ‘ 2c ° 2c? 02+ 3c? 3° 5x4 TRB. Sada ert (260) changing z into 2? L 2? 2 326 Do ee elet ean Mee opt eee ak BA as Tee er ne me 2B a Bee. « So BeBe et The student may operate like changes upon (256), and investi- gate analogous forms for higher roots, as the 4th, 5th, &c. These expansions may be employed for the extraction of the roots of numbers. Thus, let the square root of 101 be required. We have (258) 4101? = (10? 4.1)? =104+4 — sooo + teat Trev o000D = 10°0509381211. Find //102, 103, /104; ./99, ./98, /97, */1001, &c. We may put (250) under a new form, for dividing by a”, there results, (1+) =14n0-Z4n(n—1). _ Qa? + n(n —1) (n—2) +5 tans (262) or (1 + Heed Att Ws 2 4 ? x +..., putting v= cy (262) n(n —1)(n —2)+ 5-3 LOGARITHMS, 145 LOGARITHMS, Definition, Let a’ =z, (263) then is y denominated the Logarithm of x. The constant a is called the base of the system. PROPOSITION II. The Logarithm of a product, consisting of several fac- (264) tors, is equal to the sum of the Logarithms of those factors. For, let a= 2, ©, 15; + > Y=Yis Yor Y29 ove 3 then (263) ai = 2; 2 = Lay a3 = Do, &c., &c. Mirna@als Wie woiesati iar Yeti = 2) Morg elegy ’st 5 or® L(x, + 22° 23..)=(Yy:tyw+tys +...) = Lr,+ Lr, + Dr,+... Q. E. D. Cor. 1. The logarithm of the nth power of any number (265) is equal to n times the logarithm of the number itself. For, making z,= 7,=2;=..., we have L(z,+ 2, + 2, ¢...[n]) = Lr,+ Lr,+ Lr,+..., or, Lae \enLz,. (265) Cor. 2. The logarithm of the nth root of any number, (266) is equal to the nth part of the logarithm of the number itself. Cor. 3. The logarithm of a fraction is equal to the loga- (267) rithm of the numerator diminished by the logarithm of the denom- inator. For, if we divide a“ = x, by a2 = Loy zx we have ai~%=-'; Z2 Lr; or L "e. =Y,— Y= Lr, — Lr. 2 Scholium, We perceive that addition of logarithms corresponds to multiplication of numbers, subtraction to division, multiplication to involution, and division to evolution. We have, then, only to possess a “ Table of Logarithms,’’ calculated to a given base, * I, logarithm of. 10 146 LOGARITHMIC DEVELOPMENT. , say a = 10, in order to perform numerical operation with remark- able facility. ‘Thus, to obtain the cube root of 2, nothing more would be necessary than to take the logarithm of 2, divide by 3, and seek the corresponding number from the table. Again, if 5 then (265) L3 = L(2*) = zL2; Ey ae? AS whereby we are enabled to solve, with the utmost facility, a numer- ical equation in which the exponent is the unknown quantity. PROPOSITION III. A number being given, it is required to find a form by which we may calculate its logarithm. Since the relation a’ = x, gives y a function of z, let us endeavor to expand y in terms of xz. To this end we proceed to determine the derivative of y. Changing z into x+h and y into y+ 4, we have a’*F¥=er+h; subtracting Gok at, there results a ea'—a=h, or h=a"(a'—1); in which, substituting z for a’ and 1+-d for a, in order to subject a‘ to the influence of the Binomial Theorem, we have (262) 2 h=x[(1+0/'}—1J=2[1+hb+k(k—1) + = 5° 3.3 +..—1]; dividing & by each member of this equation, observing that 1,—1, cancel each other, and that & then becomes a factor common to the numerator and denominator of the second fraction, there results k 1 ¥ [b+ Ae a TS as i, sy tN R—2%) + a at from which, observing k =0 when h becomes = 0, we have ; of & 1 1 f= ]= pa Se, (0208) [s-sta—gtin [2 4+ k(k —1) (k—2) » LOGARITHMIC DEVELOPMENT, 147 putting M= 3 - : ; (269) 5 pa POMS MAin grad) a 1 Reece CST) CEs (a Lice 2 + 3 st) 4 +, ~~ 9 eve Applying the rule of (246), in order to return from the deriva- tive (268) to the primitive function, we fall upon the equation —1+-1 —1+1 from which z disappears, and nothing can be inferred—save that the logarithm of a number cannot be developed in terms of the number simply. But if we make y the logarithm of 1 + z instead of x, and repeat the above operation, from aw=l14+2, we obtain y =M(1+2z)"', or (262), y =M[14(-Io+(-I)(-1-)S4(-1)(-1-1) M y=M- + constant = my + constant, 3 = gt |= Mtl —z+4+2?—27'?+71—27'4+,-,...]; from which, returning to the function, we get y = M[z —42?+42? —42t + 427° —tr'+, —,...] + constant. In order to determine the constant, we observe that x and y van- ish together, since y=0 in a¥=1-+2, gives 14+ 2=a° =1, and .. £=0; therefore, substituting these corresponding values of y and z, we get 0=M-0- constant, .. constant =0, and L(+ 2)=y=M[e — 42° + 40°40! 4 42°20 +,—,...}, (270) a logarithmic series, in which the logarithm of any number 1+ 2 is expressed in terms of a number less by unity, z. The constant, M, depending upon the base, a, (269), is denomi- nated the Modulus of the system. Taking a different base, a, we obtain a new modulus, (-1-2). ip th | (ce) — Shy mes 148 MODULI COMPARED, and the logarithm of 1+ z, derived from dy =1-+ 2, becomes (270) L{1 + x)= M[ zr — 32° + 32° — Fat +, —, J; by which, dividing (270), there results L(l+2)_ L(1+ 2) =n} Cor. 1. The moduli in different systems are to each other (271) as the logarithms of any given number in those systems, If we make the modulus equal to unity, we have the logarithms employed by Lord Napier, a Scottish nobleman, who was the in- ventor of this admirable system of numbers; and, if we assume 10 for the base, we have the logarithms in common use, or those of Briggs. It is customary to indicate a Napierian logarithm by a . small 7, and a common logarithm by the abbreviation log. Adopt- ing this notation, and observing that 10”.=1-+ <2 gives 10'= 10, or log. 10 =1, we have (271) log. 10 M,_1 1 272) 10 © jap ee Oe We have, then, only to make the modulus one in (270), and thereby calculate the Napierian logarithm of 10, in order to deter- mine the modulus, M,_,,, of the common system; but, by a little artifice, we may convert this series into one more rapidly converg- ent, and therefore better adapted to our purpose. In (270) changing z into —z, we have L(1 — z)= M(— z— 42? —47r?’ —...), which, subtracted from (270), gives Li+2)— L1 —z) =2M(2+42?+40'4+...), (278) or (267), L(; +5) = QM(x +409 442° +427 +...).. (274) And this series will be expressed more conveniently by putting, as Borda has done, 1t+r m —n Rs » and .° ; whence Lae tn en m+n (2) =m ( (52) +22) H(SSa) +=]. em Making M=1, m=2, n=1, in order to calculate the Napierian logarithm of 2, we find U4) =22=2[4 +3(4)? +4(4)? +]; ; hence, COMPUTATION OF LOGARITHMS. 149 where we have only to divide successively by 9[= 3*] and the odd numbers 3, 5, 7, ..., as follows : 3/2+00000000 ‘66666667 ‘07407407 00823045 1|‘66666667 ‘02469136 ‘00164609 00091449} '7/:00013064 ‘00010161} 9)‘00001129 0000112911 /*00000103 ‘00000125/13)|‘00000009 00000014) 15/00000001 -. 12 = 69314718 Having thus obtained the Napierian logarithm of 2, /44=1(2 « 2) = 12+12=212= rene and putting m=5, n =4, ~jF Ol CO moomoo oO OD Ol 1=24+4. att se toa) = §22314354 ; 9 Operation. but bs BH vi M SI) 222222221 1 22222222 sist! ie iy ee 1} 274348/3| 91449 a | ; 110 = 1(2+5) =12-4 15 3 we ; — 2:30258508 : mai ee vidtaaiaa 22314354 (272), Mi, 19 = 75 = 043420448 ce m—n\' m—n\3 or log. m=log, n +-o-s6s59806[ (7 +) 4+3(e— a) +o]. Suppose, now, it were our object to construct a table extending from 1 to 10000, we should commence with 100; since, in calculat- ing up to 10000= 100 - 100 we should fall upon 200, 300, 400, &c.,=2 + 100, 3 - 100, 4 - 100, &c., whereby the logarithms of 2, 3, 4, &c., would be determined, Since log. 100 = log. (10?) = 2log. 1O=2+-1=2, log. 101 = log. 100 ++ 0186858906 (95 : ial 201 0'86858896 = 24 = 210043214 ; . (278), log. (= )- o-seasssoe| (7 — 150 COMPUTATION OF LOGARITHMS. ‘86858896 log. 102 = log. 101 + ——-— = 2:0086002 ; log. 103 = log, 102+ Sai = 20128372 ; -&e &e. > &e. After two consecutive Jogarithms have been obtained, the opera- tion may be shortened; for, putting m =n-+1, we have ‘86858896 log. (n a 1) = log. n oe “Qn+1” ‘86858896 | W+3 ’ 1°73717793 log. (n+ 2) = 2log. (n-+- 1) — log. n — ric eae pL (277) log. (n+ 2) =log. (n+ 1)+ Observe also that 4(2n + 3) is the difference of two consecutive divisors, 4(n + 1)? — 1, 4(n + 2)? — 1. ‘ 1°7371779 Making n= 100, log. 102 = 2log. 101 —log. 100 er Bae T Tam & Operation, 10001 + 200 + 40086428 oe kas — 20000000 10201 Es 426 4 1 one. —_—— .. log, 102 = 20086002 40803) 1°7371779(426 1°7371779 r= 101, oe log. 163 = 2log. 102 — log. 101 ~ 40803 + 4(20243) . Operation. 4:0172004 40803 4 — 20043214 820 : eee ie _ 417 41623) 1°7371779(417 ——— ( . log, 103 = 20128373 The student should continue the computation. Before terminating this problem, we will make one other trans- formation whereby the calculation of logarithms will be rendered far more rapid. COMPUTATION OF LOGARITHMS, 15] For small numbers, Borda puts* m=(p—1)*(p-+2) = p?— 3p+-2, n= (p+ 1)*(p—2) = p?—3p—2; a" m—n =A, and m+n = 2p?—6p; therefore, substituting in (275) and making M =1, we have (Frytp=3))*L(ams,) + pap) +4(pcq,) +] or 2U(p—1) + (p-+2)—2U(p +1) —U(p —2) is ic : =) Has) + | Making p = 5, 6, 7, 8, successively, we obtain 212 — 313 +17 = Wes +45)? ...] = 036367644171, 124-215 — 217 = 2 As + 4(a5)? +...] = 020202707317, — 412 — 15 4 413 = 2[4q + 44) 4+... ] = (012422519998, — 513 + 15 + 217 = Anti + F(z)? +... ] = 008196767203 ; from which eliminating 17, we have 512 + 215 — 613 = ‘092937995659, [2+ 315 — 513 = ‘028399474520, —412— 15+ 4/13 = 012422519998 ; eliminating 73, — 2/2 -- 15 = #223143551312, — 16/2- 715 = ‘175710498070 ; whence [2 = ‘693147180557, 15 = 1'609437912426 ; ne 110 = 2:302585093003. Had the above operation been extended sufficiently far, we should have found (272) M, _ 1 = 043429 44819 03251 82765 which, introduced into the last transformation, gives, for the com- mon logarithm, log. (p+ 2) = Alog.(p + 1) + log.(p — 2) — 2log.(p — 1) + 8685889638 [(——5] 4a (- a nt: ao sy) +a]. (279) Employing this form and imitating the above operation for the Mopicriay logarithms, 12, 13, i, 17, the student will findt * Franceur, Mathématiques pures. * The number of digits employed in the calculation should exceed, according to the nature of the operation, those which it is intended to retain in the results. 152 EXERCISES. log. 1=0, [10° = 1.] log. 2=030102 99957, log. 3=047712 12547, log. 4= 060205 99913, [log. 4 = 2log. 2} log. 5=0°69897 00043, log. 6= 077815 12504, flog. 6= log. 2+ log. 3.} log. 7=0‘84509 80400, log. 8=0°'90308 99870, flog. 8 = Slog, 2.] log. 9= 095424 25094, flog. 9 = log. 3.] log. 10 = 1:00000 00000 ; (10! = 10.] .. log. 11 = 1104139 26852, [p=9.] since | log. 11 = 2log. 10+ log. 7— 2log. 8-+ ‘86... [3414 4+-4(34y)? 4+. J 5 log. 12 = 107918 12460, [log. 12= log. 3-+-log. 4.] log, 13 = 1411394 33523, fp =11.) log, 14= log. 2+ log. 7, log. 15=log. 3+log. 5, log. 16=2log. 4, log. 17 = 123044 89214, [p = what ?] log. 18=? log. 19 = 1°27875 36010, log. 20=? log. 21 =? log. 22=? log. 23 = 136172 78360, log. 24=? leg. 256=? log. =? log. 27 =? log. 28=? log. 29 = 1:46239 79979, log. 30 =? log. 31 = 1:4913616938, log. 32=? log. 33 =? log. 34=7? log. 35 =? log. 36=2 log. 37=2 log. 38 =? log. 39=? log. 40 =? log. 50 =? log. 60=2 log. 100 =? log. 1000 =? log. 10000 =? log. +4 =? log. +¢=? log. 001 =? As p increases, the series (278) increases in convergency very rapidly ; so much so, that, when pis no greater than 102, the sec- ond term will have its first significant figure in the 18th decimal place ; and, confining ourselves to 7 decimals, the last.two only will have to be obtained by division. If we confine ourselves to seven digits, after the computation has been made up to 1000, the remaining logarithms may be readily ealculated from the table itself. Taking the differences of five consecutive POT as those of 100, 101, 102, 103, 104, and the differences of these differences or the second differences, and the third differences, we find INTERPOLATION, 153 Ist Dif.(-L) | 2d Dif. (—) | 3d Dif. (—) 20000000 20043214 20086002 20128372 20170333 We observe here that the third differences, 8, 9, are nearly equal. We are naturally led to the following problem: PROPOSITION IV. It is required to determine what function y, is of x, when y, depends upon x in such way that, if we attribute to x the particu- lar values x =0, 1, 2, 3, 4, the third differences of the corre- sponding values of the function, y,=Yo. Yiy Yo Ya: Ya Shall be equal to each other. If in the first power of z, we make ta), 1,2, 'o, 4, the first differences, 1, 1, 1, 1, are the same ; if wad, 1, 273, 4, z* gives z* = 0, 1, 4, 9, 16, the first differences are, h, 3, 0, a and the second differences 2, 2, 2, are the same; if z= 0, 1, 2, 3, 4, x® gives x? =0, 1, 8, 27, 64, the first differences are, 127 Al9, 37s the second differences are, 6, 12, 18, and the third differences, 6, 6, are constant. Hence we assume =A+ Br+Cr?+Drz', Attributing to z the particular values 0, 1, 2, 3, indicating the corresponding particular values of y, by yo, Ys Yo Ys, taking the differences as above, and denoting the first difference, y; — y, of the first differences by D,, the first difference of the second differ- ences by D,, the first of the third differences by D,, (which, for the sake of distinction, may be called the first, second, and third differences,) we have, 154 INTERPOLATION, YHA, D,=y,— y= B+ C+D, D,=C 24+ D-6, y=A+B+C+D, Y—-Y=B4t+C-34 D7; D;,= D+6; Y=A+B-24+C-44+D-8, Ce24+D.12; B+C.54+D-19; yz= A+ Be34C +e 94D « 27; ». D=tD;, C=4(D,—D,;), B=D,—4D,+4D,, and A= ; hence y, = y+(D, —$D,4+ 4D3;)z +4(D.—D;)z? +4D; «2°, (279) or y,=Y%+2[D,—4D,+4D,+ 2 +4(D,—D,+2+4D;)], = D, D; D; Ob. der ther te helt se Bake dy bee? is the function sought. For any particular series of numbers, it will be advantageous to calculate beforehand the coéfficients of the several powers of z, and to attribute to the terms their proper signs. Thus, for loga- rithms, the second and third differences, D,, D;, being minus, we have v= +2 | Di +4D.—2D,— 25+ 2 3), or 9¥,=%+2[C,—2z(C,4+2C,)], (280) putting C,=D,+4D,—4D;, C,=3(D.—D;), C;=2D;. Let us make an application by requiring the logarithm of 100345. We have nee (3 D, = 0043214, |.°. C, = 0043424, ~ 940000000, | D2 = (0000426, | C= «0000209, Yo = 200 > | D, = 0000008 ; C, = 00000014 ; log.100:345 = y, = 2+ 345] 0043424 — +345(-0000209-+ ‘345 * 00000014)]. INTERPOLATION. 155 Operation. 14 43424 ‘345 72 345 43352 115 345 ‘460 13005|6* 2095 1734)1 216|7 20946 eee 345 0014956 20000000 62°7* pe) Se 84 20014956 = log.100°345 . 10 721 * What is the logarithm of 101°79062 Ans. 2007709. Required, the logarithms of 100‘1, 1002, 100‘3, 1004, 1005, 1006, 100°7, 100‘8, 100°9, When it is required, as in this example, to interpolate an arith- metical progression of tenths, form (280) may be advantageously modified, as follows. Suppose that we have found the logarithm corresponding to z= ~ and wish the logarithm for x paid! , we have 10 10 (280) r 7 re Her sedge igi 1007" 1006 ” r+l r?+2rt) ret 3r?+ 3r+1. Yar = Yor 10 C; <= Om © is ipsatdciomas (17) SERN 1 2r+1 3r?+3r+1 —Y,= C e—_—— « —__— —. « ——__________. , Year T¥e = Gi 9 ETO “GGT TG 1000 C 4G, C; 2C, 3C; of dret=I-T 19 Too 1000 "Lioo t+") * jooo J (5!) To adapt this form to the computation of the tenths from 100 to 101, we have * Shortened multiplication. Rule? 156 INTERPOLATION, C, = 43424, | .-. 3 = = 4342:4, C, 6 Cre 209, an ae 2:09, e C; — 6 ° C,= 1; Sapo 001 ; 5 Yeons = = Ye + 4340:309 — 7r[4+18 + (r ch 1) + 003]. Making r = 0, we ire log. 100¢1 = site = log. 100 + ‘0004340 = 20004340 ; r=3, 13008 i], 4340‘309 gives 4340 gives 4186 12 Me OO TO Wee oo oe ean 4336‘123] gives 4340 |100°4) 2:0017336 are ae 8 16 ah Lia SP EM ee SON ae OOS 0) En ame Ne een am «- log.100°2 = 2:0008676 ;|log.100‘3=2:0013008 ;/100+5) 2:0021660, Adapt (281) to interpolate between 101 and 102, and compute the logarithms of 101‘1, 101°2, ..., 101°9. For returning from the logarithm to its number, we have (280) Yr — Yo C,—2(C,4+ 20C;) iaiag 2 where, it is to be observed, that C, is a very near trial divisor, though too great. We may therefore find an approximate value of x by dividing by C,, and then, having perfected the aes repeat the operation. Given the logarithm 2‘0014956 to find the corresponding num- ber. Operation. 20014956 344 | 43/3|5/2)14956(‘345 20000000 C.= 13006 = 4|3|4|24*)1495/6(+344 115 1950 1303 1734 cheek 209°459 eae 192 344 216 175 or de 216 pet 62/8 adie 17 8/4 8 ., No, 100:345 * Shortened division. "2 CHARACTERISTIC, 157 If we had divided simply by D, = 43214, we should have found ‘346, which is near the truth, and the approximation will be still nearer as we ascend above 100; so that ordinarily it will be suffi- cient to diminish the given logarithm by the tabular logarithm next below it, and divide this difference by the difference of the tabular logarithms above and below. And when greater accuracy is re- quired, the third difference may generally be neglected, whereby (280) and (282) will be reduced to Y2=Yt+2(D,+4tD,—2 4D) ea (283) Further it will be sufficient to correct the divisor by the first digit of the quotient, or the nearest to it, which may be found by inspection. The operation above becomes D,=43214 43 4 2 7/14956(‘34494 6 4)13009 4|3|3|6|3| 1947 1734 213 173 40 39 — 1 The Characteristic, or integral part of the logarithm, is not usu- ally inserted in the tables, since it can readily be determined from the relation IW'=2; which gives for y=0, 1, 2% * By ay 1, Ry. 8h on5 z=1,,10, 100, 1000, ..., °*l;, ‘Ol, ‘OOL, ...5 Hence the characteristic is always indicated by the dis- (284) tance of the first significant figure from the place of units; + if to the left, — if to the right. Thus the characteristic, or integral part of the logarithm answer- 158 EXERCISES. ing to the number 365 is 2; because 3 is two places distant from the units. Therefore, bearing (267) in mind, we have log. 365 = 256229, [See tables]. log. 365 = log. 42% = log. 365—log. 10=2'56229 —1, =1'56229, log. 3°65 = log. 264 = (2 — 2)56229 = 056229, log. 365 = log. +288, = (23 )'56229 = 156229, log. ‘0365 = log. 73$55 = (2 — 4)'56229 = 256229; d&c., &e. It will be seen from these examples that the logarithm of a num- _ ber in part or wholly decimal, is to be found in the same way as if it were integral, except the characteristic which is determined by the rule above and may be either plus or minus, while the tabular part of the logarithm is always plus. EXERCISES, ~ 1°, Multiplication (264). Operation.| Multiply by Multiply 465 266745 | 366 157, by 377. 1‘57634 | 90109 ‘035, - ‘007 034. Ans. 17530. 424379 2°. Division (267). Operation. | Divide by Divide 054 Berg099 | « SHS 4095, by 1°75. 024304 | ‘0005 ‘00789. Ans. ‘030857. 248935 3°, Involution (265). Operation. | Square 139°75. Cube 17'356. 123945 (10399)! =? 113 foal. 3 353 Ans, 522852, 3°71835 4°, Evolution (266). Operation. | Find ./365, 2/001, Find °/2 : 5/ 3010300 : ¥ nfo 148702 SeamED (75:00005)', V°, ns. Is . —- —— pine ot & 2, 02, /:002. Note, An operation should be so conducted as to restrict the EXERCISES, 159 minus sign to the characteristic. Thus, in finding the square root of ‘2, we have to divide 1‘3010300 by 2, which is readily performed by observing that 1 = 2+ 1, we have 13010300 : 2=1'6505150 ; so 33010300 : 2=(4 4. 1-3010300) > 2 = 26505150. 5°, Rule of Three. Find a fourth proportional to the Numbers, Operation. 13°79, 1‘13956 99:367, 1:99724 720°25. 2'85748 1‘:005 : ‘00356 : : 79099 : 2 2057106 3657 5892167 ee eee eee ee Ans. 51899, 3°71516 | 3379071 * 7112 ** 1000009 * ° 6°. Exponential Equations—Compound Interest. If a denote the amount of one dollar for one year at compound interest, then aa = a’ will = the amount of $1 for 2 years, a’?a = a® = amount of $1 for 3 years, a’a = a* = amount of $1 for 4 years, ..., a‘ = amount of $1 for ¢ years; .. Pa'=amount of $P for ¢ years = A, is the equation for com- pound interest. Taking the logarithms of both sides, we have log. (Pa') = log. A, or log. P + log. (a‘) = log. A, or log. P+ t log. a= log. A, the logarithmic equation for compound interest, It furnishes, also, an example of the solution of an Exponential Equation, since the exponent ¢ may be the unknown quantity, and we have ; — 108: A—log, P log. a : Application. In how long time will any sum of money double at compound interest, at 7 per centum ? PROPOSITION V. It is required to develope an exponential function in terms of the exponent ; y=’, (285) 160 EXPONENTIAL SERIES. in terms of z. Attributing to y and x the increments & and A, we have ytk=at*=a* eat, k=a*(a*—1)=a7{(1+56)*—1], [putting a=1+5}, or k= o[1+2b-+Rh—1) > +..—1]3 The 5. ae b be ; F =a [b+ (= 1) 6 B+ (b= 1) (h-2) «et ] k , D* . Decale VY y=fe = l= | le ae oy a Piened wll, —, +. ]=Aa’*, (286) : eo* 5 ge putting OM gl are 2 : 3 a2 4 2 (gee en ee ee [See (268), (269).], That is, the derivative of an exponential is found by multiplying the function itself by a constant quantity which is the reciprocal of the modulus of a system of logarithms to the same base, Taking the derivative of (286), or the second derivative of (285), we find alana \otaAae Ait x Ata? s0ql el At a, plizoitin’y a4 and making x =0, we have Yrz—9— GC = I; Yo=9 = A, Yr_9 = al Ys 0 = A®, ooo 9 whence we are assured, precisely as in the demonstration of the Binomial Theorem, that the function y = a’, is expansible in terms affected by integral additive power of z and by no others, so that the form y= a = (+ Cr+ Caz? 4+ Cyr? +...4 Cha* t+ ons is possible, and no other. Therefore taking the derivatives, we obtain j y = Aa? = Ci +14 Che 22+ Cy + 32? + C, o 4z' +... —C, making h, and, consequently, kh, z,, 2, = 0. So from yf=!f uy U SFr V'= I, 2, we find yA, = Cie fu) (ue fo) . (V'9—s52) 3 and generally Yn SY, le Sh * Deas, (295) When several variables are successively functions of each other, the continued product of their derivatives, taken in the same order of succession, will be the derivative of the first, regarded as a function of the last. Illustration. See Proposition II., Section First, y =f, z = Az”, 1 2=fyr=2". Cor. The derivatives of converse functions are recipro- (296) cals of each other,* For, let y and x be functions of each other, or y =fx and r=9¢y, then we have (Yyare) * (Le gy) = yay = 1, 1 Y y=se Illustration. In (285) let x and y change places, and compare (286) with (268). t ao OR ) WEEE ee * The chord and arc ofa circle are conversely functions of each other. FACTORS, 165 PROPOSITION IV. To find the derivative of a product, the factors of which are continuous functions of the same variable, Given y=fr; y= ut, u=/f,2,0=f,2; to find y',_,. We have yptk=(u+2) (+7) =y+vi4-u+y, . k=vityt+a; but =Wyopet 2% Ort = (uz, + 2)h, and = Dyan, tf Za OF J = (Vz + Za)h 5 k=v(u, + 2)h+ u(v, + 2)h+ (u, + 2) (ve + 2)h? ; oF ofus +2) + ule’ +2) + (ul +i) (oe +h and Yq i= [+] = (uv) = vu, + uv,’ or (uo) = ed ae) = UD fig) B In the same way, if » be = st, continuous functions of zr, we have O yelagy = a FB ee ee which, substituted above, gives, (stu) s' tw stu s , t “E wu’ (stn) pee a Le and generally — “e pope et ‘ (297) ois (fitfrf,c..) fit, fat, fex fishirfew) fi fe fe io Tuma RAE omitting the variable z, and employing only the symbols of oper- ation; fi» Sos tas ves which may be done, since they are equally ap- plicable to any quantity which may be made the independent vari- able ; thus, instead of 4/xr « 4/x = \/z, we may write 4/+3/=4/, as a general rule. We may enunciate (297) : 166 POWERS, The derivative of a product of continuous functions, divided by the product itself, is equal to the sum of the derivatives of the functions divided by these functions severally. PROPOSITION V. To find the derivative of any real power of a continuous func- tion. If in (297) we make the functions s, ¢, u,..., all the same and m in number, we find , (u") a aye _. n—1 ’ “a Fae 7 OF (w) = ne ou. (a) If y=u-", u being =f, 2, we have qu =u" su =u = I, (yur): oy 4 (4) + @ apy (297) yu pe ey a ih ok ae I ; ru ey! rue eu a = ah ee e =—y" (2) CA iesoat ats 7 (see =—ruT ew, (d) If y=fr=u", u being =f,z; 1 we have y =v", putting 0= wu", oru=v"; .*. (a) or (5), y, = me, -and a= ne", eal 1 or (296), v1, = The non? f ' ’ t m 1 ’ (295), Y, =Y, 0, * Uy = Mv" yn? Ub, 5 , ™m a , or Y.=—U t,. (c) From a comparison of (a), (d), (c), it appears that the same rule holds for the derivative of a power of a function, whatever real quantity the exponent may be ; and we may write [( fz") = n(fay (fe), (298) or (Fy) =a fy (FY bes, The rule found in the first section for the derivative of any power of a variable, holds for the power of a function. Indeed (298) embraces (245) ; for if fr = 2, then (fr) =2' = 1, and (298) reduces to (2”)' = nz". FRACTIONS, 167 PROPOSITION VI. To find the derivative of a fraction, the terms of which are con- tinuous functions of the same variable. From (297) we have (ir Alf A(E +8) fifi fe .. putting Si=Su to= (;)- (fa)'s [fn (fay = (fa + fa +h (APT but (298), [( fae] == fir fs (2) 22 Fraexia fi apfpaah, wpe ~ (299) f ek Aah eae en 1, €., d fe" AF) (fa)? The derivative of a fraction whose terms are continuous func; tions of the same variable, is equal to the denominator multiplied into the derivative of the numerator minus, the numerator multi- plied into the derivative of the denominator, divided by the square of the denominator. PROPOSITION VII. In finding the derivative of any continuous function (300) y = fx, we may replace the increments k, h, one or both, by such quantities, k,, h,, as are separately functions of k and h, and such that the final or vanishing ratios k, : k, h, : h, become that of unity. For, by hypothesis, we have fi =1-+ 2,, z, being such as to reduce to zero when & and &, become = 0; or k,=(1+2,)k, so h, = (1+ z)h ky 1+-2, & 1 1+2% oni betes! fT (Yy=se+ 2) 3 1+ 2, I+0 but when h becomes = 055 nye reduces to 1407 1, and we have lel-[el-y-- eB. 168 FUNCTION EXPANDED. PROPOSITION VIII. To find the expansion of a continuous function, such that its successive derivatives all become finite when the independent va- riable reduces to zero. Let y= fx be the function. It follows, from a process of reason- ing precisely like that employed in the demonstration of the Bino- mial Theorem, that no other than integral additive powers of z can enter into the expansion; and it only remains (and is sufficient) to see if the assumption y = Ay-+ Ar + Agr? + Azz? +... , is possible, or, what amounts to the same thing, if the coéfficients A,, A,, As Ag, ..., are determinable, and, therefore, real. Taking the successively derived functions, we find y =A, 1+ A, + 27+ A, e 3xr?7-+..., y= Age 2®el+tAze3e +A, +40 Bx?-+..., y' =Az,e3e2el+A, 2 4¢3¢ Qr+Az-+5 +40 3r7+.,., yr =Ape 4e38-2-14+A,+-5264-3 6 QrH..., &e., &c., &c., &c., Now, if in the above equations we make z= 0, and indicate the corresponding finite values of Yr Ys Yr Yoo Yi oes by Yoo Yor Yor Yo's Yos ane 5 there results y= A, yi=A,o 1, Yi=Ane Vel, Yi/=A,+3e Qo l,...3 f ae i x “uj x YF Yot tos pM ways prs 4 (73 8a aioe i 9h) This is essentially Maclaurin’s Theorem, and is very service- able in expansions, being more general than the Binomial, which it becomes simply by putting y=(a+ 2)”. + yi « £2, 2. EXERCISES, - 169 EXERCISES. (log. ry=5 . Qr ooh. +, (Ix?) = — =, [(298), (268), (245)}. es [(log. 2)*] == SE; .. (Unyty =. , (log. 2" =="; (ley = % [flee 2)" = mui vy , M n—l1 . [log. (az"+ b)] eee . [(ale + by] = ania TFG a Gt , (log. a*)' = = »Aa=MA=M. =. =1. — [(286, (296).] ey _ (lz) » [(lx)"] eee » (ufeyy = *G. [(298)] 2 wy _Mb+es %r+d-3r? +...) [Ua+ bz + cx? +...)"] = A as Sa PR [(293)] i ne eae i log. 2) am (2 24 299 (los. 7) = (7 -F). Libis (tos. el =M. mentite A abimaufat bs (3) ]- 2. ([(298) 1] a GY ]-[¢-F]. BOOK SECOND. PLANE TRIGONOMETRY. SECTION FIRST. Trizgonometrical Analysis. Construction. Describe the quadrant ABC; drop 1 the perpendiculars BX, BY, upon the radii OA, OC ; produce OB so as to intercept AT and CV, perpen- diculars drawn through the extremities of OA, OC, in T and V; then: Definition 1. The ares AB, BC, are said to be com- wt plementary to each other. BC is the complement of Fig. 52. AB and AB is the complement of BC; the are 90° — a is the com- plement of a and a is the complement of 90°—a; 45°42 and 45° — x are complementary arcs. Def. 2. The perpendicular BX is called the sine of the are AB. Hence the sine of an arc is the perpendicular let fall from one extremity of the arc upon the diameter passing through, the other extremity of the same arc. Def. 3. AT is the tangent of AB. Def. 4. OT is the secant of AB. Def. 5. BY (= OX) is the sine of BC or the cosine of AB. Def. 6. CY is the tangent of BC or the cotangent of AB. Def. 7. OV is the secant of BC or the cosecant of AB. Def. 8, AX is the verst sine of AB, Note. The abbreviations of the titles above, either with or with- out the period, are employed as symbols of the quantities them- selves. ‘Thus, if @ denote any arc less than a quadrant and } any arc not greater than 45°, the above definitions give TRIGONOMETRICAL ANALYSIS, 171 sina = cos(90° — a), cosa = sin(90° — a) ; sin(90° — a) = cosa, cos(90° — a) = sina ; (302) sin(45° + b) = cos(45° — b), cos(45°-++b) = sin(45°—d). tana = cot(90° — a), cota = tan(90° — a); &c. (303) seca = cosec(90° — a), coseca = sec(90° — a); Kc. (304) PROPOSITION I. The sum of the squares of the sine and cosine of anarc (305) is equal to the square of the radius, or to unity, when the radius is taken for the unit of the trigonometrical lines, We have OB? = BX? + OX? = BX?-+ BY?®, (fig. 52.) or sin’a + cos*a = r* = 1, when radius 7 = 1. Cor. The sine is an increasing and the cosine is a de- (306) creasing function of the arc, or the sine BX increases from 0 to r as the arc increases from 0 to 90°, while the cosine OX decreases from r to O for the same increase of the arc, See (194) and observe that the sine BX is half the chord of double the arc AB. PROPOSITION II. The tangent of an arc is to the radius as the sine to the (307) cosine ; or, the tangent is equal to the sine divided by the cosine, if the radius be taken for unity. TA * "Bx tana sina We have (aet axes par cou (fig. 52.) sina or tana = ,rm=l cosa PROPOSITION III. The radius is a mean proportional between the tangent (308) and the cotangent of an arc; or, the tangent and cotangent of an arc are reciprocals of each other when r= 1. For, by similar triangles, we have (fig. 52.) AT : AO=CO : CV, or tana : r=r: cota; tana cota = r*, = 1, when r = 1. 172 INCREMENTAL VANISHING ARC, PROPOSITION IV. The square of the secant is equal to the sum of the (309) squares of the radius and the tangent. We have ) OT? = OA?+ AT®, or sec2a =r? + tan2a, ( fig. 52.) The student may obtain other forms when wanted ; as, for in- stance, the following : The secant is to the tangent as radius to the sine, Also (310) secant X cosine = 7? = 1. PROPOSITION V. An INCREMENTAL VANISHING Arc ts to be regarded as (311) a straight line perpendicular to the radius. Let AB be the are in question; draw the tan- gents AT, BT, intersecting in T, and join OT; then will the triangles AOT, BOT, be equal, and OT will bisect the chord and arc in P and Q and be perpendicular to AB, From the similar trian- gles TAP, TOA, we have AT _OT_OQ+QT_, , QT, AP OA OQ OQ’ but QT = OT — OQ = (OA? + AT?)? — OA, which reduces to 1 [QT] = (OA? — 0?)*— OA = 0, when the are AQB becomes = 0, since then AQ=4AQB=0 and AT = tanAQ=0; therefore the ultimate ratio of the vanishing pe [AT], [AP], becomes A : Ee =14[55 pis that of unity ; v4 AT TR) [RAL and « | |=[ ae |- But (113) the are men is greater than the chord AB, and less than the broken line ATB; .°. the quotient sa is greater than A ATB ‘ AB or unity, and less than AB’ which also becomes = 1, when the arc AQB=0; DERIVATIVES OF SINE AND COSINE. 173 .", the ratio ea of the vanishing arc [AQB] to its vanishing chord [AB] cannot be less than one, and cannot be greater than one, and therefore must be =1; which proves the proposition (300), observing that the arc is perpendicular to the radius (177). Scholium. It is not stated that the vanishing arc, when employ- ed as an increment, merely may be regarded as a straight line per- pendicular to the radius, but it is proved that it must be so regarded. On the other hand, it is to be observed that we do not affirm that the arc will ever actually become a straight line, or that it will not always exceed its chord in length, but that, for the purpose pointed out, it must be so regarded, in order to reduce it to zero, and there- by to eliminate it from the function under investigation. PROPOSITION VI. To find the derivatives of the sine and cosine regarded as functions of the are. Denoting the arc by z, the sine by y, and the cosine by z, the functions may be represented by = LB gly which are the same as Fig. 54. y =sinz, z= cosz, Attributing to z the incremental vanishing arc h, (311), and to y, z, the corresponding increments, k,—7,(306), the similar triangles, whose homologous sides, taken in order, are k, h,—i; z,r=1, y, give us k Zz Vren=[Z]= Foes kes The derivative of the sine regarded as a function of (812) its arc, is equal to the cosine, radius being unity. - ' ie = Again we have oe = tay; % Ca; The derivative of the cosine regarded as a function of (313) its arc, is equal to the sine taken minus. Proceeding to the 2d, 3d, 4th, &c., derivatives, observing that (294) gives (—f)’ = —(f’) when = = — 1, we find, (312), (313), 174 DEVELOPMENT OF SINE AND COSINE. ae sinz, vu =-+2,=cosr, y; = Ze = 94 Uf = Z, = (—y.)' =e i, =—Zyy, =z —z,=+ yy y= 2, =y, =e, &e,, &c., &e, 3 1. €., Cor. The sine and its derivatives are alternately sine, (314) cosine ; sine, cosine; ..., in which the algebraical signs alternate in pairs, +, +; —,—; +,+; —,-—; .-.., and the cosine and its derivatives are alternately cosine, sine; cosine, sine; ..., in which the signs alternate in pairs, also alternating, +, —; —, +;+4+, oo eet ae PROPOSITION VII. To develope the sine and cosine in terms of the are. Let y=sin(a+2z) and z=cosin(a+ 2); then are y and z con- tinuous functions of the arc z, y= Fr, z= F, x, which it is requir- ed to determine. In (314) substituting @+-z for x, we find y=sin(a+ 2), y' =cos(a+ 2), y’ =—sin(a+ 2), y'=—cos(a+z), y*=sin(a+ 2), y’=cos(a+z), ..., which become Yo = sina, y= cosa, yj’ = — sind, y}' = — cosa, yy = sina, yj = cosa, yj = — sina, ..., when r=0; 2 .. (301), sin(a + x) = sina + cosa > — sind e — cosa 1-2 3 4 5 zw . Hb £ REDE ee Eee Gi Ere el . ‘ 1 i a4 Hs HY gimmie acon (315) kg © oi “7.3. Sa Bee : 5) : 3 ‘ m E 5 ; gt ay ..) +0080(5- 7-5 x ag’ tise Tgs4ik 1. OS38t as Boag! In like manner (314) we have Z) = COSA, Z =— Sind, Z;’= — cosa, z'= sina, z'” = cosa, ... CHANGES OF SINE AND ARC. 175 id “. (301) cos(a-+2) = cosa(1— 7" BY 4 .a.304 (316) 76 . x3 7° mila Se gts) sine pat ye 5 lta}, Making a= 0, and observing that (306) sin(@,_9) = sin0 = 0, and cos(a,_)) =r =1 there results, (315), (316), x 3 7° zi MLNS PI ETs PPL ETAT E TS Ly ee I) and 2 6 Thi 9 agp ei ee eet ee carey lie andere whee Qin «6 and these are the developments required, They were discovered by Newton.* If’we change z into — z, (317) and (318) pesigini (6 ), (6.) sin(— Bin Snag iz 57 a pn =a (ez i. oe 1, oad Ue? nat 2 satya eat ws ea = 1-5 ts = oe = COST; 1. G., Cor. The sine of an arc changes from + to — as the are (319) itself changes from + to —, but the cosine remains still + while the arc passes through the value zero, which is in accordance with (180). Nothing, however, it is to be observed, has been demon- strated in regard to arcs greater than 90°, or z > a quadrant. PROPOSITION VIII. It is required to express the sine and cosine of the sum and difference of two arcs in terms of the sines and cosines of the arcs themselves. Changing x into — 2, (315) and (316) become ; } co 2° sin(@ —zr) = sina(1— Ts +,-, a _ cosa( 2 — {2903 = oe w) x* cos(a —z) = cosa(I —“Teat ) + sina( 2 — [e203 =; -) ; * Lagrange, Lecons sur le Calcul des Fonctions. 176 FORMS OF SINE AND COSINE. with which and (315), (316), combining (317), (318), there results sin(a + xr) = sina cosz + cosa sinz, sin(a — ©) = sina cosz — cosa sinz ; (320) cos(a-+ x) = cosa cosz — sina sinz, cos(a — x) = cosa cosz + sina sing, These four forms are constantly recurring in trigonometrical an- alysis, and should therefore be committed to memory; they may be enunciated as follows: I. The sine of the sum of two arcs is equal to the sine of the first multiplied into the cosine of the second, plus the cosine of the first multiplied into the sine of the second. Il. The sine of the difference of two arcs is equal to the sine of the first multiplied into the cosine of the second, minus the co- sine of the first multiplied into the sine of the second, Ill. The cosine of the sum of two arcs is equal to the cosine of the first multiplied into the cosine of the second, minus the sine of the first multiplied into the sine of the second. IV. The cosine of the difference of two arcs is equal to the co- sine of the first multiplied into the cosine of the second, plus the sine of the first multiplied into the sine of the second. Consequences. Making a =z, we have (320) Cor. 1. sinQr = 2 sinz cosz, (321) Cor. 2. cos%x = cos*z — sin?z ; (322) but (305), b c= ics? +} sin? 2 which, combined with (322) and (321), gives Cor. 3, 1+ cos2x = 2cos?z, (323) Cor. 4. 1 — cos2z =2 sin*z ; (324) Cor. 5. 1+ sin2z = (cosx + sinz)*, (325) Cor. 6. 1—sin2r = ao ee — sinz)? ; (326) * Cor. To( 1+ sine)? +(1- sine) = 2 cosz, (327) Cor. 8. (1-+ sinc)? + (1-— sin2r)* = 2 sing. (328) What is the sine of a double arc? of half an arc? the cosine of a double arc? of half an arc? Enunciate (323), (324), (325), (326), (327), (328). Adding and subtracting forms (320), and making a+r=p,a—z =g,and .. a=+(p+q), r=4(p — q), we have Cor. 9. sinp + sing = 2 sint(p+ q) + cosd(p — q), (329) Cor. 10. sinp — sing = 2.cosh(p+ ¢) « sind(p — q), (330) —— CU a SS ay See eee” SUPPLEMENTARY ARCS. 177 Cor. 11. cosp + cosg = 2 cos#(p+ q) « cosd(p— q), (331) Cor, 12. cosg + cosp = 2 sint(p-+ q) « sint(p — q). (332) These forms are useful in the application of logarithms, by con- verting sums and differences into products and quotients, Dividing (329) by (330) we have (307), (308), Cor. 13. Piece es Oa AL gs (333) sinp —sing tand(p— q) The sum of the sines of two arcs is to their difference as the tangent of half their sum is to the tangent of half their difference. By similar processes, other forms, occasionally useful, may be developed, as cosp-+-cosg_ a) Cor, 14. Scag sebae cott(p+q) « cotd(p + q), (334) sinp-Esing , Cor. 15. ‘siaamashonih tand(p + q), 7 (335) Cor. 16, Pee ~ cotd(p+q). (336) cosg — cosp Making a = 90°, forms (320) become sin(90° + x) = cosz, sin(90° — x) = cosz ; cos(90° + xz) = —sinz, cos(90° — rz) = sinz; i. e., Cor. 17. The sines of supplementary arcs are equivalent, (337) being equal to the cosine of what one exceeds and the other falls short of 90°. Cor. 18, The cosines of supplementary arcs are numeri- (338) cally equal, but have contrary algebraical signs. Arcs are supplementary when their sum amounts to 180°, as (90° + x) + (90° — x) = 180°. In the above forms, making zr = 90°, we have sinl180° = sin(90° ++ 90°) = cos90° = 0, cos180° = cos(90° ++ 90°) =— sin90° = —1; i.e., Cor. 19. The sine of 180° is = 0, and the cosine =—1, (8339) Cor. 20. sin(180° + x) = — sinz, (340) Cor. 21. cos(180° + zr) = — cosz ; (341) Cor. 22. sin370° = — sin90° = — 1, (342) Cor. 23. cos370° = — cos90° = 0; (343) Cor, 2A. sin(370° + x) =— cosz, (344) Cor. 25. cos(370° + xr) =+ sinz. (345) 12 178 TANGENT AND COTANGENT, Scholium. The consequences evolved by these last forms (337) ... (345), are in accordance with the principle enounced in (180). Indeed, if we suppose the arc to in- crease from 0° to 360°, the sine will pass a -& through the value 0 at 180° and again at 360°, while the cosine will reduce to zero at 90° and peak: + 270°; and it is obvious that the same correlation \\~ of values will be repeated in a2d, 3d, &c., circum- Fig. 55. ference. The algebraical sign of the tangent will be determined from the relation (307). \ PROPOSITION IX. It is required to develope the tangent and cotangent of the sum and difference of two arcs in terms of the tangents and cotan- gents of the arcs themselves. Consulting (307) and (320), we obtain sin(a+ 6) | sina cosh + cosa sinb ag pee bi — + 5 cosa cosb — sina sind sina _ sind cosa ' cosb sina sind’ 1— ‘-— cosa cosb dividing numerator and denominator by cosa cosb ; tana + tanb tan(a + b) = 7 EAR AB’ (346) is one of the relations sought. So tan(a — b) = anes (347) and (308), cot(a- 5) = eee OO a eee . (348) Cor. 1. tan2a = 4 doe (349) Cor, 2. cot2a = ee : (350) Resolving equation (349) in reference to tan a and consulting (309), (310), we find SCHOLIUM, 179 4 —1+(1+tan?2a)? — — Con S. “tats ema Pee a Ree as tan2a tan2a 1 — cosa " aasinge Cor. 4. Cota = cot2a+ (1+ cot?2a)* = cot2a+ cosec2a 2 = I+ costa : (352) sina ° 9 Making @ = 45°, cos2a will=0, sin2a =1, and we shall have (351), (352), Cor. 5. tan45° = 1 = cot45°. (353) whence, making a = 45° in (346), (347), (348), we find 1 + tanb te) — ° Cor. 6. tan(45° + b) = yeaa (354) 5 . cotb= } Cor. 7. cot(45° + b) = ant ine i (355) Making 2a = 90° + w in (351), we get 1+ si Cor. 8. tan(45° + 4u) = eee (356) Scholium. Other forms, serviceable in turning sums into pro- ducts, and vice versd, may be found; for example, if we make p = 90° in (329), we get 1+ sin. g = 2sin(45° + 49)cos(45° — 49) = 2sin(45° + 47¢)sin[ 90° — (45° — 4¢)] = 2sin(45° + 4¢)sin(45° + 4¢) = Qsin2(45° + 4). (357) So 1 — sing = 2cos*(45° + 4q) = 2sin?(45° — 49) ; (358) 1 + cosp = 2cos? 4p, 1 — cosp = 2sin? 4p. (359) Combining (329), (330), (331), (332), and (321), we find sin*p — sin*g = cos*p — cos*q = sin(p+q)sin(p—gq), (360) also cos*p — sin?q = cos(p + q)cos(p — 9). (361) The student will also find sin(a+5) cotb+ cota tana-+tand. sin(a—b) cotb— cota tana —tanb’ (362) sin(a+b) _cothtcota — tana+tanb | (363 cos(a= 6) +1-+cotacoth 1+ tana tanb’ ) 180 DENOMINATE EQUATIONS, cos(a-+ 5) cothb—tana 1 — tana tand_ cos(a—b) cothb+tana 1-+ tana tanbd’ (At eo Tr ir eR a tana ee tan?(45° +- $9) ; (365) Gite a) Pe lalonaal T cot? tp ; (366) 1+sing _ sin?(45°-+49) 1 —sing _ sin?(45° —iq) | (367) Ll+cosp — cos*4p =” ~1—cosg — sin*3g 0’ tana + tanb = sles?) ; cota + coth = See ; (368) cosa cos b sina sind Bw tana + cotb = poate cota + tanb = aE A (369) cosa sind sind cosb PROPOSITION X. All denominate equations are homogeneous. (370) A denominate equation is one involving denominate quantities, such as length, surface, volume, weight, time, velocity, and the like, referred indeed to a unit of measure, but distinguished from abstract quantities or mere numbers, By clearing of fractions and transposing, it is evident that any abstract or numerical equation may be represented by abc... [n factors] + dgb.C, ... [n, factors] + a3b;c3... [m3] +... = 0; for, if there were any powers they would be embraced in the pro- ducts of equal factors, such as ab = aa = a?, abc=aaa=a’...; a, * b, = d,* d= a3,..., &c.; and we may suppose the equation freed from radicals by involution. Now a, 6, ¢c,..., da b,..., being numbers, may represent the quotients of any denominate quantities divided by their unit of measure, or we may have Es OL ee OF Fy ~ MM? roo ie tt, 12> VT’ eee gy thus, if A =a line 15 feet in length and the unit of measure be one A 15 feet "M8 feet values in the equation above, we have At RS ain Agote. M ° M e Mu eee [+35 e ik oe [Nq] +... = 9, or ABC .,.[n]+A,.B.C, ... [m] + M™™-+ A,B; .. [ns é M3 a i = 0, yard =the number 5; and so on, Substituting these UNIT OF MEASURE, 181 clearing of fractions, on the supposition that the equation has been arranged so thatn > n, >; >.... The last equation is homoge- neous, being of the nth degree, or containing n factors in each term, and, as it is denominate, the proposition is demonstrated. This theorem may be serviceable to those not yet well practised in algebra, by detecting errors. For, if we begin a problem with a denominate equation, all the following equations being denomi- nate, will be homogeneous, and if any one, as that containing the result, want this homogeneity, it is an index of error in the op- eration, But a more important application is the restoring of a quantity which has disappeared from a denominate equation by being as- sumed as the unit of measure. Thus, if M = 1, the above equation becomes ABC «...[n]+ A,BoC, + ... [ng] + A,B,C; +... [ng] +... = 0, and the homogeneity disappears ; to restore it, however, it is obvi- ously necessary and sufficient to introduce the unit of measure, M, as a factor, into each term affected by an exponent which is the de- ficiency of the term in degree. Suppose we are to restore the radius in (317) ; the first member, sinz, is of the first degree, every term of the second must be the same, which requires a 2 x' 1.2.3 T.. or 1. 78 The first equation in (320) becomes r sin(a + x) = sina cosz + cosa sinz, when the radius is restored. So (346).when radius = 7, is r*(tana + tanb) r? — tana tand ° +35 1e sinz=2r— tan(a + b) = It is recommended to the student, as an exercise, to restore the radius in all the preceding forms, and to inspect the geometrical equations which have occurred in regard to their homogeneity. PROPOSITION XI. It is required to develope the arc in terms of its tangent. Let the arc AB, as it is the function, be indicated by y and its tangent AT, being the independent variable, by x; it is required 182 COMPUTATION OF T. to find the function y = fz, that y is of x. Give to x the vanishing increment h, to y the cor- responding increment hk, and draw J through T perpendicular to TO and terminating in the secant drawn through the extremities of k and h. . Simi- lar triangles give us ( ita [l= eMaees LE oa cee ae . T om we" 4 eee tere whence, returning to the function (246) we have Fig, 56. PO et ae x! y=Er ges roe ote shims arb constant; but z and y vanish together, 0=0- constant, .. constant = 0; = 2" YREo sie cach hei pa maveipe tie yee (371) which is the required relation. When radius = I, (271) becomes y = tany — 4 tan®y + 4 tan’y —4 tan’y+,—,.... (372) PROPOSITION XII. To compute the semicircumference Tt (pi), when radius is made unity. For this purpose Machin puts tany = 4; is (B49) any =, ye oat Ze A 120° 1—()? 119” but (353), tan45° = r=1, tan4dy — tai45” 1 : prot eA Gd TTS F ei hopiesto g Saale ehainiga din 1 +-tan4y tan45° 9-239” whence, by substitution in (372), we ce are y tany=+4 =t- +(4)° a tan4dy = tan2(2y) = os bees $(z45)? Hoy trues br = 45° = dy — (Ay — 45°) = 408 — 4) yy ad —[as5— tats)? +, —, we} and arc(4y— 45°) tan dy—159) - COMPUTATION OF T. Operation. + Terms. | 1°, 4 = «2/+200000000000000 | 3°. (£)5=(4)3(4)?="008-‘04—+00032) 000064000000000 5°, ‘000000512/+000000056888889 7, 0000000008192) 000000000063016 9°. ‘00000000000131072/000000000000077 ‘200064056951982 — Terms, 2°, (4)3 = (4) (4)? = 2 + 04 = 008|002666666666667 4°, ‘00001 28)‘000001828571429 6°. ‘00000002048 | ‘000000001861818 8°, ‘000000000032768 *000000000002185 10°. ‘0000000000000524288) 000000000000003 ‘002668497102102 ‘197395559849880 4 4ft—4(4)2 +, —,...] = +. /*789582239399520 (2 004184100418410 dif, + |*0000000244 16592 — |*000000000000256 4n= |785398163397446 mT = 3)‘1415926535897/[ 84] or 7 =3)/'141592653589793, correcting the last digits by an extension of the work. Cor. 4 = 90° = 14570796326794896, An = 30° = 0'523598775598299, 10° = 0/174532925199433, 1° = 0:017453292519943, 6’ = 0°1 = 0:001 745329251994. — [sts —4(a3s)* +> aie ifs PROPOSITION XIII. t To compute the trigonometrical lines. Combining (374) with (317) and (318) we obtain (0174533)? sinl° = (0174533) — ~~ |: pel ed NW © Ol = —,.. = 01745, 183 The order of computation is in- dicated by 1°, 2°, 3°, .... (373) (374) 184 TRIGONOMETRICAL LINES. (0174533)? 1.3 +,—... = ‘99985, and cos1°=1— Hence, (315), (316), sin2° = sin(1° + 1°) = ‘01745 [i-S ae —~——+,-, | + 99985} (‘01 heres penser .. |= 03490, 2.3 cose? = 99995 [1 — =P saat sl — 01745 [ (1...) it pee he. —, » |= ‘99939 ; sin8° = sin(2° + 1°) = 05234, cos3° = cos(2° -+ 1°) = ‘99863 ; &c. &c. &e. The processes just indicated may be advantageously used, the first for the computation of the sine or cosine of a small arc when a large number of decimal places is required; the second for in- terpolating between sines already calculated and set down in a table ; but, as two multiplications in each operation are requisite, a better method may be employed in making up a table, where a single constant multiplrer will be sufficient. In (329), making p= (m+ n)a, ¢ = (m—n)a, we have sin(m + n)a + sin(m — n)a = 2 sinma cosna, (375) a form that will be occasionally serviceable. If, for instance, we make m= 1, n= 1, we get sin2a = 2 sina cosa, a form already obtained ; m = 2, n= 1, gives sinda = 2 sin2a cosa — sina = 4 sina cos*a — sina = 4 sina(1 — sin’a) — sina = 3 sina —4sin’a, - The student may put m=3, n=1; m=4,n=1; &e.; m=83, n=2; m=4, n=2; m=4, n=3; S&e.; and find the results, If we make a = 1, we get sin(m + n) + sin(m — 2) = 2 sinm cosn, which might have been obtained by adding in (320); making n= 1° and putting r = the constant multiplier 2 cos1°, we have sin(m-+ 1) =r sinm — sin(m— 1), TRIGONOMETRICAL LINES, 185 a convenient form for computing the sines. Making m= 1°, 2°, 3°, ..., we have sin2° = rsinl1° — 0, sin3° = rsin2° — sinl°, sin4° = rsin3° — sin2°, &e, &c. It is recommended to the student to find r correct to 7 or 8 places, and execute the computation just indicated, employing the method of shortened multiplication. We proceed to show in what way certain sines may be express- ed in finite terms. In (327) and (328), making z = 45°, we have (i+ 1jz+ (1 — 1)? = 2 cos45°, .. cos45° = Abe = a : ee and =(1+1)?—(1—1)? =2 sin45°, .. sin45° = eae, Cae Ze LR In (331) making p = (m+ n)a, g =(m—n)a, we have cos(m + n)a-+ cos(m — n)a = 2 cosma cosna, (376) cos(1-+ 1)a-+ cos(1 —1)a = 2 cosa cosa, or cos2a = 2 cos*a — 1; cos(2+ 1l)a=2 cosa cosa — cos(2 — 1)a, or cosda = 4 cos*a — 3 cosa, If, in the last form, we put a = 30°, there results cos90° = 4 cos*30° — 3 cos30°, or 0 = 4 cos°30° — 3 cos30°. ‘ 0 = 4 cos*30° — 3, sin60° = cos30° = xe ; es cos60° = sin30° = (1 — sin?60°)? =. Again, 4 cos*18° —3 cos18° = cos3 + 18° = cosd4° = sin36° = 2 sinI8° cos18°, 4 cos?18° — 3 = 2sin18°, ia 4(1 — sin?18°) —3 = 2 sin18°, or 4 — 4 sin?18° — 3 = 2 sin18°, sin?18° + 4sin18° =1, cos72° = sin18° = da . 3 rt 4 (1042+ 5t)* sin72° = cos18° = (1 — sin*18) 4 186 TABLE OF SINES. In (328), making z = 15°, and observing that the minus sign is to be employed when z < 45°, we have (1+4)2? —(1 —4)? =2 sin 15°, 1 a cos75° = sinl5° = —— (,/3— 1). via nae The same form will give us the sine of 9°, then sine 3° = sin(18° — 15°) = sin18° cos15° — cos18° sin15°, and the cosine of 3° being known, the following table of sines and cosines may be calculated (320)* ; 3?+1 re | 1 sin 3° = cos 87° = 8. Qi (5 —1)-—3— 6+ 5+)’. sin fe ox dive Sines ee Ty eee (6 — 5)? 8 4. 2 1 1 1 sin Se ses Bees ppt olor pSaeb')t 3 1 ; Ons 0 — (5s — sin 12° = cos 78° = 3 © I+ 7H sin 15° = cos 75° = a (3* — 1). sin 18° = cos 72° = is (5¢ — 1). 4 pay Zit] Oo 0 — + e as fe sin 21° = cos 69 5. yO sos Magnan 5! — 5? sin DAcwaibonibooe t= (ee an he I « 5 — 5 8 4.2 i | 1 tae sin 27° = cos GSS st eo (5¢ — 1) +4(5 + 54)?, sin 30° = cos 60° = d. 3 cm sin 33° = cos 57° = er (5'—1)+ al (5 + Bi), : 1 ne sin 36° = cos = So oe (5 — 54)”, 3+1 3] 4 sin 39° = cos 51° = & oi (5' ++ 1) — a (Ga By * Library of Useful Knowledge. TABLE OF SINES: 187 tiv 3? sin 42° = cos 48° = — 4(5' — 1) to (5+ 5')’, sin 45° = cos 45° = * » (6 1) (Bt sin 48° = cos 42° = =i or sale Bi)’. sin 51° = cos 39° = a sin 54° = cos 36° = +9 +1). sin 67° =Weuaiggo Se Bek (5§—1)+ ms zie : (5-45)? 3} sin 60° = cos 30° = 3 ; 1 4 sin 63° = cos 27° = ae (5! — ae vy sin 66° = cos 24° = 4(5' + 1) + ——._ (6— 5h)?, 4.2 3 2 — yi sin 69° = cos 21° = Bah (8-41) + on (6 —5#)?. Bo sin 72° = cos 18° = a (5 +58)". sin 75° = cos 15° = - (3#+- 1). 2 sin 81° = cos 9° = a (5!-+1)+4(5 — 54)?. sin 78° = cos 12° = $(5'— 1) + cele, unary a . 1? sin 84° = cos 6 Pry shh) cagregy (or 5) i sin 87° = cos oe pelea —1y 4st) =: (5 +58)". sin 90° = cos 0° = 1, Other sines may be interpolated by (315) or by the method of differences, and the tangents will be found by dividing the sine by the cosine. It is recommended to the student to execute several of the com- putations indicated above, carrying out the work to ten or fifteen decimal places, and employing a shortened method of extracting 188 LOGARITHMIC SINES, &C, the square root as well as in multiplying and dividing. It will be advantageous to free the denominators of radicals; for example, let it be required to find the sine of 75°. Operation. sin75° = as (3?-+ 1) = co Saale 2.2 3)1°7320508076 2 = 14142135624 1 27320508076 27)200 28284271248 189 9899495937 | — 424264069 343)1100 28284271 ih 707107 3462) 7100 eae 6924 4 cE te. 8 3/4|6|4/0|5) 1760000 2 ear 0°9659258511 = sin75°. 27712 How many digits may be de- 74 pended on? 263 242 21 21 But a table of sines is comparatively of little importance, as the logarithms of these numbers are generally preferable in practice. PROPOSITION XIV. To compute the logarithmic sines and tangents. Restricting (317) to the fourth power of xz, we have sing = 2(1 — 42? + +4572"), .. log, sing = log. r++ log. (1 —t2? + +2") = log. ¢ +-M[(—40*4+44:2")'!— H(—42%)"]_ [(270)] ot, Seog gir x*(1 zt =) = . 7 6 . 30 ’ LOGARITHMIC SINEs, &C., 189 log. sing = log. z — No. to $2‘8596331 + 2log. zr + log.[1 + No. to (2log. 2 — 1:477)]}. (378) As an example, let it be required to find the logarithm of the sine of 5°, ' We have (374) xr = 5° = 08726646 ; log. + = 29408474 2log. x = 3:8816948 [ 3882 28596331 | 1:477 ——EEE log. 15000254 = 00000103 | 4'405 47413382 LNo. ‘000254 No. 00005512 log. sin5° = 29402962 Form (378) should'not be employed when the arc exceeds 5°, and the last term, log. [1 + No. to (2log. 2 — 1'447)], may be omit- ted if the arc be less than 3°, Imitating the process above, we find log. cosz = — No. to $1'3367543 ++ 2log, 2 + log.[1 + No. to (2log. x — 0778) ]?. (379) Operating as in the last example, we obtain log. cos5° = — 0:0016558 = 19983442, The logarithmic tangent will be found from the relation e = —_——,, or log, tan = log. sin— log. cos cosine ’ 8 8 8 , and the logarithmic cotangent results from cos ; couse » OF log. cot = log. cos — log. sin. Thus log. sin5° = 29402962, log. cos5° = 1‘9983442 ; fs log. tan5° = 2+9419520, and log. cotS° = 140580480. In order to avoid minus characteristics, 10 is usually added; thus in most tables we find log. sind° = 8*9402960. 190 ARC FUNCTION OF SINE. Dividing the first of (320) by sina, there results atl = cosz + cosa sinz = cosz(1 + cota tanz), .. log. sin(a+ 2) = log, sina + log, cosz + M(cota tanz —tcot?a tan’z +, —,...). (380) By a similar process we find log. cos(a + x) = log. cosa + log, cosr — M(tana tanz + 4tan’a tan’z+,—,...). (381) Let a= 5° and x = 0°] = 67, then + log. sina=2‘9402962 -log. M =1‘6377843 (1) + log. cosr=1‘9999993 | log. cota =1'0580482 (2) + No. =0:0086639 | log. tanr=3‘2418778 ) (3) — LNo. =0:0000864 39377103 + 14No, =00000011 42376363 2[(1)-+(2)]+(1) , log.sin5°1=2'9488741 65375623. 3[(1)+(2)]-++(1) _ Calculate the log. cos5° 1, In order to avoid the accumulation of errors, the computations should be recommenced from new points of departure, for which purpose the above table of sines and cosines to every 3° may be employed. PROPOSITION XV. To develope the arc in terms of its sine. We have , ; k 1 1 in Lz ih cai | ORES oy, 4 y or (250), = (1) *# + (2) (I) + (— 2?) 1 or PEM Es we a Deemer set into 4a F301 6:2 b 1 taBia 2 1 : 2 2 6 BS eee Mae 2) Tes 2 143 sas 5: 2 2 ' 2 1 8 RESOLUTION OF EQUATIONS. 19] 1+1.-2 1 ysothedeatpaege TOs 141-2 142.2 1 a. g§. ee oS (COU CLO 7 eas 2 2 Gaebae ow eek te Ne oe 2 143-3 Ay Sua Meh a |, 4 1 9 i gg ee vi a Me ee) where no constant is to be added, since z and y vanish together. PROPOSITION XVI. To resolve the equation. asinz + b cosr = ¢. : sinz b Putting tanz = =—, (383) cosz a : sinz c we have sing + * coszr = —, cOsz a * ; *c¢ COSZ sinz cosz -+cosz sinz = b] : , CCOSZ or sin(c-+2z) = . (384) (383) makes known z, then (x-+ 2) is determined by (384) and finally z. For an example, let /216 « sing + ./72 + cosz = 12; we find x = 15°. PROPOSITION XVII. To resolve the equation. sin(z +k) = m sin(x + 1), We have (362), sin(z + sin(t +k) _ _ sin[x +4(k+ 1) +4(k —1)] sin(z+1) ~~ sinfa-+ 4(k + 1) —4(k—1)] _ tanfe+4(k+1)]-+ tand(k—1) | ~ tanfa +4(k+ 1)] — tant(k — 1)’ ) tanf7+4(k + 1)] = Sa , tana! —k), l) or tan[z +4(k + A = rn: + v) tand(Z » fos putting (354) = tanv. 192 PLANE TRIGONOMETRY. The second form will be preferable, when m is such a quantity as to be most readily computed by logarithms. ‘What will the equations become when k=0? when /=0? The student may form an example for himself PROPOSITION XVIII. To resolve the quadratic equation, 2 apr = qq °° when p and q are such as to require logarithmic tables. We have c+p==+(p?+q)'=+(p?+p? tan*v)? [putting g= p? tan’v] =-+ p(1+ tan?v)§ =+ p seen, a acljaiediate em 2 or r+ p Sarak tanv Phas (385,) Example. Given 25'35sin86 ‘25° 2 © 357sin86‘25° an 4 Qe a aaa © = See eo a eee sinl61‘75° sin7d‘5°sin161°75° to find zx. SECTION SECOND. Resolution of Triangles and Mensuration of Heights and Distances. PROPOSITION I. The length of a line and its inclination to a second line being given, to find its PROJECTION upon that line. : Let a be the given line and / the line upon which its projec- 2 tion is to be made; drop the =... JU a perpendiculars p, p, from the Gl) rZ iti fe o extremities of @ upon; then CEP y*) 2 a,, the portion of J intercepted Fig. 58. between these perpendiculars, is called the projection of a upon l. Through the extremity of a nearest /, draw a; parallel to a, and terminating in p,, then a,=a; Also produce a to intersect J, PLANE TRIGONOMETRY. 193 making the angle (a,/); from the angular point and on the pro- duction of a, measure off the radius, r= 1, and from the extrem- ity of r drop the perpendicular intercepting the cosine, cos(a,/) ; then Gg inl) it COS( db): 1; or ligy 30.2%. COS(Gst) 2 4.3 dy = a cos(d,l), i..€., “The projection of a line is found by multiplying the (386) line into the cosine of its inclination to the line upon which it ts projected. PROPOSITION II. To find an equation as simple as possible that shall embrace the relation existing between the sides and angles of a triangle. Let the sides of any triangle be denoted by a, C b, c, and the angles respectively opposite by A, ‘ j B, C. Then, dropping a perpendicular from C, we have (386) dz = acosB, Fig. 59. and b, = beosA ; s c=a,+ b,= acosB + bcosA, i. e., Either side of a triangle is equal to the sum of the pro- (387) ducts formed by multiplying the two remaining sides into the cosines of their respective inclinations to the first mentioned line. This proposition, obviously little else than a corollary from (386), may be regarded as the fundamental theorem in the resolution of triangles ; since it gives at once the equations, a=beosC'-+ ccosB, b = acosC + ccosA, } (387,) c=acosB + bcos ; from which, by elimination, all possible relations among the sides and angles may be drawn. If one of the angles, as A, become greater than 90°, the corresponding side, b,= bcosA, will be minus (338) .. c=a,—b, = acosB — bcos, and the theorem still holds good. 13 194 PLANE TRIGONOMETRY. In order to find an equation embracing but the single angle A and the sides a, 6, c, eliminating cos C between the first and sec- ond of (387,), and multiplying the third by c, we have a* —b* =accosB—becosA, and c* —accosB+ becosA; 424 c? —a?+b*=2bccosA, or 6?+¢?—a*+2becosA; 1. e. ° PROPOSITION III. The sum of the squares of any two sides of a triangle (388) ts equal to the squares of the third side increased by the double pro- duct of those two sides multiplied into the cosine of the angle which they include. a? +b* —c*+2abcosC a? +¢*—b*+2accosB (388, ) ; b? +c —a*4+2becosA J Cor.* a* +6%+c? =2abcosC+ 2accosB+ 2becosA. (389) PROPOSITION IV. To transform (388) so as to be convenient for the logarithmic com- putation of A. Combining (388) and (324) we have 2.44 MQyne ead. =cosA=1—2 sin?4A, 2be a®—(b?+c?) a*—(b—c)? Pic rte gets Si i sae lc 2be 2be _(a+b—c) (a—b+c). ‘i 2bc _ [(at+b—e) (ate—b)]4 a a a | 20 6 ae a eye (390) putting hA=4(a+b-+c). So costA= irae ‘. (391) * Ixxpress in words. PLANE TRIGONOMETRY. 195 and .*, tan, $A = eo. (392) a) Also, taking the double product of (390) and (391), we have (321), ; 2[h(h — a) (h— b)(h—c)}* sinA = he ; (393) sind _ 2[h(h—a) (hb) (h—c) “fet abc sinB 2[h(h—}) (h—a) (h—c)}* so SE b bac Snes = SRS ora tod. 3,2,9i0.A ‘s sinB ; i. ¢. PROPOSITION V. The sides of a triangle are to each other as the sines (394) of the opposite angles. If the angle C = 90°, or B be the complement of A, then ~ a sinA sind —— = = -—— = tan A, or 6b snB cosA ? Cor. a =5b tanA, (395) Again (394) gives (40,3°) a+b: a—6::snA-+sinB : sinA —sinB, or (333) a+b: a—b:: tan}(A+ B) : tant(A—B); ie. PROPOSITION VI. The sum of any two sides of a triangle is to their dif- (396) ference, as the tangent of the half sum of the angles opposite to the tangent of half their difference. The abuve theorems are adequate to the solution of all problems in Plane Trigonometry ; and these problems may be reduced to one or other of the four following cases : Case. I. The angles and one side of a triangle being given, to find the remaining parts. Rule. As the sine of the angle opposite the given side, Is to the sine of the angle opposite the required side ; So is the given side ‘ To the required side. 196 PLANE TRIGONOMETRY. 1°. As an example under this case, let it be requir- ed to find the distance, z, of an object rendered inaccessible by the intervention of a river. For this purpose I measure a base line of 10 chains, and “j taking the angles at its extremities, I find them to Fig. 60. be, the one 80°, the other 70°, We have the angle at the object = 180° — (80° + 70°) = 30°; sin30° log. sin30° = 1'69897 [—] “sin70° log. sin70°=1':97299 [-+] 10 log. 10= 100000 [+] mv Aals.s, log.caus ~ 127402 *, @ =18'794 chs. Case II. Two sides and an angle opposite one of them being given to find the remaining parts, Rule. As the side opposite the given angle, Is to the side opposite the required angle ; So is the sine of the given angle To the sine of the required angle. 2°, To illustrate Case II., in the triangle ABC, let AB= 1356 chs., BC = 7 chs., and the angle A = 25‘3°; required Z C. Operation. 7 0‘84510 13°56 113226 sin25‘*3° 163079 ~. sinC, 191795 C = 55‘877°, or = 180° — 55‘877 = 124'123°. The ambiguity of the angle C will be illustrated by dropping the perpendicular BP upon AC, and calculating its length; we have sin90° 100000 »gin25*3° 163079 _ 136. 1:13226 mink? Taine 076305 BP = 5'795; whence, since BP is less than BC, taking the point C’ in AC and on the side of P opposite to C, so that PC’ shall = PC, and join- PLANE TRIGONOMETRY. 197 ing BC’, we have BC’= BC. Therefore the two. triangles, ABC, ABC’, are alike compatible with the conditions of the problem ; and the angles BC’A, BCA [= BC'C], are supplementary. If we had taken A = 445°, we should have found log, sinC = 0413282, and, as a consequence, the sine of C greater than radius, which is impossible. The cause of impossibility will be manifest by com- puting the perpendicular BP, which, being found = 95043, shows that BC is too short, when A = 44‘5°, to form a triangle. The student should make all the computations here indicated. 3°, In order to ascertain the altitude, LT, of a Te tower which I am prevented from approaching, I take a station, A, in the same horizontal plane with its foot, L, and observe the angle of elevation, TAL, at the top of the tower= 35°; then measuring BAX AB = 118 feet directly back from A and in a line CAA with L, I find the angle of elevation TBA = 25° | What is the altitude of the tower, and what was my distance from its foot when at the first station. Ans. LT = 157°74, 4°, An individual, in order to fix the position of a certain point, P, in a harbor, selects a convenient place on shore and measures a base line, AB = 19709 rods, and finds the angles, PAB = 100‘13°, PBA = 37°.. Required AP, BP. . The student will compute AP, BP, and then verify by taking AP, or BP, for the known side and calculating AB. 5°. In order to determine the dis- tance between two places, A and B, situated on opposite sides of a hill, and their relative altitudes, I measure the horizontal line AC = 15 rods, and take the angles of elevation PAK = 37:9°, PCK = 30°, P being a flag on the sum- Fig. 62. mit. I then measure the base, BD = 11 rods, and take the angles of elevation PBL = 40°, PDM = 31‘14°, also the angle of depres- sion BDM = 25°. Case III. Two sides and the included angle given, to determine the remaining parts. Rule. As the sum of the two given sides Is to their difference ; So is the tangent of the half sum of the opposite angles To the tangent of their half difference. 198 PLANE TRIGONOMETRY, 6°. A surveyor, wishing to determine the side AB of a field, ren- dered incapable of direct measurement by reason of an intervening morass, runs the line AC, south 38° west, = '7‘75 chains, then CB, south 25‘8° east, 1015 chains. Produce AC in K and draw the meridian CS; then SCK =88° and BCS = 28'8°, 2(CAB -++ CBA) = 4BCK = 319°, 1015-1 7°75 1'25285 10:15 — 7°%5 038021 tan31‘90 1°79410 ~ tand(A—B) * 292146 1(A—B)= 4° 46’ 46” and 4(A +B) =31° 54’ 00"; i" A = 36° 40’ 46”, The angle A being determined the solution will be readily fin- ished, and we shall find AB, S 1° 19' 14” W, 15246 chs. 7°. Given the following courses, 1°. AB, N 46° E, 35 chs. 2°. BC, N 20° W, 55 chs. 3°, CD, S 35° W, 45 chs., to determine DA. [Verify by employing the computed value of DA to: find AB already known. | Case IV. The sides given, to determine the angles. Rule 1. log. sintA=}{log. (A— b)-+log. (h — c)—(log. b+log. c)]. 2. log. cosgA=t[log. h+log. (h—a)—(log. b+log. c)]. 3. log. tantA=t[log. (A—b)-+log. (h—c)—log. h—log. (h—a)]. 4, log. sinA={log. A+-log. (h—a)+log. (A—b }rlog. (A—c)] +log. 2—(log. b+-log. c). 8°. Let it be required to find the angles of a triangular field, the sides of which are 3001, 2672, 199 feet. — PLANE TRIGONOMETRY. 199 First Operation. a = 30025 | h —b = 115°7%5 206352 f+] b = 26775 | h —c = 18450 226600 [+] c = 19900 b = 267°75 2,42773 [—1] 2)76700| ¢ = 199-00 229885 ] 38350 2)1+60294 log. sintA= 1'80147 1A = 39° 16’ 40:5” A= 78° 33’ 21”. Second Operation. h = 38350 258377 h—b=115°% 206352 a = 30025 2,47748 c = 19900 2:29885 2)1‘87096 log. cosk}B= 198548 B=60° 55’ 40”, a Third Operation. h—a= 83°25 1:92038 h—} = 115° 206352 h = 38350 2°58377 h—c = 18450 226600 2)1+13413 log. tant$C = 1556706 C = 40° 30’ 42” But B= 60° 55’ 40” and A = 78° 33.21; aS A+ B+ C=179° 59’ 43” diff. from 180° 00’ 00” by 0° 00’ 17’ = the sum of errors. The third operation for the computation of C is obviously un- 200 PLANE TRIGONOMETRY. necessary, unless we wish to test the accuracy of the work, We have employed three methods in order to illustrate the rules ; sometimes one and sometimes another will be preferable, according to the numbers, 9°, Required to determine the angles of a quadrilateral field from the following data: AB=56, BC = 76, CD =87, DA = 43, BD = 67. 10°. Given two sides of a triangle 367'23, 273 chains, and the difference of the opposite angles 15‘7°, to determine the triangle. 11°. Given the sum of two arcs and the ratio of their sines, to determine the arcs. Let the arcs be denoted by u, v, their sum by e, and their ratio by r; we have u+v=e, and sinv=r sinu, sinv tanw we - =r, which put = tanw = ; sinw Aare i. 22 ‘ sinu-+sinv 1+r_ 1-+tanw — + sinu—sinv l—r 1—tanw’ tand(utv) +r qc ee = tan(45° w 1—r tante tan}(u—v) = —— « tante = 2 x ) ea at. 7 “> tan(45° + w) : which makes known 4(u—vd), but H(u+v) =e, whence w and v are finally determined. The student may make an application of the above in the solu- tion of the following problem, taken from Davies’ Legendre : 12°. ‘From a station, P, there can be seen three objects, A, B, and C, whose distances from each other are known, viz., AB=800, AC = 600, and BC = 400 yards. There are also measured the hor- izontal angles, APC = 33° 45’, BPC = 22° 30’. It is required from these data to determine the three distances PA, PC, and PB.’’ The angles CAP and CBP will be the w and v of the eleventh. The student will make the computations, and devise means to sat- isfy himself of the correctness of his results. 13°. Wishing to ascertain the distance, AP, to an inaccessible object, P, also invisible from A, I measure to the right and left the QUADRATURE. 201 equal lines AB= AC = 21°37 chains, and the angles, BAC = 113° 12, ABP = 65° 36’, ACP = 89° 5. How might the principle of this problem be applied to deter- mine the distance of the moon, her zenith distances being observed by two astronomers, one at St. Petersburg, and the other at the Cape of Good Hope ? SECTION THIRD. Quadrature of the Cirele, the Ellipse, and Parabola. PROPOSITION I. To find the area of actrcular sector in terms of its radius and arc. The sector y is obviously a function of its arc Q zx, the radius, r, being a constant quantity. Giv- Jo ing to z and y the vanishing increments, A and k, asi we have (311), ps mele 3 [kh] =4r -[h]; Fig. 64. rr ys [=] =$7r sors": ., (246), y =4trx-+ constant, but You = 03 ; Gard, les The circular sector is measured by half the product of (3897) its radius and arc, Cor. The area of a circle is equal to the product of the (398) radius multiplied into its semicircumference. Scholium. The celebrated Problem of the Quadrature of the Circle* is evidently reduced to the following proposition: PROPOSITION II. The diameter of a circle being given, it is required to find the circumference. * See Montucla, ‘‘ Histoire des Recherches sur la Quadrature du Cercle.” 202 QUADRATURE. If in (3871) we make z =7, there results y = +tcircumference = (1—4-+4—4t+5 —s wu)? 5 (circumference), =4(1—4+4—F+, —, «.) © 27. So (ctrcumference),.= 4(1 —4+4—t+5 5 en) © Br for any other radius, r, ; : (circumference), 2r_ Tr. ; (circumference), rz 12’ and if 2r, = 2, and .*. (378), (circ.),. = 2m, we have circumference), 2r : icing ee me, =—~=T17 5 1. €., Qr 2 The circumference of any circle bears to its diameter (399) a constant ratio. (circumference), =T + 2r = 3'1415926535897993 « 2r. Cor. 1. The arcs of similar sectors are to each other as (400) the radii of the respective circles of which they form like parts, For, if uw, w,, denote the arcs of similar sectors, or u like parts, as the nth, of the circumferences of which 7 the radii are r, 7., we have NU=T ¢ Br, * and NUg = Te aN. 3 ‘ 2 i r Fig. 65. iin ME Eliminating the semicircumference from (398) by aid of (399), there results, Cor. 2. Area of circle = tr?; or, the circle is measured (401) by the square of the radius multiplied into 7. Cor. 3. Circles and the like parts of circles, as similar (402) sectors and segments, are to each other as the squares of their proportional lines—such are the radii, diameters, circumferences, similar arcs and their chords, sines, tangents, secants, and versed- sines. For let S-+-s and S,-++s, be similar sectors with their equal angles made vertical, s, s,, the similar seg- ments cut off from the sectors by the chords c, ¢,, T, Tx being the radii, Then S+s, S,-+s,, being like parts of their respective circles (?) if n(S-+s) represent the first, n(S,-+ s,) will represent the second circle ; and we shall have (401) QUADRATURE. 203 a(S -+-s)=tr?, and n(S, + s,) = tr,” ; nS+s) S+s_ r? S hig tN A = -—, and (160) = —; n( St S:) S+s, 1,” ( ) S, ‘ S+s S,+s, Ay Sivoujao® ol Sf oe lag 8, Oy aise MAUS coe Shacks 4 2 _., S#+s _ ntS+s) s, 8,” *. S,+% n(S,-+ 2)’ which shows that the segments, sectors, and circles are propor- tional. Further, denoting the arcs of the sectors by u, u,., and ., the circumferences by nu, nu,, we have Be. te Oe sinw — tanw secu NU, UU, . CC, Bing, tau,” sect, Nero eT faye (ag) Ue ee BIg n(Sits) 8, Te (nr)? (nw)? ue oP sin’ Q. E.-D. PROPOSITION III. An Incremental Vanishing Arc of any continuous (403) curve, is to be regarded as a straight line. Let z be such an are, c its chord, and t, t,, tan- zat gents at its extremities, P, P., that is, coinciding c p. in direction with z at P, P., and terminating in Fig. 67. : their point of intersection. We have (387) = tcos(t,c) + t,c0s(t,C) ; but it is obvious from the implied condition of continuity, that as the arc z decreases to its vanishing state and the points P, P,, ap- proach to coincidence, that the tangents t, ¢,, chang- a 26; ta ing their directions by insensible degrees, will Me ‘ ‘ Fig. 68. come to form one and the same straight line, and the angles, (¢,c), (t,c), decreasing to zero, their cosines become, [cos(t,c)]=1, [cos(t,,c)] = 1; oe [e] =[4]+ [4]; SPH and (113), Zz ay : Q. E. D. Scholium. Itis obvious that if the curve were ~# other than continuous, no such conclusion as the above would result. we 204 QUADRATURE. PROPOSITION IV. To find the Derivative of the Segmental Area of any contin- uous curve referred to rectangular cobrdinates. Let Y be segmental area in question, and K its increment ; we have (403), (146), | y “ine . a Fig. 70. [K]=(h] Hy y+(a)s - [F ]=9+ 408, but i =05 y'= [=|- y, which = fz ; i. e., The derivative of the segmental area of any continu- (404) ous curve is equal to the ordinate of that curve regarded as a function of the abscissa. [Y' =y = fx.] PROPOSITION V. To find the area of the Ellipse. Let Y be an elliptical segment embraced by the semi- minor axis and the abscissa 2; we have (404), (203), (250), ial dG) G8) fet ) 2.3 eee) ep ah — 1) ip ° =< ween ig eae ._— — e Gee): ¥=0(2 i 3a s" Bal 2.3 az apshat (405) the area sought, no constant being added, since Y,_, =0. If we make } =a, there results ae aly e411) (FQ) = —t-. ae 6 toa beak a Y. a2 yee 2 5a! 2.3 x7 wget? gd np (406) the corresponding segment of the circle circumscribing the ellipse ; Fig. 72. QUADRATURE, 205 b a, Fed Xe it 02 Coe aoe a Y,. [Enunciate. ] (407) Making z =a and multiplying by 4, we find 2(z— 1) (Ellipse), ,= a(1 BHD pe | rn) sik eae a any m)aes (409) Ellipse : Circumscribing Circle :: 6: a@:: 2b : 2a :: Minor Axis : Major Axis. (410) (Ellipse),, , = Ma (Circumsc, Cir.) == eta?=abert. (411) (Circle), = mb? ; . (411), (Liltpse),,.2° (Circle), £: a:; 6. (412) PROPOSITION VI. To find the area of the Parabola. We have (232) y’ = y = QUyra® 3 $+1 Y=23p7 . it = 2+ Wprrte = tyr; ie, (413) 2 The Parabola is two-thirds the circumscribing rect- (413,) angle. Scholium. It is obvious that the quadrature of the circle and ellipse can only be obtained approximately, while that of the parabola is exact. Fig. 74. 206 PROXIMATE AREAS. PROPOSITION VII. To find the proximate area of any continuous curve. Suppose the equation of the curve, referred to rectangular co- ordinates, to be y= Ay+ A,r t+ Anz? + Agxi +... 5 (414) then Yo=y=A,+ A,r+ Apr?-+ Azz? +... Y=A.r+4A zr? +4A,7?+14A,7' +... + constant, but, if the area Y become =0 when x=0, which condition is always admissible, since the origin may be taken at pleasure, we have Y= A.r+4Aj,r?+4A rz? +4A,7!t+... (415) But, since three points, P,, P,, P., determine with considerable accuracy a curve of moderate extent, we will take the foot of the first ordinate, Yo, as the origin, and the abscissas, 2, =h, x, = 2h, so that the corresponding ordinates, y, Y,, Y2y shall be equally distant from each other; .*. mak- ing z=0, h, 2h, we have (414) Yo= Ay, =A,+ A, + h-+ Ay + h’, Yo = Ay + A, ¢ 2h-+ Ay » 4h? ; “ Yi—Yo=A,+h+ A, ae and Yo—-Y= " oh-+ A, A —2y, + Yo = Ay ¢ 2h? ; but when z= a we have ae Y= A, wate 4h? +44, « 8h? = 2h(A, + Ash + 4A, « h*); ae = PAY +3 * HY2— 2% + y)], or = $h(ty + 24, +44). (416) lf we pat ¢ the same method of admeasurement and notation, we have = 3h pe cain. Y,=#h (44. + 2y,+ 44), Sh (44, + 2y;+ ty), &c. &c. Y;, — Bh(4Y 202 + 2Y2n—1 + SY on) 3 EXERCISES, 207 oe Yop ¥yt Voto + You. = F4(Syo + 21 + Yot 2ya+ ys Ht Syst vee + Yonot 2Yan—1 + FY 2n)« (417) This beautiful and useful theorem is due to Simpson, The stu- dent should enunciate it in common language. EXERCISES. 1°, Required the diameter of a circle having ten linear chains in circumference to every square chain in area, 2°. A square plate of silver, 3 inches on the side, is worth $4. What is the value of the greatest circle that can be cut from it? 3°. Had the plate in 2° been an equilateral triangle, what would have been its value? 4°, The two sides including the right angle of a right angled triangle, are three and four rods; what is the area of the circum- scribing circle? 5°, Determine a circle circumscribing an isosceles triangle, the two equal sides of which, including an angle of 36°, are 15°15 chs, each ? 6°. The equal sides of an isosceles triangle embrace an angle of 473°, and the area of the inscribed circle is one acre. Determine the triangle. 7°. A circular plate of brass, 20 inches in diameter, is worth $3‘75. What is the value of the three greatest and equal circles that can be cut from it? 8°. Required the area of a circular segment embraced by an arc and its chord, the length of which is 5‘87 chs., and the breadth 1:35? 9°, The dimensions of an elliptical fish-pond are 10 and 15 rods. What is its area? 10°. The ordinate of a parabolic segment is 3 chains, and the corresponding abscissa 7 chains. Required the area. 11°, Required the area of an elliptical segment embraced between the semiminor axis and an ordinate = 5‘657 chains, and having a breadth of one chain. 12°. Wishing to ascertain the cross section of a river 100 yards from water’s edge to water’s edge, I take soundings every 10 yards, and find them to be in yards: Y= 0, y, = 12, y, = 2041, y; = 253, y, = 284, ys = 29'9, yg = 2959, Y, = 261, ys = 209, y, = 128, y,) = 9. BOOK THIRD. SURVEYING. SECTION FIRST. Description and Use of Instruments. To determine the boundaries of lands, to delineate them in maps, and to compute their areas, constitute the Art of Surveying. The instruments employed in determining the boundaries of lands are, the Chain, for measuring the lengths of lines, the Sur- veyor’s Cross to determine right angles, and the Azimuth Com- pass, or the Theodolite, for fixing the inclinations of lines to each other and to the meridian. The Chain. Gunter’s chain, as has already been observed, is 4 rods or 66 feet in length, and centesimally divided by a hundred links, each, con- sequently, equal to 792 inches. It is a maxim in land surveying that every instrument, whether for measuring lines or angles, must be used in a horizontal po- sition; for it is the base, or the projection of the field upon the same horizontal plane, that is desired, The projections of the bounding lines are usually obtained by carrying the chain in a horizontal position, as represented in fig. 77, where 1, 2, 3,..., are the successive positions of the chain. When the inclination of the ground is too great to admit of a whole chain, a half or a quarter may be taken, and in all cases the proper position of the elevated extremity should be determined by a plumb line or by the dropping of a stone. CHAIN, 209 The chief points to be attended to in chaining are, 1°, to keep the chain in a horizontal position ; 2°, to avoid straying from the line ; 3°, to record without error the number of chains. The first condition will be secured by the surveyor supporting, when necessary, the middle of the chain, and directing the elevation or depression of its extremities. The second condition will be readily attained by the foreman fixing his attention upon the flag-staff or object of sight, thus draw- ing the chain constantly in line. His march will be corrected by the hindman, who will cry out “ rieuT !”’ “ LEFT!” as the occasion may require. Error in record will be guarded against by employing ten iron pins which should be turned at the top into a small ring, and tied with a piece of red flannel, the better to be seen. The foreman takes the ten pins and draws out the chain, the hindman, as it is near being stretched, cries ** Down !”’ when the foreman, giving the chain a wave to bring its parts in line, pulls it tight and puts down a pin. Marching on he repeats the same operation, until, coming out empty-handed, he puts his foot upon the extremity of the chain to secure it in place, and cries ‘TALLY oNE!’’ and the hindman responds ‘‘ tally one!” that the number may be fixed in the mem- ory, also recording it in some way, as by a notch in a stick or a pebble put in the pocket, if thought necessary. He then, quitting the hind end of the chain, marches up to the foreman, who counts the pins to assure himself of the reception of the ten; when, stretch- ing on, the second tally is executed like the first. A field may be surveyed by the chain alone, as illustrated by the subjoined Field Notes. Contour. AB = 237, BC=4‘67, CD=5‘00, DE=4'98, EA=367, Diagonals. BD = 4:83, BE = 5°25 chs, Required the angles of the pentagon. The question naturally arises: ought we not to measure the in- clined plane rather than its horizontal projection, since the surface of the former exceeds that of the latter? The answer to this ques- tion must be given in the negative, and for two reasons; 1°, a uni- formity in surveying different lands is desirable, that they may be the more readily compared with each other, and it is obvious that this uniformity can be attained only by reducing them to their horizontal projections ; 2°, the real value of a field cannot exceed 14 210 THE COMPASS, that of its horizontal projection, since no more soil will rest on an inclined plane than on its horizontal projection, and the same num- ber of plants will stand upon the one as on the other. For let abcd represent a vertical section of the soil of an inclined plane; it is only equal to its hor- izontal projection efgh, the vertical depth ad being supposed equal to eh; and trees will grow as thick together on the inclined as on the horizontal plane. The Surveyor’s Cross. The Cross has been already described and the method of using it in surveying pointed out. [See Book Second, Section First. ] The student may now employ the chain and cross to survey a small field, and then compute its angles. The Compass. The Surveyor’s or Azimuth Compass consists of : a horizontal circle to which are attached sight-vanes, and a magnetic needle delicately balanced on its cen- tre, by which the vanes may be directed to any point of the horizon so as to determine the inclination of lines to the magnetic meridian, and, consequently, to each other. The degrees, marked on the limb of the instrument, are numbered from the north and south points, N, and S., both ways to the east and west, designated by the opposite letters, W. and E., for a reason that will presently appear. To use the compass, set it firmly upon its staff (better and usu- ally a tripod), furnished with a ball and socket joint, capable of being loosened or tightened at pleasure, by the aid of which and two. spirit levels, placed at right angles to each other on the face of the instrument, the limb is to be brought into a horizontal posi- tion, When this is effected will be known by the bubbles remain- ing in the middle of the levels while the instrument is made to revolve on its axis. The needle is now to be let carefully down upon its pivot by a screw in the under side. See thatit plays with its points just skimming along the graduated edge of the limb, Turn the vanes into the required direction by sighting at a staff wound with a red flag, and held vertically in line by an assistant stationed at a suitable distance. Observe if the needle settles with Fig. 79. THE VERNIER, 211 a free motion, describing nearly equal arcs, slowly decreasing, on each side of a definite point, and if it finally rests at that point. Should there be any doubt as to this, the needle must be agitated, either by the attraction of a knife, or by gently tapping the tripod with the fingers, and be permitted to settle a second time. To in- sure a correct position of the needle is the principal difficulty in operating with the compass. In order to prevent mistakes, that sight should be turned toward the flag staff which will bring the north end of the needle into the part of the compass marked N., or with the fleur de luce. The bearing will then be read off by the forward end of the needle, using the letters it stands between, as in the figure, N. 30° E., if the course be northerly, or S. 30° W., if in the opposite direction ; both ends, however, are to be observed in estimating the amount of the angle. Back sights should be taken at each station in order to verify the bearings. The Vernier or Nonius isa slip of metal, fitted to slide upon the graduated limb of an instrument, and to serve the purpose of an extended and impracticable subdivision. If x denote the value of a division on the vernier, of which 2 cover 21 divisions of the instrument, we have nmz=n+t1, z=l1t- 9 n fen n 17 from which it appears that the distance, x —1, of the first division of the vernier from the first division of the instrument, will be an nth part of the unit of graduation—the distance, 2z — 2, of the second division of the vernier from the second division of the limb will be two nths of the unit of graduation, and so on; so that, by sliding the vernier along the limb, we shall be enabled to measure spaces to the nth part of the smallest divisions of the instrument. Thus, if a scale [eG Ee bb pd have a division of 10ths, and [Jo |) [2 [3 [a [5 |e |7 [s [9 [wd 11 of these be covered by a Fig. 80. 212 THE THEODOLITE. vernier of 10 divisions, we have n= 10, and .. r—1 =); of +4, Qr —2=-+4, of yj, ...3 so that, by sliding the vernier along the scale to make the points, J, 1; 2, 2; 3, 3, ..., agree in succession, there results the measures ‘O01 ; ‘02; ‘03, .... In like manner, if 30 divisions of the vernier attached to the compass, cover 29 half- degree divisions of the instrument, we shall be enabled to measure angles to the minute. In some compasses the vernier is attached to the extremity of the needle, and, being carried along by it, the degrees and parts of a degree are read by simple inspection ; in other instruments it is on the outside of the limb, and fastened by a clamp-screw from below, which must. be loosened, when the ver- nier will be driven by a tangent screw adapted to the purpose. The Theodolite. The Theodolite consists essentially of a verti- [ Frontispiece. | cal and horizontal circle, for the purpose of measuring angles in altitude and azimuth. It has, like the compass, spirit levels, by the aid of which and screws, the azimuth circle may be brought into a horizontal position. When this is accurately accomplished, the theodolite is ready to measure any horizontal angle having the angular point at its centre, provided all the parts of the instrument have been carefully adjusted, To this end, direct both the upper and Jower telescopes (the first attached to the vertical circle, the second to the axis below the horizontal) to the same mark situated at a distance in one of the sides of the angle. Observe the posi- tion of the vernier upon the limb of the azimuth circle, reading the degrees and parts of a degree by one or more microscopes, fitted to this end—unclamp the upper telescope and direct it to a mark in the second side of the angle, clamping and finishing the motion by aid of the tangent screw. Observe, by the lower telescope, whether the azimuth circle has suffered any displacement by the motion required in making the second observation; if no such de- rangement has happened, the difference of the first and second read- ings will be the measure of the angle in question. In order to secure greater accuracy, the axis of the azimuth circle may be unclamped, the upper telescope brought back to the first mark, carrying the azimuth circle along with it—the azimuth again clamped, and the angle measured a second time. This operation repeated as often as desirable, the whole amount of are passed over divided by the THE THEODOLITE. 213 number of observations, will give the angle required with a corre- sponding degree of exactness. Wescarcely need say that the eye- glass must be drawn out or pushed in till the cross wires, indicat- ing the line of sight—line of collimation, as it is called—shall be seen distinctly, and that the object glass is to be moved in like manner till the mark becomes well defined. Ina similar way, a vertical angle will be measured by the vertical circle, having pre- viously brought the telescope to a horizontal position and observed the reading, which should then be zero. When this horizontality shall be accomplished, will be determined by a level attached to the upper telescope, or, if no such level exist, by those already mentioned. There are several adjustments, either permanently fixed by the in- strument maker, or, for the execution of which, he furnishes means in screws and parts capable of being detached from each other. There are five lines that should be respectively perpendicular to each other, viz., the vertical axis, or axis of the azimuth circle, the horizontal axis, or axis of the vertical circle, the horizontal line, or line of collimation when the vertical circle indicates zero, the ver- tical wire, and the horizontal wire—or Z (Ay A.) = (H,A,) = (Hy An) = (bs As) = (0,4) =(h, Hf) = (0,1) = 90°; also the circles should be perpendicular to their ¥ axes. The method of testing these adjustments consists principally in reversing the lines (73) ; for which purpose the telescope and horizontal axis, one or both, may be lifted from their ys or supports, the object Fig. 81. and eyeglasses change places, or the vernier plate carrying the vertical circle and the telescope, revolved 180°. Also the wires, h, v, which form by their intersection the line of col- ic limation, being attached to a ring, may be moved to the right or left, elevated or depressed by screws from without. If we direct the line H to a distant and well-defined mark, and, when the telescope is reversed, find the sight upon the same mark, we may be assured that the line of collimation, H, is perpendicular to the horizontal axis A,. This adjustment perfected, the horizontal axis, A,, will be perpendicular to the vertical, A,, when, passing the line of collimation, Fig. 81s. 214 VARIATION OF THE NEEDLE. H, through two marks having a considerable angu- lar distance, and turning the vernier plate 180°, H continues to pass through the same points. The vertical wire v will be in a plane perpen- dicular to the horizontal axis A,, when, moving the vertical circle backward and forward, a distant and well-defined mark continues accurately on v. In order to know whether the horizontal wire, h, be in its proper position, or if the line of collimation, H, when the vertical circle indicates zero, be perpendicular to the vertical axis, A,, it is only necessary to reverse the telescope, and that the wire h is in a plane parallel to the azimuth circle, will be determined by a backward and forward motion of the vernier plate. When these adjustments shall have been perfected, by often repeating them one after another in a different order, whether the levels are parallel to the plane of the azi- mauth circle will be known by leveling this circle, making the vertical circle indicate zero, if its telescope have-a level attached to it, re- volving the vernier plate and seeing if the bubbles continue in the middle. Whether the level which may be attached to the telescope be perpendicular to the axis A,, will be known by bringing A, over two of the leveling screws, and then, by aid of these screws flinging A, out of level, or by revolving the telescope in its Ys, if it be capa- ble of such a motion. If, on any occasion, it be desired to make the vertical circle coin- cide with the greatest possible accuracy with a vertical plane, we may suspend a plumb line before the telescope and observe when the line of collimation traces this line. The Variation of the Magnetic Needle may be conveniently de- termined with the theodolite by the process of equal altitudes. Let the magnetic bearing of the sun before noon at a determinate altitude be e°, and at the same altitude after noon f°, and suppose x = variation; then will the angular distances of the sun from the true meridian be, before noon, e+ 2, afternoon, f= z, but these distances are equal, f-e a" The equal altitudes may be taken at any corresponding hours be- fore and after noon, and in any season of the year, but the most favorable time is about the 21st of June, when the sun, being near the summer solstice, will change his declination but a few seconds Fs te ee oa A LEVELING, 215 between the observations ; and, if these determinations be made several successive days, both before and after the 21st, the errors, lying in opposite directions, will tend to balance each other; thus, suppose the values of z for June 18, 19, 20, 21, 22, 23, 24, to be DLiy Lay V33 Ly, V5. Ve. B73 then GS peakais Mi eta aL oe 4 The observations may be made several times during the same day, but the best hours will be when the sun is in a position nearly east or west. The Art of Leveling consists in finding the difference of eleva- tion between two places: and is not only necessary in the con- struction of railroads, aqueducts, and canals, but is useful for laying the foundations of edifices, and for other like purposes. The op- eration may be readily executed by aid of the theodolite, or, more conveniently, with the Leveling Instrument made expressly for the purpose, and consisting of a telescope, level, and tripod, being in all respects similar to the theodolite, except in not possessing the graduated circles. The instrument is to be firmly planted midway between two stations, situated at a convenient distance, and its tel- escope made to revolve accurately in the horizontal plane, when the depressions of the first and sec- ond stations are to be noted by sighting at a graduated staff held vertically on these two points in succession. In like manner the second station is to be compared with a third, the third with a fourth, and so on, to the last. For an example, suppose it required to find the elevation of the fountain, A, above a dwelling house at B, from the following notes, in which the back altitudes are marked + and the forward ones —. 216 PLOTTING. | Ans. 10°29 feet elevation. SECTION SECOND. Plotting. PROBLEM I. Througn a given point to draw a lineperpendicular to a given line. Join the given point, P, and any convenient point, Q, of the given line ; with the middle, O, of the line PQ as a centre, describe the semicircum- ference PP’Q, cutting the given line in P’; the line drawn through P, P’, will be the perpendicular ag required, Why? , If the given point be P’ in the given line, set one foot in P’ and the other in any convenient point, O, out of the line, deseribe a circle and draw the diameter QOP ; PP’ will be the perpendicular required. Other methods of drawing perpendiculars may be employed, as in- dicated in the figures, but the “* Right-angle”’ is preferable in practice. Fig. 83. Fig. 833. Fig. 834. Fig, 839. PLOTTING. 217 PROBLEM II. Through a given point to draw a line parallel to a given line. Set one foot of the dividers in the given point, P, and with any convenient centre describe the circumference PBAX, cutting the given line in A and B; take the chord PB and apply it from A upon the circumference at X ; PX will be the Fig. 84. parallel required. (Why?) Do the same thing with the Right- angle and Straightedge. PROBLEM Iii. From a definite point in a given line to make an angle equal to a given angle. : Around the given angle, A, with any convenient x B radius, AB, describe the arc BC; around the given age point, P, with the same radius, describe the are © a2 QR, Q being a point in the given line; apply the & \ chord BC in QR; QPR will be the required an- P Q gle. Fig. 85. PROBLEM IV. To construct a triangle, having given its three sides. Draw an indefinite line, KL, in the required posi- Q tion, and apply one of the given sides, C, from K to , NI L; with the other sides, A, B, as radii, describe around the centres, K, L, arcs intersecting in Q; eaana Viana KQL will be the triangle required. Fig. 86. Scholium I. This problem enables us to plot a field when it is surveyed by the chain, that is, when its diagonals are known, either by actual measurement or by computation. The student will find it a profitable exercise to plot the pentagon given in the preceding section under the chain. Scholium II. The preceding Graphical Problems give us the six following Problems of Construction, whereby any geometri- cal problem, solved algebraically, may be executed in a geometri- cal way. 218 PLOTTING, PROBLEM V. To construct the sum of two lines, zr=a+b, Lay off 6 upon the production of a. Ch mere ote We ae Fig. 87 PROBLEM VI. To construct the difference of two lines, r=a—b, Lay off 6 from one extremity of a towards the other, as P. a2 from the right hand towards the left in the figure ; the re- Fig. 88. mainder will be = z. We observe that if 6 exceed a, x will be drawn to the .P _ left instead of to the right (180), and that_z will be mi- ' % Fig. 882. nus instead of plus in the equation z = a — 8, PROBLEM VII. To construct the square root of the sum of the squares of two lines, x = (a?-+5?)2, Make aand 6 the sides of a right angled triangle; z /| : Ww ib will be the hypothenuse. 4 a Fig. 89. PROBLEM VIII. To construct the square root of the difference of the squares of two lines. x = (a? —b*)2, On the greater line, a, describe a semicircle, and from the extremity of a lay off the chord 5; z will aT be the chord joining the extremities of @ and d. Fig. 90. PROBLEM IX. To construct a fourth proportional, be @:0 30: 2) Or ax = bc, or 2 = —, a PLOTTING. 219 Inscribe the chord }-+c in a circle of suitable ra- dius, then with a in the dividers and one foot in the junction of 4, c, cut the circumference ; the inter- section will give the position of a, such that its pro- duction, intercepted by the opposite part of the circumference, will be z. This problem may also be solved as indicated by fig, 23,. Scholium. If c be made equal to 8, or if a : b:: 6: a, the problem reduces to that of constructing a Third Proportional. Fig. 91. PROBLEM X. To construct a mean proportional. 1 Nee ape ie sat c, or x = (bc)?. In the above, make 6-++c a diameter, and z per- pendicular to b+ c; then will a of fig. 91 become =z in fig, 92. b Scholium. This problem solves also VIIL., since wy x = (a? — b*)? =[(a +b) (a—)]?, Fig. 92. where xz is a mean proportional between a+b) and a—b; again, the eighth may be advantageously employed in executing the pres- ent—seeing that Q — r\2eas x= (be) =[(~$*) -("$)] : for b is the sum of bei be 2 2 and c the difference of Bene Sp ere Scholium. It is to be observed that an equation, in order to be capable of construction, must be homogeneous (370) ; and, conse- quently, that it must not embrace any unlike quantities, such as lines and surfaces. Further, all geometrical quantities are to be expressed in lines, as these are the only magnitudes which we can measure directly ; thus, a surface S by a square, a®, whose side is a; a solid V by a cube, a*, having the side a; a ratio r by two . m . lines, m, n, r= Teil Such an equation as z*-+ 2azr = b cannot be constructed, not being homogeneous. 220 PLOTTING. As an exercise, let it be required to construct the general quad- ratic equation z? + 2axr = be, where a, 5, c, are known lines, We shall find v=—at[a?+ dep, which may be put under the form 1 e=—a%[a(a+=)]"; *, 1°, construct “ by Prob. IX, ; 2°, construct a an by Prob. V. ; + 3°, construct [=(« +3)] ; by Prob. X. ; 4°, construct 7 = — as+| a (« + Nils by Probs. V. and VI. As a second exercise, let it be required to divide a line into Ex- treme and Mean Ratio, i. e., so that the whole line, AB, shall be to its greater part, AX, as this greater part, AX, is to the less, XB. If the parts be denoted by a+ z and a— @, we shall find t= — 20 + fa(at-atatata))*, or = — 2a + [2a Q14)?, The student will find it a useful exercise to construct the problems which he has already solved, as those at the end of Section Third, Book Second. He should also endeavor to combine, as much as possible, elegance and simplicity in the arrangement and execution of the several parts of a complicated construction. ‘Theorems oth- er than those involved in the foregoing problems, as (182); (185), may frequently be advantageously employed, PROBLEM XI. To divide the Gy RIeT epee of a circle into equal parts, as de- grees. First Method. With the required radius describe a circumference, to which the dividers, unaltered, will apply six times, since the chord of 60° is = to radius. Next bisect the arc of 60° thus found ei- ther by intersecting arcs described from its extremi- ties and a line drawn to the centre, or by trial ; Fig. 93. ° PLOTTING. 221 in like manner bisect this are of 30°, which must be divided into three parts by trial; and, lastly, the same method. Second Method. Having gra- duated a quadrant of large ra- dius, as above, transfer the chords to a single line. This line, engraved upon a ruler, is denominated the Line of Chords. the arc 99 » 2 of 5° is to be divided by 00 &% Chords Nf Fig. 94. To make use of it, we have only to describe a circumference with the chord of 60° as radius, and then to this circumference apply the chord of the required arc, also taken from the Line of Chords, Third Method, Lay off the arc from a scale of equal parts, by aid of the following Table of Chords. Chord 2x = 2sinz, r=1., 30° 0‘347 0365 0535 0°/0-00010«174 1°!10:018|0«192 0‘518/0'684 0*700)|0‘861 2°|10°035/0‘209/0'382/0'5510'71 7/0877 3°!/0°052/0'226|0'399 0'568)0733)|0+892 4°|/0:070|0:244/0416 0:585)0749/|0'908 5°||0087|0'261 |0°433 0601 |0+765)|\0:924 6°||0°105)0+278)0°450 0:618)0782)/0:939 7°|'05122|0‘296|0‘467 0'635/0'798 8°|/0°140)0°313|0°454 0-651 /0+814 '||0°157|05330 0:501 0°668 0*829 0+845 0:954 0‘970 0984 Bley hi 26 12 Xx Six Fig. 95. 80° | 90° | 1000] 1*147|1‘286|1‘414 1015) 1*161) 15299) 1 +426 1*030) 1‘176|1+312)1°439) 1:045] 1199) 1'325) 15451 1*060) 1204/1 +338) 1463 1075) 1'218)1‘351)1°475 1089) 1°231|1‘364/1+486 1'377)1‘498 1118) 1259) 1‘389)1+509 16133) 1°272)1+402/1'521 This method will be found particularly convenient for mechanics who have at hand a ruler graduated into inches and 8ths, or, better, 10ths. Thus, with a radius of 20 inches, the chord of 57° 45’ is 20(0:954 + 2 » 16) = 19°32 inches. Fourth Method. Find how many times the chord of 1° is con- tained in radius, We have chord 1° = 2sin#°® ; log. chord 1° = 03010300 + 3:9408419 = 22418719 ; 222 PLOTTING, rad.[=1} ,, - and radius = 57‘2965 chord 1°, or = 573 chord 1° nearly. Hence a circumference described with a radius of 573 taken from a scale of equal parts, will be readily subdivided, the chord of a degree being one of these parts. Circles, semicircles, and short rulers, made of brass or ivory and graduated on their edges, are also employed for laying down an- gles. A particular description of these instruments, as well as of others, usually contained in a surveyor’s case, such as steel points and pens for drawing blank lines and describing circumferences in ink, is unnecessary, as their use will be obvious on inspection.* But the first method, which requires only a pair of dividers, will be found quite as accurate as any, and perhaps as expeditious, We scarcely need say that all exact drawings should be made witha steel point and afterward inked so far as required. A very conve- nient and good pen for describing circumferences may be had by thrusting one leg of the dividers through a common pen, having cut for the purpose a gash downward in front, just above the clear part, and another upward on the back side, a little above the point. PROBLEM XII. To plot a field, having given the lengths and bearings of its sides. As an example, let it be required to plot the first of the follow- ing Field Notes. Draw a vertical line through the middle of the paper for a me- ridian, the top of which is to be regarded as north and the bottom as south, then the right hand will be marked E., and the left W. * A pair of dividers having three legs will be found very convenient in copying. See Lerebours’ Catalogue. PLOTTING, 223 Next, with one foot of the dividers centrally situated in the meridian NS, describe a circle as large as the paper will conveniently admit of, when the bearings are to be laid off, by the preceding problem, upon the circumference from the north and south points, Na = 365° towards E., Nd = 15‘3° towards W., Sc = 46°, towards W, Laying the perpendicu- lar of the rightangle upon the centre, O, and the point of the circumfer- ence, a, and applying the straight- Fig. 96. edge to its base, transfer the line Oa to a convenient position, a, and, having constructed or chosen a scale of equal parts, lay off a =15*75 chains from the first station, 1, to the second, 2; this done, b is to be drawn parallel to Ob through the second station, 2, and measured off from the same scale, when, the third station, 3, being thus determined, the side c is to be laid down in like man- ner; and lastly d. The Diagonal Scale is con- venient for laying down lines, also for graduating the circum- ference, either by the third or - fourth method. The Sector is likewise frequently employed, on 10 account of the ,facility with which its unit is varied to suit the dimen- sions required in a given plot, sim- ply by causing its branches to ap- io proach or recede like the parts of Fig. 93. a common jointed rule. Thus, if the sector be opened so that from 10 on one side to 10 on the other shall = 1 inch, then from 20 to 20 will = 2 inches, from 21 to 21 will = 2‘1, andso on. A sectoral scale may be constructed on paper for the occasion, and to any re- quired unit, say 10 to the half inch, by first drawing an are witha radius = 4 inch, then running off from the centre, with any conve- nient opening of the dividers, a line of 10 equal parts and from the extremity drawing a secant to the arc, to which the same parts may be applied. 224 PLOTTING, A very small pair of dividers going with a screw, or, simply, a fork cut in metal by a file, or even in hard wood with a knife, will be found convenient. PROBLEM XIII To reduce or enlarge a plot. From any point conveniently situated, draw lines passing through the several stations—then draw, in succession and terminating in these lines, parallels to the sides of the plot. The student will exercise himself in plotting all the following fields, measuring the last sides and their bearings. Field Notes. I. N 365° E | 15°75 chs. N 153° W | 20/00 S 46° W |.30°25 S 58°24° E | 20°781 N 80° 45‘53 chs. N28 65°23 N 86° 57°86 S 32° 50‘00 S 46° 53' 23” 50‘82 S 48*4° WwW S 785° W | 15 N 313° W | 2% N 35‘6° E | 33 SnG6OFes 4 E | 23 S 4° 43'54" E | 23°764 LAST SIDE, 225 IV. N 69:0° W N 284° W N 316° N 568° S 671° S 190° V. In order to determine a line rendered inaccessible by interven- ing declivities, I trace the line ABCDEFG through a neighboring ravine, and find AB, N 23° E, 25 chs.; BC, N, 31 chs.; CD, N 5° W, 10 chs.; DE, N 12° W, 15 chs.; EF, N 10° E, 35 chs., to the top of the ravine, thence to the second extremity, G, of the required line, S 45° W, 51‘87 chs. Required, AG. VI. Find where the meridian, passing through the middle point of the side a of IV., will intersect the opposite part of the perimeter. Ans. At a point in e distant 55 links from the extremity of d. SECTION THIRD. Computation of Areas. PROPOSITION I. If the sides of any polygon be projected on the same (A418) line, the sum of these projections, taken in order with their pro- per signs, will obviously be equal to zero. Thus if the sides AB, BC, CD, DE, EA, of the polygon ABCDEA, be projected on the line LL, in ab, bc, cd, de, ea, we evidently have ab +be+cd+(—de)+(—ea)= 09. And, generally, if we regard the perimeter as described by a point revolving about the polygon 15 226 LAST SIDE. in a constant direction, as ABCDEA, then will the projection of this point upon any line given in position, as LL, describe, by its motion, the projection of the perimeter, which projection, increas- ing and diminishing, will obviously become nothing when the re- volving point returns to its first position, as A, Scholium. The principle enunciated in (180), applied to this problem, serves to distinguish the plus and minus projections, which will be found to be measured to the right or left, corre- sponding with the motion of the revolving point. Thus, as the point revolves through ABC, the projection, measured towards d, decreases, becoming = 0, when the point arrives at D, and then reappears, measured in the contrary direction, as the point returns through DEA. It follows that if a, b, c,..., 7, &, J, denote the sides of a polygon taken in order;:6'y Biy,C 4) p25 9.5 Koad yA MCIr, Pron jections north and south, a”, b", c", ...5 j', k’, Ul’, the corresponding projections east and west, and ma meridian line; then will the sum of the meridional projections, taken with their proper signs, be = 9, as also the sum of the projections at right angles to the me- ridian ; and we shall find a+ B+ c++ j +h +1 =0, : or acos(a,m) +- bcos(b,m) + ccos(c,m) +-.. + jcos(7,m) + keos(k, 2pm l we (419) since a = acos(a,a’) =acos(a,m), b' = cos(b,m), l' = lcos(l,m) ; and a’ +b" 4+e"4+..497°+k"4 1 =0, or asin(a,m) + dsin(b,m) + csin(c,m) +.. te paint gem ) + ksin(k,m) + iain( l,m)=0, > (420) since a =acos(da,a’) = asin(a,a’) = asin(a,m), Ll" = Isin.(1,m) ; asin(a,m) + bsin(d,m) +-...-+ ksin(k,m) acos(a,m) + bcos(b,m) +... -+ kcos(k,m) (421) _ = isin(tm), = tan(J,m) ~ — Icos(l,m) writ dd by which (J,m) becomes known ; SIMILAR POLYGONS. 227 then (420) 7 = a sin(a,m)+-b sin(b,m) +... k sin(k,m) (422) — sin(/,m) and the last side, 7, is completely determined—that is. its bearing and length are drawn from the bearings and lengths of the other sides. Enunciate the above forms. How is the denominator of (421) formed from the numerator ? Cor. 1. If the sides of a polygon, except one, vary insuch (423) way as to preserve their mutual ratios and inclinations constant, then will this excepted side bear to any one of the others a ratio and inclination also constant.* For, denoting the constant ratios which 4, c, ..., k, bear to a by Ty Tey oo Ty OF putting Va? de Cm fap i, de= F,0; which is in accordance with the condition that any one of the sides, a, b,c, ..., k, shall have a constant ratio to any other, since the ratio of 5 to k, for instance, is 6: k=r,a: 7T,A4=7, : Ty we have asin(a,m) -+ ar,sin(b,m)-+-ar,sin(c,m) +... + ar,sin(k,m) acos(a,m) + ar,cos(b,m) + ar,cos(c,m) +... + ancos(k,m) ’ Grpiitat( dan jons sin(a,m) + 7sin(b,m) + 7,sin(c,m) 4-...+ 7,8in(k,m) ; : cos(a,m)-+ 7,cos(b,m)-+7r,cos(c,m)-+...+7%c08(k,m) ’ which fraction is obviously a constant quantity, being independent of a, or of the absolute length of any of the sides, a, b, c,..., k, and depending only upon the constant quantities, 7, 7,5 ...5 Ti sin(a,m), sin(b,m), ..., sin(k,m), cos(a,m), cos(b,m), ... , cos(k,m). The tan(/,m) being constant, the 7 (/,m) is constant; Zs, (l,a), (2,5), (2,c), ... 5 (1,4), are constant ; asin(a,m) + ar, sin(b,m) +... + ar, sin(k,m) — sin(/,m) =(constant)- a. Q. E, D. Polygons, which, like the above, have the same number of sides proportional in the same order and their homologous angles equal, are said to be similar. Cor. 2. In similar polygons, like diagonals are to each (424) other as the homologous sides ; for the diagonal is obviously in the same condition as a last side, J. Cor, 3. The perimeters, or their like portions, in similar (425) polygons, are to each other as homologous sides or diagonals, tan(/,m) = * See Variation, Part 1, Book 1. 228 SIMILAR POLYGONS, For,: a=a/b2=70%c=ra, dS 7, s.¢ k= ra,l=r0; give a+b+c+d+..tk4+la=(l4tntrotrat-. bret ri). Lemma, Similar polygons may be described by the (426) revolution of variable radii vectores, preserving a constant ratio to each other and the same angular motion. For let the poles be made common in O and the radii OP, Op, depart from the same _ [ fig. 99.] axis of angular motion, OaA, then will OpP be a straight line for corresponding points of the polygons; and we have, since by hy- pothesis OP bears a constant ratio to Op and OA, Oa, OB, Ob, OC, Oc, ..., are corresponding positions of OP, Op, OP : Op=O04A 2 Da=O0B8 1: OF =00 s Ces, whence POA, pOa,— POB, p0Ob,— AOB, a0b,— BOC, bOc,—-... , are similar triangles, and AP, ap,— AB, ab,— ABC, abc,—..., are similar parts of the perimeters of similar polygons. Cor. 4. In similar polygons, the perimeters, or their (427) like portions, are to each other as the corresponding radii vectores. Cor. 5. Similar curves, regarded as defined by a descrip- (428) tion altogether analogous to (426), or their like portions, are to each other as their corresponding radii vectores. For (427), de- pending only upon the condition that the radii vectores maintain a constant ratio, is independent of the number and magnitude of the sides of the polygons, which therefore may be made to coincide with similar curves by less than any assignable quantities, On this principle is constructed the Pantograph, useful for copying maps, or any kind of plane figures, whatever may be their outlines. It consists of a parallelogram of rulers, ABCp, jointed at the angles and having two of its sides, BA, BC, sufficiently produced to admit of pins being inserted in them at P and O, in a straight line with the intervening cor- Fig. 1012. ner, p. If O be fixed, the points, P, p, will describe similar curves; so that an exact copy will be obtained, either diminished or enlarged, according as the pencil is placed at p or P—and the relative magnitudes of the figures may be varied by changing the pins, A, C. If p be fixed, P and O will describe similar figures, which may bear any ratio to each other, not excepting that of equality. Scholium I. There will be no difficulty in determining (429) EXERCISES, 229 the direction of the last side, 7, if we observe that, on arriving at its first extremity, our progress will have been farther north than south, or our LatirupE north, when the sum of the northings shall exceed that of the southings, or when a +6 -+...+k' or acos(a,m)+...+h cos(k,m)> 0, i. e., +; and v, v.; also that our position will be found to the east of the first station, or our Departure east, when the sum of the eastings shall exceed that of the westings, or when a’ +-b"4+-...+k" or asin(a,m) +... +h sin(k,m)> 0, and ». v. Scholium I. A part of the tabular labor may be saved by (430) making the projections upon and perpendicular to one of the sides ; thus, if we assume a as a false meridian, or make (a,m) =O, (421) becomes 6 sin(b,a) +... + & sin(k,a) a-+b cos(b,a) +... +k cos(k,a) _ bsin(b,a)-++ ... +4 sin(é,a) ‘- > Sinll,a)x oy dons Scholium ITT, By eliminating between (420) and (419) any (431) two sides, as k and 7, may be determined when all the other quan- tities are given; but it will be better to operate as if 1 were a me- ridian, when (l,m) will=0, (a,m) = (a,l), (b,m) = (0,1), (c,m) = (c,l), and from (420), (419), there will result a sin(a,l)+5 sin(b,l) +c sin(c,l) +... +7 sin(7,) vasa Dain(kyl) —l=a cos(a,l) + 6b cos(d,l) +c cos(c,l)+...+% cos(k,/). tan(l,a) = and (422), l 9 EXERCISES. 1°. Required the last side in I. Operation. 15°75 sin 365° — 20 sin 15*3° — 30°25 sin 46° 15°75 cos 365° + 20 cos 153° — 30°25 cos 46° [N, +] [N, +] (s, —]. = tan(d,m). 230 EXERCISES. logs. nos’ 1‘19728 130103 148073 logs. sins 1*77439 142140 1,85693 logs. coss 1:90518 198433 184177 log. a’ = 097167. log. b” = 0'72243 log. c’ = 133766 ~ » log. a = 1510246 log. b' = 1'28536 log. ci = 1532250 a’ =a sin(a,m)=-+ 93685 a =a cos(a,m) =+ 12661 b = 5 sin(b,m) = — 5775 b’= b cos(b,m) = + 19*291 ce’ = c sin(c,m) = — 21°7600 c’ = c cos(c,m) = — 21014 numerator = — 17‘6690 denominator = -+ 10938 . eastings < westings. .. northings > southings. log. 17669 = 1:24721 log. 17669 = 1°24721 log. 10°938 = 103894 log. sin 58:24° = 1:92955 log. tan (d,m) = 020827 log. d= 131766 -, (d,m) is S. 58*24° E. d= 2081 If, in accordance with the second scholium, we assume a for the meridian, whose bearing is N 365° E, the NW and SE bearings will be increased by 36‘5°, and the NE and SW diminished by the same angle; hence Field Notes, No. I., will be transformed into a N, 000° E, 15°75 chs. b N, 51‘8° W, 2000 c SB, 952, 30°25 d The student will calculate d according to this table. 2°, Calculate e of IT by (421) and (422). 3°. Calculate f of III twice, by projecting first on 5 then on c, and balance the errors by adding together and dividing by 2. 4°, Calculate g of IV and verify by a different method. 5°. Find AG in V. 6°. Calculate the diagonal drawn from the first extremity of a in IV to the second extremity of c. 7°. Solve VI, finding also the length, ?, of the intercepted me- ridian. [1 mtersects e: why 2] AREAS, 231 — 115 sin69°—17 sin28*4°--11 sin31‘6°-+15 sin56‘8° = 7 = 055, — sin67‘1° l= 115 cos69° + 17 cos28*4° + 11 cos31‘6° + 15 cos56‘S° — 055 cos67‘1° = 36°444. 8°. Required the length, 7, and second point of intersection with the perimeter, of a line running from Z (a,b) of LV, N 25° W. k PROPOSITION II. It is required to find the area of a polygon in terms of the sides and the angles which these sides make with each other. The easiest way of ascertaining the : area of any polygon (a, 0, c, ..., 7, &, L), is obviously to divide it into triangles . (A,B), (Ay Cy (Ay dy ony (As j)s (Ay Bs and then to compute these triangles. But the double area of a triangle is had at once by taking the product of its base and altitude. ‘Therefore from -A, the common vertex of all the triangles, and upon their bases, 5, c, d, ..., 7, k, pro- duced, if necessary, let fall the corresponding perpendiculars, Pos Pes Pas +++ > Pj» Px; there results 2triangle (A,b) = dp,, 2tr(A,c) = cp,, Ztr(A,d) = dp,, &e., «&c., 2tr(Ay) =jpp 2tr(A,k) = kp, But p, is obviously the projection of a upon the line of p,, p, the sum of the projections of a and } upon the line of p., p, the sum of the projections of a, 6, c, upon the line of p,, ..., p; the sum of the projections of a, b, c, ...¢, upon the line of p,, p, the sum of the projections of a, b, c, ..., k, upon the line of p, ; .". Py = @ COS(a,p,) =a sin(a,b), since Z (a,p,) + (a,b) = 90° ; so p,=a@sin(d,c)+5 sin(d,c), pa=a sin(a,d) + b sin(b,d) +c sin(c,d), &e., &c., &c., &e., p; = a sin(a,j) +b sin(b,j) +c sin(c,j) +... +4 sin(iy), p, =a sin(a,k) +b sin(b,k) +c sin(c,k) +... +7 sin(j,/) ; 232° AREAS, therefore, substituting above, and adding, there results, [ adh st ab sin(a,b) (432) + 2(A,c) +- ac sin(a,c) + be sin(8,c) QP=4 + 2A,d) = + ad sin(a,d) 4+ bd sin(b,d) + cd sin(c,d) + &c. +&e, &c, &e., | + 2(A,-) | {+ ak sin(a,k) + bk sin(b,k) +... +k sin(7,-) ; . The double area of any polygon is equal to the sum of the products of its sides, save one, multiplied, two and two, into the sines of the angles formed by the sides belonging to the several products. Cor. 1. The double area of a triangle is equal to the pro- (433) duct of two of its sides multiplied into the sine of the angle inclu- ded by them. 2tr(a,b,c) = ad sin(a,6). Cor, 2, The area of a paralJlelogram is equal to the pro- (434) duct of its dimensions multiplied into the sine of the included an- gle (433). Cor. 3. When two triangles have an angle of the one (435) supplementary to an angle of the other, the triangles are to each other as the products of the sides about the supplementary angles (433). Cor. 4, Combining (433) and (393), or 2tr(a,b,c) = ab sin(a,b) Q[h(h — a) (h—b) (h—c)]*. ab 5 and sin(a,b) = there results tr(a,b,c) = [h(h — a) (h—b) (h—c)]23 we. (436) The area of a triangle is equal to the square root of the contin- ued product of the half sum of the three sides and the three remainders formed by diminishing this half sum by the sides severally. This furnishes a rule very convenient for the applica- tion of logarithms. Cor. 5, Similar polygons and their like segments and sec- (437) tors are to each other as the squares of their homologous lines, whether sides, diagonals, or radii vectores, For, preserving the notation under (423) and substituting in (432) there results AREAS. " 2338 ¢ | r, sin(a,d) 7 +r, sin(a,c) + 7,7, sin(d,c) ae = J + r,sin(a,d) + ryrz sin(d,d) + r,rq sin(c,d) + a*, (438) | + &c., &c., &, | | + r,sin(a,k)+ r.r,sin(b,k)-+...+ rr, sin(7,k) | Cor. 6. If the polygon is equilateral, that is Cae pac Spy Be Sh or fp. Slee = 7, > 7, > 1, we have + sin(a,d) + sin(d,d) + sin(c,d) a”, (439) + &c., &c., &c., wt sin(a,k) + sin(b,k) =... + sin(7,£) Cor. 7. If the polygon be regular, that is, have its angles as well as its sides equal, putting Zz (a,b) =€, and therefore, (a,c) =e, (b,c) =e, (ajd) = Se, (b,d) = 2e; (cd) = e, &c., &c., &ec., (a,k) =(n— 1)e, (b,k) =(n — 2)e, (k= (ty 'B)e sce (75h) es n-+1 denoting the number of sides of the polygon, there results, (439), sin(a,b) + sin(a,c) + sin(d,c) = Fig. 1022. [ sine 7 | -+ sine + sine 2P,,.., =< +sinde + sine + sine “at, + &c., &c., &c., | +sin(n—1)e+sin(n —2)e+sin(n—3)e+ ... a or 2P,..,=[(n— l1)sine + (n — 2)sin2e + (n —3)sin3e + ... + sin(n — 1)e] « a?. Cor. 8. Making n = 2, we find . 360° P,= +| @- 1)sin | «a? = 45in60° + a? = 32(1a)?, (441) 0 for the area of the Equilateral Triangle. Cor. 9. Making n = 8, we find P, = 3[2sin90° + sin180°] + a? = a?, (442) for the area of the Square, as already known. 234 AREAS. © Cor. 10. Making n =4, we find P; = $8in36°(1 -++ 6cos36°) « a? I 1,4 (6+3 e 5?) (2 oh — 2-0 52) lie for the area of the Regular Pentagon. Cor. 11. For n= 5, or the Regular Hexagon, P,= a? + 3sin60° = a? « oye (444) a’, (443) Cor. 12, Similar plane figures, whether bounded by (445) straight lines or curves, are to each other as the squares of their homologous ares, chords and radii vectores, (437,) (426). Scholium I. It is obvious, from the course of the de- (446) monstration of (432), that the angles must be all estimated in the same direction, either from the right round to the left, or from the left to the right. And the rules for the algebraical signs of the trigonometrical lines, as everywhere else, are here also to be rigor- ously observed, Thus, sin211° = sin(180° + 31°) =— sin31°. For the computation of the area of the first field, we have OD ab sin(a,b) [2(A,d) J + ac sin(a,c) + bc sin(d,c) ; [2(A,c)] es a Z (a,b) = 36'5° + 153° = 51'8°, (a,c) = 180° + 365° — 46° = 170°5°, WwW E (5,c) = 180° — (15°3° + 46°) = 118*7°. Fig. 1023. Operation. 51‘8° | 1189534 | 239365 a, c | 170*5° | 1*21761 | 1689562 b, c | 118*7° | 1:94307 | 2*72383 a | 15°75 | 1:19728 | 24755 [2(A,d)] b | 2000 | 130103 | 7864 ae A,c)] 30°25 | 1°48073 | 529°45 2P = 855'64 P = 427,82 sq. chs. = 42782 acres. AREAS, 235 Operation for P of II. N 180° b ce aN 80° W Ww E bN 2° W L, cN 86° E , dS 32° E Fig. 1024. 3:463 1495 | 290:502/290:502=2(4,b) 2804 3513 | - 63°731 3576 5612 | 377+191/440:922=2(.4.c) 3:228 3412 — 169177 b, d| 150°/T698 9700] 3212 3874 | 163075 c,d) 62°'T945 9349] 3407 2833 | 255437 249335-2( 4,d) 45+53|1+658 2977/44] [4 980°759 a 5b |65'23)1'814 4474). + ay 4 90379 ss c |57‘86/1°762 3784 379 acres = P d 50°00 a,b) 78°)1990 4044 a, Cc 166°/1383 6752 b, c| 88°'T999 7354 a, d| 228°/T871 0735 Scholium II, The operation may be materially shortened (447) by making a diagonal the last or excepted side, and computing the polygon in two distinct parts; also gross errors will be detected, and the slight ones, always incident to mathematical tables and instrumental admeasurements, corrected, by changing the diagonal. Thus, if we imagine a diagonal joining the extremities of a and c in II, we shall find (a,b,c) = 365‘711 acres, (d,e) = 124669 do. ; a (a,b,c,d,e) = 490380. Again, let the diagonal join the extremities of 5 and d, and there results, (b,c,d) = 397'852, (e,a) = ~ 92629, ate (a,b,c,d,e) = 490381 ; so (c,d,e) = 345‘131, (a,b) = 145:251, (a,b,c,d,e) = 490382 ; aS 3(a,b,c,d,e) = 1471'143, and (a,b,c,d,e) = 490-381. 236 AREAS, Scholium III. When the sides of the polygon are numer- (448) ous, it will be expedient to divide it into parts by one or more diagonals determined by computation or by actual measurement. EXERCISES, 1°, Calculate the area (a,b,c,d) of I by calculating the parts (a,b), (c,d). 2°, The same by computing (d,c) + (d,a). 3°, Balance errors by taking the half sum of 1° and 2°, 4°, Find the area of I by excepting d. 5°. The same by excepting a. 6°. The same by excepting 6. 7°. The same by excepting c. 8°, Balance errors. 9°, The area of II may be calculated by several of the following methods : (a,b,c) + (d,e), (b,c,d) + (e,a), (c,d,e)-+ (a,b), (d,e,a) + (b,c), (e,a,b) + (c,d) ; (a,b,c,d), (b,c,d,e), (c,d,e,a), (d,e,a,b), (€,a,6,c). 10°. Employ some of the fifteen methods following for the computation of the area of III. (a,b,c) + (d,e,f), (b,¢,d) + (e.f@), (c,d,e) + (F145) 5 (a,b,c,d) + (e,f')y (b,c ae) )s pie d,e.f) + (a,b), (d,e,f,) + (5,¢), (e.f,4,6) )» (f,4,6,¢) + (d,e) 5 (a,b,c,d,e), (5,¢,d, a (c,d ‘ok a), An e,f,a,b), (e,f,a,5,c), (f,a,b,c,d). 11°, Write out all the methods for the computation of the area of IV, and execute a number of them. 12°, Calculate the area lying on the west of the meridian which passes through the middle point of the side a of IV. Ans. 44‘812 acres. 13°. Compute, by the aid of logarithms and (436), the area of a triangle the sides of which are 13‘334, 15°75, 16°024 chains. 14°. Find the area of the pentagon given under ‘“ the Chain.” AREAS, 237 PROPOSITION III. To cut off from a polygon a given area, PBCDX, by (449) a line, PX, running from a given point, P, in one of the sides, as AB, Compute the triangles PBC, PCD, PDE.,..., till two consecutive areas, PBCD, PBCDE, be found, the first less, the second greater than the required area, PBCDX; then will it be known on what side, DE, the point X must fall. The areas DPE, DPX, being known, the one by computation and the other by hypothesis, we have the proportion DPE DE DPX DX for calculating DX, when XP will be determined by the process for a last side, As an example, let it be required to cut off 200 acres from the left of II, by a line running from the first station A. By referring to the computation above, we find the point X will fall on c, and the operation may be put down as follows : tr(A,c) 220461 sq. chs. 3'343332 tr(A,z) 54749 sq. chs. 2°738376 _¢ 57:86 chs. 1°762378 ~ 2 . 2 = 14'369 chs. 1‘157422 PROPOSITION IV. To lay out a given area in a triangle of a given form. Let A, B, C, denote the angles severally opposite the sides a,b, c,; then (432), (337), 2P = ab sin(a,b) = ab sin, 2P = be sin(b,c) = be sinA, 2P =ca sin(c,a) = ca sinB ; op — sinB sinC he ge (450) sinA Cor. 1. If the triangle be isosceles, or C = B, whence 238 AREAS, 180°=A+B+C=A-+2B, B=C=90°—+#A, sinB = sinC = sin(90° — 4A) = cosfA, sinBsinC costA costA _ costA costA sinA sinA ~ sind A cossA there results 2 P = cotha. (451) Cor. 2. If the isosceles triangle P be applied m times about its vertex A, to form* the regular polygon, P,,, we have, nme Eo m@r= re cottA, ma? _ 180° or Pa = ae eo (452) since mA = 360°. What will (452) become for m = 3, 4, 5, 6, 7, 8, ... 2 For illustration, let it be required to lay out ten acres in a trian- gle whose angles are, A = 45°, B = 87°, We have (450) 200 sin45° 14. sin87° ae | : a = 13‘80:chs, For the side of an octagon of 15 acres, we find 8 a. 150 = a te cot 8 or 75 = a? cot224° ; + > o=| we | = [75 tan224°]? = 31°07 chs. PROPOSITION V. To lay out a trapezoid of a given area, P, on a z base, a, also given, knowing the inclinations of the » ad be ° ° ce oblique sides, xX, y, to a. Fig. 104, We have ra sin(z,a) + ry sin(z,y) + ay sin(a,y) = 2P, ay sin(a,y) + az sin(a,z) + yz sin(y,z) = 2P, zx sin(z,z) + za sin(z,a) + rasin(z,a) = 2P ; * The student will construct the figure, AREAS, 239 but sin(y,z) = sin(a,y), sin(z,x) = sin(z,a), sin(a,z) = sin(z,a) = 0; the second and third equations reduce to y(a + z)sin(a,y) = 2P, x(a-+ z)sin(z,a) = 2P ; (453) eosin{a,a) a so (454) by which the first equation reduces to J ‘ sin(a,y) 2P ; sin(a,7¥) aflaety [2 Saleen sa sin(z,a) sin(z,y) ” ee) sin(a,y) 2P sin(a,y) sin(z,y) ? sin(z,a) " sin(z,y) ted, this equation may be reduced as an ordinary quadratic, but better by (385,), observing that The coéfficients a « , being calcula- ik q” sin(a,y) A tanv eg p=a4e STIGHI NE (456) ype J sin(a,y) | 1= Sin(aa) * sin(ry) > (286) and x thus becoming known (454) determines y, when (453) gives (a+ z) and, consequently, z. Let it be required to cut off 15 acres by a line parallel to 6 of I. Making } = a= 20, the operation will be as follows, xz N 365° E id.y pbs LISe7° a N 153° W (x,y) = 1705° y S 46° W (z,a)= 518° log. sin(a,y) = 1943072 log. sin(z,y) = 1‘217609 log. sin(a,y) — log. sin(z,y) = 0°725463 log. 2P = 2477121 log. sin(z,a) = 1‘895343 log. 2P — log. sin(z,a) = 2581778 log. g = 3307241 —_—_— log. g+ = 1653620 240 AREAS, " log. a= 1'301030 log. p = 2026493 log, tanv = log. gt — log. p = 1,627127 log. cosv = 1964237 —- log.(z + p) = log. p — log. cosv = 2062256 z+ 1063 =+115*4 Ty OA _ sin(z,a) ie ice “sin(ay) 7 = 4° 2P eis Z = 21595 It may sometimes be desirable to execute the above problem without the aid of tables, or any other angular instrument than the cross. For this purpose measure in any conve- z nient position a line, p, perpendicular to a, and through its extremity a parallel, b, determining the trapezoid (a,5), which should differ as little as one may judge from (a,c), the area to be laid out. The area of the trapezoid (b,z) thus becoming known, it remains only to find its breadth, z; and for this purpose we have a(b + 2) jt tem: - (24 Fig. 1042. trap « (b,z) = a bn ee 2bp _ 2p trap(d, z) ° 2 a a & aA cdl rh eee ied Taek (457) PROPOSITION VI. To cut of a given area,XABCDEFY, Y from a given field, by a line, YX, running in a given direction. Having ascertained what.sides YX ©< v will intersect,* find AZ parallel to XY, Fig. 105. * This may be done, if preferred, by dividing the plotted field ABCDEF into trian- gles, taking the measures of their bases and perpendiculars from a scale of equal parts, and finding the sum of their areas; which, however, must not be employed any far- ther; for when FZ, ZA, have been found by calculation, or by actual measurement in the field, the area of ABCDEFZ must be accurately computed. = -.' --_ -* — - AREAS. 241 and compute the area ABCDEFZ; when the area of the trapezoid XAZY becoming known its parts will be found by Proposition V. PROPOSITION VII. Given the length and direction of the line AB, intersecting the lines AK, BL, also given in di- rection ; to draw a line, KL, in a given direction and intersecting AB in I, so as to make, on oppo- site sides of AB, the areas IAK, IBL, equal. CC? Representing the lines IA, IB, by a-+z2, a—z, Fig. 106. and the angles IAK, ILB, IKA, IBL, by e, f, m, n, we have (450) (a+ x)? sine sinz sinm e ae sine \2 sinn “ (a@+27)(-— =(a—2 a (Sam) (a (ae . ak e L e e 1 (Ss) (= ) , ( sine se) _ \sinf sinm} | = sinm sinn} |. : a : 1 e ° sinn\* re sine \2 i+ sine sinf \? sin sinm sinm sinn (a — x)? sinn sinz | = 2tr([AK) = 2tr(IBL) = sinf 1 — tanv or (354), z = ———_-. - a = atan(45° — v), eat 1 + tanv ( i (458) . sine sinf \7 l putting tanv = (Sinn sie ° J Scholium I. We observe that (458), being well adapted to log- arithmic computations, may be advantageously combined with (449) for the solution of Propositions V and VI, also in changing the direction of a line between two farms. Thus, let it be required to cut off a given area, KUVL, by a line, LK, running in a given direction. In the line VL take any point, B, and by (449) determine the point A in the line of UK, such that AUVB may = the required area, KUVL, and apply (458). It will obviously facilitate the operation if it be convenient to re- duce B to coincidence with V. The same process will change AB, the division line of two farms, into a second required position, KL. Scholium II. The intelligent student will not find any difficulty in applying the propositions now given to the straightening of a broken boundary between estates. : 16 242 AREAS, PROPOSITION VIII. Through a point, P, given in position in a giv- en polygon, to run aline cutting off a required area, S+8S’. By calculation or actual measurement, determine Z8\ any convenient line passing through the given point Fig. 107. and separated by it into the parts a, 6; also compute the included area S+S”; it follows that, since S-+S'’ and S+S" are given quantities, their difference, S’'—S" becomes known. But we have a’? sinmsint _ 2S! and 5? sinn sinz 4 sin(m-+i) sin(n +7) a? sinm sint 5b? sinn sinz ; A . le "yaaa 2) Gna e —S a which put=c, a? sinm sinz b? sinn sinz 28" ‘ or Se eS Fe a a ee sinm cos? + cosm sin? sinn cos? + cosz sinz a* sinm ' 6? sinn or ACE? Wy ih Seng TIMES Bike. a) sinm cote + cosm sinn cotz-+ cosn Ai a* sinm sinn cott + a? sinm cosn — b? sinm sinn coti — b? cosm sinn =c sinm sinn cot*z + c(sinm cosn + cosm sinn)cott + ¢ cosm cosn ; cot?z + pa an) + Aes a S| cotz sinm sinn c a? cotn—b? cotm = —__________——_— cotm cotn, c i b+a)(b— a . a*cotn or, cot?7 + [ cotn + | cotz = ‘Saat if it be convenient to make m = 90°. (459) Prana tiebname Woke ontinals SOLID GEOMETRY, SPHERICAL GEOMETRY, AND NAVIGATION. wy wre AR ; _ ea @ om , vu We me f » i aghe pe pebyguat, ty ue ves Po rained Sli eae aE | i A, a eedidig ot: oreehs oetgrioondie’ et er bers ney. hs ery a ee che ge tan girin lt Fin er krin Grn mete: stress eliieinta Wk EQST ye eee | ae: cht N eA an wuss “Nano, idtemaantte \earisKoun ) ax199, oth tt POORLY ti OR Boe .* ell jo eee RO F petitbicte StS eee ok BOOK FIRST. SOLID GEOMETRY. SECTION FIRST. Planes. PROPOSITION I. Three points not in the same gr arete line determine the (460) position of a plane. For, let the plane, P, pass through two of the points, as A, B; then, revolving upon these points, it will become fixed or determined in position, when it shall contain the third point, C, Fig. 108. Cor. 1. Two intersecting lines, as AC, BC, Aa ee (461) the position of a plane. Cor. 2. A triangle, as ABC, is always in the same plane (462) and determines its position. Cor. 3. A plane is determined in position by two paral- (463) lel lines, as AP, BQ. Cor. 4. Two planes cannot intersect each other in more (464) lines than one. For if A and B be any points common to the planes P and P,, it is obvious from the definition of a plane that the straight line AB will lie wholly in both planes, and will therefore be a line of intersection, Now the planes, P, P,, cannot have a second line of intersection, as ACB; since this hypothesis would reduce the planes to coincidence (460), the three points, A, C, B, not in the same straight line, becoming common to P and P,. 246 PERPENDICULAR PLANES, Cor. 5. The intersections of planes are straight lines. (465) Cor. 6. Planes coinciding in three points, not inthe same (466) straight line, coincide throughout. PROPOSITION II. A line drawn through the intersection of two other lines (467) and perpendicular to both of them, will be perpendicular to their plane. For let p be the perpendicular, a, a, equal por- tions of the intersecting lines, b, b, the hypothe- nuses to 7p, a, -- p, a, which will therefore be equal, and let d be any line drawn through the angle (a, a) and terminating in the line m-7n joining the extremities of a, a, and divided by d into the Fig. 109. parts m, n ; it only remains to show that e, joining the extremities of p, d, isa Sai We have (137) — d?=mn= b?—e?; , aging lade D., (133). Cot. 1. ieee one side of a right angle be made a fixed axis (468) of revolution, the other side, in revolving, will describe a plane. For, if the right angle (a, p) revolve about p as an axis, a will be found constantly in the plane of (a, a). Cor. 2, Of oblique lines drawn from any point in a per- (469) pendicular to a plane and terminating in this plane, the more dis- tant will be the greater, those equally distant will be equal and ter- minate in the circumference of the same circle having the foot of the perpendicular for centre. Cor, 3. Planes which are perpendicular to the same (470) straight line are parallel to each other; and, con- versely, if a line be perpendicular to one of two par- alle] planes, it will be perpendicular to the other also. Fig. 1092. Cor. 4. If from any point without a plane, a perpendiec- (471) ular be dropped upon the plane, and from the foot of this perpen- dicular a second perpendicular be let fall upon any line in the plane the line joining the first and last-mentioned points, will be perpen- dicular to the line drawn in the plane. For if m =n, d and e will both be perpendicular to m + n. PARALLEL PLANES. Q47 Cor. 5. Through the same point, either without or within (472) a plane, but a single perpendicular can be drawn, For let e be any line intersecting the perpendicular, p; e is ob- viously inclined to d and therefore to the plane (a, a). Cor. 6. A plane passing through a perpendicular to a (478) second plane, is perpendicular to the same. For let d revolve to take up a position perpendicular to a; then (p, d) being a right angle, the plane (p, @) is said to be perpendic- ular to the plane (a, a, d). i Cor. 7. The intersection of two planes perpendicular to (474) a third, is perpendicular to the same plane. Cor. 8. Lines perpendicular to the same plane are paral- (475) lel to each other. For let p, ps, ps3, .-., be lines perpendicular to the plane, P, and a, b,..., the lines joining the points in which the perpendiculars intersect the plane; it follows from (473), (474), that the planes (p, @), (pz, 5), will intersect in p,; whence 7, pr, being perpendicular to a, ps, p3, to b, ... 5 Py Pos Pas are parallel to each other. Cor. 9. A line parallel to a perpendicular to a plane is (476) itself a perpendicular to the same plane. Cor. 10, Lines parallel to the same line situated any (477) way in space, are parallel to each other. For they will be perpendicular to the same plane. Fig. 109s. PROPOSITION III. Tf a plane cut parallel planes, the lines of intersection (478) will be parallel. For, if the intersections m, n, of the plane, P, with the parallel planes, M, N, were not parallel, but met on being produced, then would M, N, cut each other in the same point, which is contrary to the hypothesis, Yeas Fig. 110. Cor. 1. The segments, as r, s, t, of parallel lines inter- (479) cepted by parallel planes, are equal. Cor. 2. Conversely, two planes intercepting equal seg- (480) 248 PARALLEL PLANES, ments of three parallel lines not situated in the same plane, are - parallel. Cor. 3. Parallel planes are everywhere equally distant. (481) [Let 7, s, t, be perpendicular to M, N.] Cor. 4. Two angles, having their sides parallel and open- (482) ing in the same direction, are equal, and their planes are parallel. For, let the sides AB, AC, of the angle A, be paral- B lel respectively to the sides ad, ac, of the angle a, and 4 open in the same direction ; draw Bd, Cc, parallel to Aa, then will the quadrilaterals, Ab, Ac, Bc, be par- allelograms, and the sides of the triangles BAC, bac, severally equal,— .. 7 A=a; but Aa =Bb=Ce, .. (480) the plane BAC will be parallel to the plane bac. Cor. 5. A Dihedral angle, or the angle which one plane (483) makes with another, is measured by the inclination of two perpen- diculars drawn through the same point in its edge, one in each side, For, make A a perpendicular to the planes BAC, bac, then will ’ Aa be perpendicular to AB, AC, ab, ac, and the dihedral angle BAac will be measured by the plane angle BAC = bac; from which it follows that the point A, through which the perpendiculars AB, AC, are drawn, may be taken anywhere in the edge, Aa, of the di- hedral angle. Cor. 6. The segments of lines intercepted by parallel (484) planes are proportional. For, let ABC, abc, be any lines whatever, pierc- ing the parallel planes, P, P, P, in the points A, B, C, a, 6, c; and through B draw mBn parallel to abc, piercing the planes in m,n. Since A, B, C, : | m, n, are in the same plane, and mA parallel to Cn, ie g Fig. 1103. we have AB : BC=mB: Bn=ab: be. (4 e Fig. 110e. PROPOSITION IV. If a line passing through a fixed point revolve in any (485) manner so as constantly to intersect two parallel planes, the figures thus described will be similar. SIMILAR SOLIDS. 249 For, let VaA, VrX, VyY,..., be positions of the revolving line passing through the fixed point, V, and piercing the parallel planes in A, a, X, z, Y, y, ...; drop the perpendicular VpP, piercing the planes in P, p, and join PA, PX, pa, pz. The ra- dii vectores PX, pz, have the constant ratio Vp: Vp, and Z APX=apz; .°. (426) the plane figures AXY, ..., GF, ... 4 are similar, Definition. The solid (V, AXY ...) is denominated a Cone, when the perimeter AXY ... is wholly curvilinear, and it becomes a Pyr- amid when AXY ... is made up of straight lines. The cone is cir- cular when its base, AXY ..., is a circle, and right if the perpen- dicular fall in the centre of the base. Cor. 1. In similar cones [the pyramid is to be included] (486) the altitudes, radii vectores, like chords and generating lines for corresponding positions, are proportional ; and the bases are as the squares of these lines, ThowiVPos Vo = PX xcs = chordXyY « chdzy =) VX Vz and base(AXY ,..) : bs(azy...) =(PX)? : (pr)? =... Cor. 2. If the vertex, V, be carried to an infinite distance, (487) the lines Aa, Xz, Yy, ..., will become parallel, and the figure ary ., will = AXY.,... Under these conditions the solid (AXY ..., ary ...) is denominated, a Cylinder when the perimeter AXY ,.. is a curve, and a Prism when AXY ... is a polygon; and these magni- tudes are said to be right or oblique according as the sides are perpendicular to or inclined to the bases. The cylinder and prism are also distinguished by their bases. When the base is a parallel- ogram it is obvious that all the other faces will be parallelograms and those opposite to each other equal, in which case the prism is called a Parallelopipedon ; and the Cube is a right parallelopipe- don of equal faces. x Fig. 111. SECTION SECOND. Surfaces of Solids. PROPOSITION I. The surface of a Polyhedron, that is, any solid bounded (488) by planes, may be found by computing the areas of its several faces. PROPOSITION II. The conver surface of a Right Circular Coneis meas- (489) ured by its slant height multiplied into the semicircumference of its base. Let y be the convex surface included between any two positions of its Generatriz, 1, and x the inter- cepted portion of the circular base ; then, since y is obviously a continuous function of z, giving to y, z, h the vanishing increments [k], [h], we have, (311), Fig. 112. (148), *, (246)* y= fiz; Yx— circumference = sl(circumference of base). Q. E. D. Cor 1. The circular conical sector is measured by its (490) slant height multiplied into half its base. y= diz. Cor. 2. The frustrumental surface of the right circular (491) cone is measured by its slant height multiplied into the half sum of its bases. For let VAB, Vad, be conical sectors having the same vertical angle, V, and, consequently ABda the frustrumental surface in question, we have al_\é A B Fig. 112e. * No constant is to be added, since y and x vanish together. SURFACES OF REVOLUTION, 251 ABba = VAB — Vab = VA «- 3AB— Va « 4ab = (Va-+aA) - sAB— Va « tab =aA +» tAB+ Va « +(AB— abd) _ aA(AB + ab) | play Spee: du ion VA AB d VA—Va_ AB—ab Va. “eee al > x eV SA! lui Ghined abe aA or Va rq: euaey: Cor. 3. The convex surface of the Right Circular Cylin- (492) der, is equal to its height, multiplied into the circumference of its base; for ab becomes = AB. Fig. 1123. PROPOSITION III. If a continuous curve referred to rectangular coérdi- (493) nates, revolve about the axis of abscissas, the derivative of the surface thus generated, regarded as a function of the correspond- ing abscissa, will be equal to the circumference described by the ordinate multiplied into the square root of unity increased by the square of the derivative of the ordinate also regarded as a func- tion of the abscissa. For let the surface Z be described by the revolu- M tion of any continuous curve, z, around the axis, z, k and M, m, k, h, the vanishing increments of Z, 2 Y, 23 then L7= r= [Fz] ’ | but (491), M=m[ty+ r(y+4)], a = (h? + k?)* « (2y+k); rs Ea 7 Chea Te Qy= 2ry(1 + Ea i¥ as or, Z7-r, = 2ry(l+y3_p~)*. Q. E. D. It follows that, in order to determine the surface generated by a particular curve, we have only to eliminate y and y’ by aid of its equation, and to return to the function, Q52 SPHERICAL ZONES. PROPOSITION IV. A Spherical Zone is equal to its altitude multiplied (494) into the circumference of the sphere. Let Z be a zone generated by the arc, z, of a circle revolving about a diameter, which we will assume as the axis of z, the origin being at the centre and the radius= 7. We have ytot=r, CY Ae Ree Lt rit) Qyk +k? + 2Qch +h? =0, ko ath. ho Rytk , x Oe a es ee Y y= (r2—2?)* a ene y 9 and y a = yp? z*\4 seine ae 2 Zz.n=2y(14+% = Qn(y? + 22)? = Qnr; S, ZL = Wrz, where there is no constant to be added if we make the surface be- gin at the axis of y.. Now let Z,, 2,, Z,, 2, be corresponding values of Zand x; we have zone(Z, — Z,) = zoneZ, — zoneZ, = (%2—2,) + Te Qr. Q.E. D. Cor. 1. A spherical zone is equal to the convex surface (495) of the circumscribing cylinder, described by the rev- olution of a rectangle with a radius equal to that of ‘ the sphere. Surface described by S = sur. described by C. Fig. 114. Cor. 2. The surface of the sphere is equal to the convex (496) surface of the circumscribing cylinder, ' Or, to four Great Circles. aes Sur. sphere = 2Z,_, = 4tr’. = . Fig. 1149. Cor. 3. The surfaces of spheres are to each other as the (497) squares of their radii [7], or diameters [2r], or circumferences [w+ 2r]. RECTANGULAR PARALLELOPIPEDONS, 253 EXERCISES, 1°, What will be the expense of gilding a globe 5 ft. in diameter at $1‘50 per square foot ? 2°. What is the surface of the earth and of each of its zones, reckoning it as a sphere of 8000 miles in diameter, and the Obliquity of the Ecliptic at 23° 28°? SECTION THIRD. Volumes. PROPOSITION I. Rectangular Parallelopipedons are to each other as the (498) products of their three dimensions, First, suppose their corresponding edges OA, A OB, OC, oa, 0b, oc, to be commensurable; that is, that OA being divided into m equal parts, oa B contain an exact number, m’, of the same parts, or that : g, OA =z, 04 = mF, (ir = the com. measure ] and OB= ny, 0b = n'y, [y = measure of OB & ob] NS i and OC =rz, oc=rz; ia then will the partial rectangular parallelopipedons, 4 formed by passing planes through the points of di- Fig. 115. vision parallel to the faces AB, AC, BC, ad, ac, bc, be all equal, since any one will be capable of superposition upon any other (487). It follows that the solids OABC, oabc, will contain severally — mnr, m'n'r’, partial and equal rectangular parallelopipedons, and [lt ig will consequently be to each other as mur to m'n'r’ ; OABC mnr _ mrenyerz OA+OB- OC cabo Oomnr mae nysrz =§=604+0b+ 0c ’ and the proposition is proved for the case in which the correspond- ing edges are commensurable. Next, let the dimensions OA, OB, OC, oa, 0b, oc, be any what- ever, and put 254. RIGHT PRISMS. OA= A, OB= B, OC=C; oa=a, 0b=b, oc=c. Now, if we increase a, b, c, by z, y, Z, so as to become commen- surable with A, B, C, and construct the parallelopipedon on (a+ 2), (6+), (c+), from what has already been proved, there results parallelopipedon [(a@ + 2), (6+ y), (¢+2)]_ (a+2)(6+y)(¢+2), parallelopipedon [ A, B, C] om ABC 4 par’dn [a, 8, c] solid [z, y, 2] _ abe ert par’dn [A, B,C] ' par’dn [ A, B,C] ~ ABC ABC’ .. (63), pard’n [A, B, C] : par’dn [a, b,c] :: ABC : abe. Q. E. D. Cor. 1. The rectangular parallelopipedon is measured (499) by the product of its three dimensions, provided the cube, whose edge is the linear unit, be assumed as the unit of solidity. For, from par’dn [a, b,.¢)lal) aenb = ¢ = 1, we have par’dn [A, B, C]= ABC. Cor. 2. The right prism with a right angled triangular (500) base, is measured by its base multiplied into its altitude. For the diagonal plane divides the rectangular parallelopipedon into two rectangular prisms, ca- pable of superposition. Fig. 1189. 4 Cor. 3, Any right prism with a triangular base is equal (501) to its base multiplied into its altitude. For the prism may be split into two, having right angled triangles for bases. i ‘ i) 1 1 h os G Fig. 1153. Cor, 4. Any right prism is measured by the product of (502) its base and altitude. For the solid may be divided into prisms, having ieee: triangular bases. ; . Fig. 1154. Cor. 5. The right cylinder is measured by the product of (503) its base and altitude. For (502) is obviously independent of the number and magni- tude of the sides, PYRAMIDS, 255 PROPOSITION II. The pyramid is equal to one-third of the product of its (504) base and altitude, and the cone has a like measure. In the first place, suppose the pyramid, y, to have a triangular base, z, to which one of the edges, 2, is per- pendicular; and, for the purpose of finding the func- 4, tion y=fz, give to y, z,z, the corresponding incre- ments k, h, 7. Since the prisms constructed with the rh altitude A, and upon the bases z, z +7, will be inscrib- YF ed in and circumscribe the solid, k, we have Fig. 116. kz ae ho 0's sot 9 8s Yap = Bal =z=a7"; (486) y=d >. 42° =42 + an* = 472, where no constant is to be added, because y,_, =0, and the prop- osition is proved for this particular case. Next, let a right angled triangle, revolving about its per- pendicular, p, and its base, varying in any way whatever, de- scribe, the one a cone or pyramid, y, the other its base, x; giving to y and xz the vanishing increments, & and A, from what has just been proved, we find [kK]=4p[h], «. y =4p3 . Yy = 3pr 3 and the proposition is demonstrated for all cones and pyramids, in which the perpendicular falls within the base. Lastly, let the base, u, be any whatever, and the perpendicular, p, fall upon its production; then, joining the foot of p, with two points of a con- : tiguous portion of the perimeter so as to form a the base, v, of a second pyramid or cone (p, v), we have Fig. 1162. Fig. 1163. cone (p,u-++ v) =4p(u+ rv), and cone (p,v) =4pv; “ cone (p,u)=4pu. Q. E. D. PROPOSITION III The Frustrum of a cone or pyramid is measured by (505) one-third of the product of its altitude, multiplied into the sum of its bases augmented by a mean proportional between them. 256 VOLUMES. Let the base AXY... =A, (fig. 111) and 04 <= Bs the altitude, VP=2-+4, and Vp=2;3 there results (504), solid [(z +a@),A]=3(r+a)A, solid [7,B] =4zB ; frustrum [A,B] =4(¢+ a)A—47B =t[aA+2(A—B)]; A Sai aad 2 ee a PANS Ce ge CEE aie ee 4 2(A—B) = te 5 (A 4+ BEY(AE — BS); “. frustrum [A,B.a] = 4a[A + CARYL + a (505) Cor. The prism or cylinder, whether right or oblique, is (506) measured by the product of its base and altitude. For B= A, gives ta[A + (AB)! + B] = Aa. PROPOSITION IV. If a solid, V, be generated by the motion of a plane, U, (507) varying according to the law of continuity, and 7 remaining constantly similar to itself, and per- pendicular to the axis of x; then will the deriv- ative of V, regarding V as a function of x, be equal to the generating plane, U, also regarded as a function of x, or ‘og 4 Ua 7 rT. For, from a little reflection, it will be evident that the incremen- tal solid [U,U,,h] must be measured as in (505), or that Veen = (ee ] = +(U+(UU,)§ + U,] =a fet ifr + f(e+h)t?+fle+h)] =tfe+fetfrj=fe=U. @Q. E. D. Cor, 1. For any solid, generated by the revolution of a (508) curve about the axis of z, the ordinate, y, describing the plane, U, we have ELLIPSOIDs, 257 Vivre = U= ny? =n( fr)’, Cor, 2. For any volume embraced between the surfaces (509) described by the revolution of two curves, or the two branches of the same curve, U being described by the difference of the corresponding ordinates, y, le y,, we have Vi, = ty? — ry? = 1(y? — yo®) = TY + Y2)(Y — 2). Cor. 3. For the ellipsoidal frustrum, estimated from Fig. 1172. its equator, or from z = 0, we find b 13 Fig. 1173. Vian» — (ate — 40%), (510) . 2 4 since, V{,=17 ST « = (a?—z?), Cor. 4, The corresponding frustrum of the circumscribing sphere, is V, = 1(a®°x — 42°), (511) since (510) becomes (511) when 6=a. Cor. 5. Prolate Ellipsoid = 2V,_, = 47ab? (512) = 3 « 2a » 7h? = #(circumscribed cylinder) PE | = 2(inscribed double cone), =. Fig. 1174. Cor. 6. Sphere, siiue—a= $70? = $ « 2a + Ta? (513) = 4(circumscribed cylinder) = inscribed double cone) =a « 4na? = ta(surface of sphere). Scholium, The last relation might have been found by imagin- ing the sphere filled with pyramids, having their common vertex at its centre, and their bases resting on its surface. Cor. 7. The prolate ellipsoid and its circumscribing (514) sphere, and their frustrums corresponding to the same abscissa, are to each other as the square of the minor to the square of the major axis. Prolate ellipsoid : sphére, = V, : V,= 6? : a’. Fig. 1175. Cor. 8. Analogous relations may be found for the Ob- (515) 17 258 SIMILAR SOLIDS. late Ellipsoid, described by the revolution of the el- lipse about its minor axis by changing a into bd, and b into a. a2 Thus V,,= 1 « Be .. Oblate Ellipsoid = 3 « 2b + ma? = 2(circumse. cyl.) = 24+ 2b + ma? = Ainscribed double cone) ; Sphere,.ains—5 = 3 ° 2b « 1b? .. Oblate Ellipsoid : Inscribed Sphere = V,, : V,,=a? : 5. (6° — tzr°), ee =1(0?2 — 42°); Fig. 1176. Cor. 9. Common Paraboloid V = tpx? = 44a « try? (516) = 4(circumscribing cylinder), cree ae since Vi= ny? =T « 2pzr; Fig. 1177. and, for the paraboloid whose equation is y?=A,+ Art Agr? +..., we have V=T2r(A,+4A ct +4Agr? +...) + Vio. 213 PROPOSITION V. Similar solids are to each other as the cubes of thetr (517) like dimensions. We understand by similar solids those in which al] like dimen- sions are proportional; and, consequently, the sections through ~such dimensions similar. The proposition becomes evident for the solids already investi- gated by making A and B vary as a? in (505), and 6 as a in (512) and (515), or by putting A=ca?, B=c,a*, b=c,a, the cs being constant, whence Similar Cones, Pyramids, and their similar Frustrums, conse- quently similar Parallelopipedons, also similar Ellipsoids and Spheres vary as 4fc+ (ce,)'+c,] « a3, 403 « a, or as a°, or as the cubes of any like dimensions. Next let any two similar perimeters, whether rectilin- (fig. 99.) ear or curvilinear, be similarly placed and revolve about any line OaA of corresponding radii vectores as an axis of x, the origin be- ing at O; the ordinates y, y,, of the extremities P, p, of any other corresponding radii vectores, will describe planes U, U,, terminating EXERCISES, 259 like frustrums of the two solids estimated from O where z = 0, and this whether the radii y, y,, remain constant for a given value of x or vary in any way whatever, that is, whether the perimeters ABC ..., abc ..., continue of the same magnitude or be variable, only that they preserve their similarity. Therefore we have te Voreth, 6 eee nti ee as “. (507) eV 2 Veeifrustrum[ 7) «ff 02)] = 42° :,403-O0P? : On’, and the proposition is proved for this more general case. Lastly, let the similar solids be any whatever, and assume any two like diameters for the axes of x, 2, the origins dividing them proportionally. It follows from the definition of similar solids given above, that the generating planes U, U,, perpendicular to z, Zz, Will have corresponding positions in their respective solids V, V., when any diameter in U is to a like diameter in U, as the ab- scissa 2 terminating in U is to the abscissa x, terminating in U,, and that U and U, will be similar figures ; s Oe eh s see) es But from (300) it is manifest that (507) is applicable in this case also ; VU 28a Uy 0, "> 2. od) Vs V,2if 0) Uy SH 40% beg 2? : 2p. Q, ED. EXERCIS&S. 1°. A cylindrical cistern, capped with a hemispherical dome, is 15 feet deep and 10 in diameter, Required its capacity. 2°, A hollow cylinder, the side of which is one inch thick, is set into a cubical box, touching its sides and equalling it in height, and within the cylinder is placed a hollow sphere also an inch in thickness and tangent to the cylinder, What is the capacity of the sphere, it requiring just 10 gallons of water to fill the space between the box and the cylinder ? 3°, What is the capacity of a cask, regarded as a frustrum of an ellipsoid, the bung diameter being 30 inches, the head diameters 25 each, and the length 40 inches; and how much will it differ from its inscribed double conical frustrum ? 4°, What is the capacity of a paraboloidal cistern, having the dimensions in 1°? 260 EXERCISES. 5°, Wishing to ascertain the weight of a marble column 30 feet high, I take the semidiameters at the elevations 0, 10, 20, 30 feet, and find them to be 3, 4, 3, 2, feet; the specific gravity of marble is 2°7, water being 1, and a cubic foot of water weighs 1000 oz. avoirdupois. What is the weight? 6°, The earth may be regarded as an oblate spheroid, generated by the revolution of an ellipse about its shorter diameter of 7899‘170 miles, while the equatorial diameter is 7925‘648, accord- ing to Sir J. F. W. Herschel. Required the excess of volume over the inscribed sphere, and the quantity of water on the surface; al- lowing the sea to be to the land as 4 to 3, and its mean depth to be 22 miles. 7°, What must be the dimensions of a tub to hold 10 cubic feet, the depth and two diameters being as the numbers 4, 5, 62 8°. What must be the dimensions of a paraboloidal kettle to contain 12 gallons, the diameter across the top being to the depth as 7 to 8? BOOK SECOND. SPHERICAL GEOMETRY. SECTION FIRST. Spherical Trigonometry. PROPOSITION I. A Spuere, being a solid bounded by a curve surface (518) everywhere equally distant from a point within, called the cen- tre, may be generated by the revolution of a semicircle about tts diameter. For, let the semicircumference, PeEP’, revolve about its diameter, PoOP’; it follows that every point, e, in PeEP’ will, while describing the circumference, ee’e”’ .. » Maintain a constant distance, Oe=O0el=VE =... =OP'= OP = OR = OF = ..., a2 from the centre, O; whence the surface 1S described will be that of a sphere. T Fig. 118. Cor. 1. The radius, OF, perpendicular to the axis, (519) POP’, will describe a Great Circle, the circumference of which, EEE’ ... , denominated the Equator, will be everywhere 90° dis- tant from its poles, P, P’. Cor. 2. Any other perpendicular, oe, will describe a (520) Small Circle, also perpendicular to the same axis, and having the same poles, P, P’. Cor. 3, Every section of a sphere by a plane is acircle. (521) Cor, 4, The perpendicular, P’T, through the extremity (522) 262 LUNES. of the diameter, POP’, will generate a plane (P’,TT'T” ... ) tan- gent to the sphere. Hence, a plane passing through the extremity of any radius, and perpendicular to it, is tangent to the sphere ; and, conversely, a tangent plane is perpendicular to the radius, drawn to the point of contact. Cor. 5. All great circles mutually bisect each other; as (523) the Meridians, PEP’, PE'P’, since they have a common diameter, REY, Cor. 6. Every great circle bisects the sphere. (524) Cor. 7. A small circle divides the sphere into unequal (525) parts, and is less the more distant it is from the centre. Cor. 8. A Lune, or the spherical surface embraced by (526) two meridians, is to the whole surface of the sphere as its equato- rial arc to the total equator; thus, Lune PEP’EP : Sph. Surface = arcEE’ : 360°, or Lune, >. 40r? 2° eB Onesie e° é = eq.are. Lune. = 695 * al E = rad.sph. Cor. 9. The Spherical Wedge or Ungula, PEP'E‘O, is (527) to the whole sphere as its equatorial arc is to the total equator ; Ungula,, 2.4 o7r* = e°_: 860°, . Cor. 10. Every meridian is perpendicular to its equator; (528) as PE'P’ to EEE’ .... Cor. 11. A Spherical Angle is identical with that em- (529) braced by the tangents to its sides, and is measured by the arc of its equator intercepted by these sides; as the angle EP’E’ = TP’T’ measured by EE’, Definition. A Spherical Triangle is the surface em- (530) braced by three arcs of great circles, (526) PROPOSITION II. A spherical triangle ts equal to the sum of its three (531) angles diminished by a semicircumference, multiplied into the square of the radius of the sphere. Produce one side, AB, of the spherical triangle, T, so as to form the circumference, ABPQ, inter- sected in P and Q by the productions of the other sides, AC, BC; then, by the addition of the trian- gles, X, Y, Z, to T, there will be formed the lunes Fig. 119 TRIANGLE, 263 T + X, T+, having the angles A, B, and to the surface T+ Z, which may be readily shown by (523) to be equal to a lune with the angle, C ;* and, from (526), there results, A? , ne Ste P+ X= oe 3 tr, T+ Y= » SPO mrt PLEX VER) a eT po A° + B° + C° — 180° 180° Cor. 1. The sum of the three angles of a spherical tri- (532) angle is always greater than two right angles and less than six ; since for the existence of T’ we must obviously have Pe: At BHO 7 pet Cor. 2. Similar spherical triangles are to each other as (533) the squares of the radii of their respective spheres, or as the squares of their homologous sides. fo) Tr, Py tyke Tr? ; 2.12 s otr?=(A+B+C—n)r*, (531) Cor. 3. A spherical polygon is measured by the sum of (534) its angles diminished by as many semicircumferences as it has sides save two, multiplied into the square of the radius of the sphere. For the number of sides being n, the polygon may be divided into (n— 2) triangles, T, T., T;, ..., whose angles A, B,C; Ay B,, C,; As, B;, Cy;..., make up the angles of the polygon, and there results (T+ T,+ Ty; +... )ne = (A+B+ C—t)+(A-+ B+ C,—t)... Jr’, or P,=[S —(n—2)r] + r*?, S=sum of angles. * For, producing the arcs CP, CQ, till they meet in R, and drawing the diameters (?) AOP, BOQ, COR, we have only to show that the triangle PRQ, which com- pletes the lune CPRQ, is equalto 4 ACB. In order to this, take the point S, equally distant from A, B, C, or the pole of the small circle passing through these points, and draw the diameter SOT; then the arcs SA, SB, SC, TP, TQ, TR, will be all equal; also(?) Z ASB=PTQ, BSC=QTR, ASC=PTR, and it may therefore be : shown by superposition that Fig. 1192. A QTP = ASB, RTQ =BSC, RTP = ASC; APQR=QTP-+RTQ—RTP =ASB+ BSC —ASC=ABC. Q. E. D. 264 FUNDAMENTAL THEOREM PROPOSITION III. The cosine of any side of a spherical triangle is equal (535) to the product of the cosines of the other two sides increased by that of their sines multiplied into the cosine of the angle opposite the first-mentioned side. For, denoting the angles of the triangle by A, B, C, and the sides respectively opposite by a, b, c, let the radius of the sphere OA = OB = OC, be ta- ken for unity, and draw the tangents, tand, tanc, terminat- - ing in the productions of OC, OB, at Q, P, and join PQ=m; then will OQ =secd, OP =secc, and the spherical angle A will = Z PAQ, also Z BOC will be meas- ured by a, (?), and from (388) there results, Fig. 120. sec*b ++ sec*c = m? + 2secb secc cosa, and tan?b + tan*c = m?-+- 2tand tanc cosA; 1+ 1 = 2secb secc cosa — 2tanb tanc cosA, 1 1 sind sinc or = —— e —— e cosa — —— - —— « cosA; cosh cosc cosh cose whence cosa = cosb cosc + sind sinc cosA ; SO cosb = cosa cosc + sina sinc cosB, (535) and cosc = cosa cosb + sina sinb cosC. Cor. 1. If two spherical triangles have two sides of the (536) one equal to two sides of the other each to each but the included angles unequal, the remaining sides will be unequal and the greater side will be opposite the greater angle; for, as A increases, cosA decreases and, consequently, cosa decreases or a increases, there- fore A and a increase or decrease together. Cor. 2. Two spherical triangles are equal: (537) 1°, When two sides and the included angle of the one are re- spectively equal to the two sides and the included angle of the oth- — er; and therefore, FOR A SIDE, 265 2°, When the sides of the one are severally § 4 equal to the sides of the other. Far, 1°, if b= 65.66 yA til ~~ then a=a;\. B=B,C=C'’ B B! . tr(ABC) =(A' BC) ; ‘ ley i sig mer ys Cc’ 2°, if gas 2 > Roe bon Fig. 1209. then A= A’; .. tr(ABC = (A'BC)). Cor. 3. The arc joining the vertex and the middle point (538) of the base of an isosceles spherical triangle, is perpendicular to the base, and bisects the vertical angle. Cor. 4. We may adapt (535), which is the fundamental theorem of spherical trigonometry, to logarithmic computation, when a side is required, by putting sinb cos A ge eagee cosb ; cos for then (535) becomes ; sinu cosa = cosb cosc + cosb sinc « cos Cc Ss e e = —— (cose cosu + sinc sinw) ; cosu and there results cosb cos(c—u) tanw = tand cos.A4, cose = ——__—____—_; Costu cosa cos(c—v sO tanv = tana cosB, cosb = Ae Got (539) cosv cosa cos(b — w) and tanw = tana cosC, cosc = ——_____——- CcOsw Cor. 5. When an angle is required, the transformations of Plane Trigonometry may be imitated, and we have 2sin*4+A = 1—cosA my, cosa — cosb cosc a } - sind sinc cosh cosc + sind sinc — cosa ae sind sinc cos(b — c) — cosa mt sinb sinc Qsint(a +b —c) « sint(a—b+c) i sind sinc 266 FOR AN ANGLE. dia = Ee sint(a+b—c) « ae sind sinc sin(h sin(h —c)]4. = [Ree = sinc sae |? ; nt B= ps (b+ a—c) + sint(b+c— ai? sina sinc (540) sin(d — a) sin(h— c) ie oy Spore pee sind sinc th and — sink C= hese (c-+a— 6) « sing(e-+b — a+ sind sinb sf eget — a) sin(h— b)]4 i sina sind | : And by a like process we shall find eas Tae sin(h — al + ah sind sinc sink sin(h—b)]4 sing sinc sink sin(h — c)]+ sing sinb (541) cost B = [ cost C= [ sin(h — b)sin(h — c)q2 7 sink sin(h—a) J , sin(h — a)sin(h — c)74 sinh sin(h —d) ’ (542) sin(h — a)sin(h — he sink sin(h — c) oe tant A= = tan4_B = [ tant C = [= Cor. 6. The sum of any two sides of a spherical triangle (548) is greater than the third side. Cor.7. If from any point within a spherical triangle, arcs (544) of great circles be drawn to the extremities of either side, the sum of the including sides will be. greater than the sum of the included arcs. [See Plane Geometry. ] Cor. 8. A chain of spherical arcs is less the nearer it lies (545) to the arc joining its extremities. Cor. 9. The arc of a great circle is the shortest distance (546) from point to point on a spherical surface. Cor. 10. The sum of any two of the plane angles that (547) make up a solid angle, is greater than the third angle. For, if about the solid angle, O, a sphere be described, REGULAR POLYHEDRONS, 267 the plane angles AOB, BOC, AOC, will be measured by AB, BC, AC, the arcs of the spherical triangle ABC, Cor. 11. The sum of all the plane angles that make up (548) a solid angle, is less than four right angles. For, let a, b, c, ... , 7, k, 1, be the arcs of the spherical polygon by which the plane angles are measured, and produce aq, ¢, till they meet, forming the arcs,a+m, c+n; the new polygon whose sides are a+ m,c-+n, d,..., 1, will have a greater peri- meter, consisting, however, of a number of sides less by one, than the original polygon, And this reduction may be continued till a triangle is obtained whose perimeter, less than a circumference, will be greater than that of the polygon. Cor. 12. There can exist only (549) Five Regular Polyhedrons. Three—Tetrahedron, Octahedron, Icosahedron, whose faces are equilateral triangles, [360° : 60° =6]. One—Hexahedron, squares, [360° : 90° = 4]. One—Dodecahedron, regular pentagons, These figures may be formed of | | Os pasteboard, the lines of folding being cut half through : Fig. 1203. Cor. 13. If the three plane angles that constitute two (550) solid angles be respectively equal, the homologous planes will be equally inclined to each other. PROPOSITION IV. In any spherical triangle the sines of the sides are to (551) each other as the sines of the angles respectively opposite. Taking the double products of (540) and (541), observing that Qsint A costA = sinA, we have 268 SIDES AND ANGLES. sinA = : : 3 sind sinc cigs ain inh — a) sini oa) oal oma (552) sing sinc « 4 eg A VP kde oe 5s | Tat aa ee eS a a ee sina sinb es YL ee Me: Le ee ‘* sinB sind’ sinC sinc’ sinC sinc’ Cor. 1. In a spherical triangle the greater side is opposite (553) the greater angle, and v. v. Cor. 2. The perpendicular arc is the shortest distance (554) from any given point to an are also given in position. Cor. 3. The angles opposite the equal sides of an isos- (555) celes spherical triangle, are equal, and vice versd. Cor, 4, An equilateral spherical triangle is equiangular, (556) and v. v. PROPOSITION V. To eliminate from equations (535) the cosine of each side in succession. Substituting the value of cosc from the third in the second, we find cosb = cosa(cosa cosb-+ sina sinb cosC) + sina sinc cosB = cos*a cosb-+ sina cosa sinb cosC' + sina sinc cosB; “. (1 —cos*a)cosb = sina cosa sinb cosC’+ sina sine cosB or sin*a cosb=sina cosa sinb cosC + sina sinc cosB, and sina cosb = cosa sind cosC’+ sinc cosB, 7 so sina cosc = cosa sinc cosbh + sinb cosC, sinb cosc = cosd sinc cosA-+ sina cosC ; sinb cosa = cosb sina cosC +- sinc cosA, (557) sinc cosa = cosc sina cosB-+ sind cosA, sinc cosb = cosc sinb cosA + sina cosB. | Vote. Instead of going through independent operations, all the above forms may be obtained from the first by a simple change of letters ; thus, changing 0b, B, into c, C, and vice versd, the first becomes the second, which in turn gives the third by commuting a, A, and b, B, and so on. FORMS, 269 PROPOSITION VI. From (557) to eliminate a sine, in order that no more than two sides may be embraced in each equation. sinc * sinC sinbsinC From (551) we have —- = ——., or sinc = —- which sub- Scoae, sind sinB’ snB ” ee stituted in the first of (657), gives . ‘ sinb sinC sina cosb = cosa sinb cos C + ———.— « cosB, sinB : cosh cosB > sing e ——=cosa cosC : e sinC; * sind a sinB : ae sina cotb = cosa cosC' + cotB sinC, sind cotc = cosa cosB+ cotC sinB, sinb cotc = cosb cosA + cotC sinA ; sind cota = cosdb cosC' + cotA sinC, sinc cota = cosc cosB + cotA sin B, | sinc cotb = cosc cosA -++ cotB sinA. PROPOSITION VII. From (558) to eliminate a side. From the fourth and first by aid of (551) we have cosh : cota= = al" cosC + cotA sinC « sind fi sind sinA = coth co cotA sinC « ———_ Y sor sina sin B’ ‘ 1 and cotb = cota cosC + cotB sinC + ——_; sina ; 1 cota = cota cos*C + cotB sinC cosC « sina ‘ sinA cotA sin C @ a = sin A sinB’ ; ; cotBsinCcosC | cotAsinA sinC and cota(1 — cos*C) = ———_— SACRE tog ae sind sing sin.b cosa, cosB sinC cosC . cosA_ sinAsin’ or - e sin? C = —_ « ——_ + —— a ——— 3 sind sin PB sina sinA sina sinB .. sinB sinC cosa =cosB cosC + cos; 270 POLAR TRIANGLES. .. cos(180°—A) = cosB cosC' +sin B sinC cos(180°—a), cos(180°—B) = cosA cosC +sinA sinC’ cos(180°—S), 0 cos(180° —C) = cosA cosB +sinA sinB cos(180°—c). Cor. 1. If in these equations we substitute A = 180° —a, a = 180° — A, B= 180° —b, b = 180° — B, C= 180° — ¢; c = 180° — 0; there results, cosa = cosb cose + sinb sine cosA, cos) = cosa cos¢ + sina sing cosB, cos¢ = cosa cosh -+ sina sinb cosO ; which are of the same form with (535), and therefore belong to a triangle whose sides are a, b, ¢, and angles A, B, 0; the relation of these two triangles, ABC, ABC, is A+a=180°=A- (569) cosB, = sinA, sind = cotC, cotc, cos C’.= sin A, sinc = cotB, cotd. It will be observed that the quadrant, a, is regarded as not sepa- rating B, C. For illustration, let it be required to find B when a and c are given, and b= 90°. We have cos(180° — B) = cosB,= cota cote. Cor, 2. Napier’s Rules may be employed in the solution (570) of Isosceles Spherical Triangles, by observing that the circular parts will be the equal side, equal angle, half unequal angle, and complement of half unequal side. For, letc =a; then joining B and the middle point of b, we shall have formed two right angled triangles, in each of which the parts will be (538), (568), a, A,+B, (45). ; cosa = cotA cotsB, 7 cosA = cota cot(4d),, cos(4b), = sina sintB, + (570) cos$B=sinA sin(4d),. | Cor. 3. Napier’s Rules may be employed in the solution (571) of Oblique Triangles, by dropping a perpendicular so that two of the known parts shall fall in one of the right angled triangles thus formed; there will result also, Bowditch’s Rules. 1°, The cosines of the parts opposite the perpendicular, are pro- portional to the sines of those adjacent. 2°, The cosines of the parts adjacent to the perpendicular, are proportional to the cotangents of the parts opposite. For, let the angle, C, be separated into the parts, M, N, by the perpendicular, p, dividing the oppo- site side, c, into the corresponding parts, m, n, subjacent to a, b, and we find, Fig. 1212. cosa = sinp, sinm,, cosb = sinp, sinn, ; (1°) cosB = sinp, sinM, cosa =sinp, sinN ; cosm, = cotp, cotB, cosn, = cotp, cotA ; (2°) cosM = cotp, cota, cosN = cotp, cotd. CASES IN SPHERICAL TRIGONOMETRY. 275 Cases in Spherical Trigonometry. Given. Forms for Solution. Pi I, |'Two sides and included angle. | (535) ; (539) ; (566) ; (571). Il. |Twoanglesand included side. | (559) ; (564) ; (567) ; (571). 535) and (384) ; (551), [?]. = |T'wo angles and opposite side, |(559) and (384) ; (551). ( IL. |'Two angles and included side. |( ILI.|T'wo sides and opposite angle. | ( ( (8 V. |Three sides. 35); (540); (541); (542); (552). VI/Three angles. (559) ; (561); (562); (563). | Right Triangles. (668). | Quadrantal Triangles. (569). [Isosceles Triangles. (570). Scholium J. There will bea choice in forms, not only on account of logarithmic operations, but also for the purpose of avoiding the cosines of very small ares, and the sines of those dif- fering little from a quadrant. Scholium II. Problems pertaining to spherical trigonometry will generally find their easiest solution by constructing the trian- gle so that one of its angles shall be at the pole of the sphere. Latitudes and Longitudes. Places. Latitude. Longitude. Boston (State House), Mass., 4Q°Q1'2257'N,| 71° 4° OW. Chicago, II1., 42 0 O N.| 87 35°) OW. Canton, China, 23 8 9 N./113 1654 E. Cape Good Hope (Obs.), Africa, 33 56 3 S.) 18 2845 E. Cape Horn, S. America, 55 58 41 =S.| 67 1053 W. Cincinnati (Fort Washington), Ohio,)39 5 54 N.| 84 27 OW. Greenwich (Obs.), Eng., 51 28 39 N.| O 0 0 New York (City Hall), N. Y., 40 42 40 N. 74 1 SW. Paris, (Obs.), France, 48 50 13 N.| 2 2024 E, Philadelphia (Ind’ce Hall), Pa., 39 56 59 N.| 75 954 W. Rome (Roman Col.), Italy, 41 53 52 N,| 12 2840 E. Washington (Capitol), D. C., 38 53 23 N.| 77 1 24W. 276 EXERCISES, 1°, Required the distance and direction from New York to Greenwich. [69*2 miles to 1°.] Polar distance of New York = 49° 17’ 20", Polar distance of Greenwich = 38° 31 21"; Difference of longitude dt e aahiae Saas = die the problem, therefore, belongs to Cases I., III. (539), tanu = tan49° 17’ 20” 0:065262 suming o as the origin of a system of oblique co- ordinates, (575) becomes, (A+2z)(A-2z)_ yy cos*(M,m) os?(P,p)’ putting m-+2=2A; but R being the radius of the circle and B the projection of that R which is parallel to P, we find A = Reos(M,m), B+ Reos(P,p); ORTHOGRAPHIC PROJECTION. 28] R? R? es ae (4° — 2") = Fe °y; 2 2 or as ees |, (576) 3°. If R be assumed in such position that its | projection, A, or, for the sake of distinction, a, Sata shall be parallel] to it and consequently 6, the new ree value of B, perpendicular to a, there results Ot AP op a =Rceos(M,m) = Reos0° = R, * b = Reos(P,p) = Reosl, eu I being the inclination of the circle to the plane of projection; we have also goo a o tae which is the equation of an ellipse of which the axes are 2a, 2h; and the proposition is demonstrated, since when J = 0, b becomes parallel to and equal to R =a, and when [= 90°, b becoming = 0, the ellipse vanishes in a straight line. Cor. 1. The ellipse may be referred to a system of ob- (579) lique codérdinates such that its equation (576) shall be of the same form with that obtained for rectangular axes (578), and 2A, 2B, mutually bisecting each other and all chords drawn parallel to them, are denominated Conjugate Diameters, of which the axes 2a, 20, are but particular values. Cor. 2. If two systems of parallel chords intersect each (580) other in an ellipse, the products of their segments will be propor- tional (575); and this property may be extended to the case in which the points of intersection lie without the curve. Cor. 3. An elliptical arc, MmApP, being given, the (581) centre, O, may be found by bisecting AOB drawn through the middle points of any parallel chords, MP, mp. N = iy (578) PZ = A Fig. 122%. Cor. 4. A conjugate, CD, to any diameter, AB, may (582) be found by drawing CD through the middle points of AB and PN a chord parallel to AB. 282 ORTHOGRAPHIC PROJECTION. Cor. 5. A parallelogram, MPNQ, may be described in (583) an ellipse by drawing chords, MQ, PN; MP, QN, parallel to the conjugates AB, CD. Cor. 6. The diagonals MN, PQ, mutually bisecting each (584) other in O, are obviously diameters, and P may be so chosen that MN shall be the conjugate axis = 2a, in which case PM, PN, are said to be Supplementary Chords. Cor. 7%. The tangents drawn through the extremities (585) A, B, C, D, of the conjugate diameters, are parallel to the supple- mentary chords, and to the corresponding diameters ; and, by their intersections, constitute a parallelogram circumscribing the ellipse. PROPOSITION II. To make an Orthographic Projection of the Sphere. With the radius OA, equal to that of the re- quired map, describe the meridian AZP,H,B P.H,, passing through the north and south poles, P,, P,, and, consequently, perpendicular to the horizon seen edgewise in H,H.. Take the arc H,P, measuring the elevation of the pole, equal to the latitude of the place Z over which the observer is supposed to be situated at a very great [in- finite] distance. Having graduated the meridian and drawn the parallels of latitude ACB, which will be perpendicular to the axis P,OP., drop their extremities and centres, A, C, in the projections Fig. 123. a, c, also project the elevated pole P, in 2p, Hn Transfer the points a, c, p,, to the central me- ie ws ridian H,OH, of the map, and through cdrawa ”™ perpendicular to H,H,, intersecting the circum- ference in m, m.,; cm and ca will be the semima- jor and semiminor axes of the elliptical parallel H, of latitude mam,, which may be described by the Fig. 1232. first exercise under the ellipse. Next, for the meridians, let H,P,H,, = 4 be the angle which any required meridian, P,H, makes with the vertical meridian P,H,,, and H,H, = H, the corresponding arc intercepted on the horizon, further P,,H, will = 1, the latitude of the place upon whose horizon the projection is made; therefore, by Napier’s Rules, P,,H,,H, being a right angle, we have GNOMONIC PROJECTION. 233 cosl,= coth cotH.,, or tan H = sinl tanh. (586) Having laid off H,H, = H found by calculation, Hn and drawn the diameter H,O, and Oz perpendic- ular to it, transfer the semimajor axis H,O to the straight edge of a thin ruler or slip of paper, and, having applied this line in p,z, and marked its point, y, of intersection with OH,, proceed to de- Hi scribe the meridian H,p, according to the first exer- Fig. 1249. cise under the ellipse.* | The work now described is to be combined in a single figure, when the features of the spherical surface will be laid down according to the projec- tions of the meridians and parallels of latitude. Fig. 125. PROPOSITION III. To make the Gnomonic Projection when the dial is horizontal. Let H,H,H, be the horizontal face of the dial, O its (fig. 124) centre and OP, the style elevated according to the latitude of the place ; the shadow at noon will fall upon the north point, H,, and its positions, OH,, for all other hours will be determined by (586), where / = the hour angle. PROPOSITION IV. To make the Gnomonic Projection of the Style upon a vertical south plane. The construction will be similar to the preceding, (fig. 124) observing that, tan(XII)T = cos/ tanh (587) * An instrument very convenient for this purpose may be constructed, consisting of a slender ruler carrying a pen or pencil in its extremity and furnished with two moveable pins, one to glide in a groove cut in the stem of a wooden T and the other along its top. 284 STEREOGRAPHIC PROJECTION, PROPOSITION V. If an oblique circular cone be truncated by a plane at (588) right angles to that plane which passes through the axis and is arpan ater to the base, the section will, in general, be an el- lipse. Let AoOBy be the truncating plane perpendic- Vv ular to the plane, AMVBN, passing through the axis at right angles to the base; draw oy = y per- B pendicular to AB, and pass the plane, MNy, par- N allel to the base, it will be a circle, and the com- mon ordinate, oy =y, will be perpendicular to Fig. 126. the diameter, MoN ; (oy)(oy) = (oM)(oN). oM _ sinoAM oN sinoBN. oA sinoMA’* oB sinoNB’ , puting OA — OB = a, Go=z,-and b= yl}, sinoAM — sinoBN a ~ sinoMA° sinoNB keicentt) s sinoAM , sinoBN But ee py ae By es I Mee ~sinoMA ° sinoNB é 2 _ = 1—-—,,, the equation of an ellipse. Cor. The section becomes a circle when the angles, (589) which the truncating plane and circular base make with the sides of the cone, are equal and contrary. For, when the angle oAM = oNB, then Z oBN = oMA, and the equation becomes yy? =a? — 2’, that of a circle, PROPOSITION VI. The Stereographic Projection of any circle of the (590) sphere is, in general, itself a circle. In the stereographic projection every point of the spherical sur- face is thrown upon an equatorial plane, denominated the primi- tive circle, by the intersection with this plane of a line drawn through the point and the eye situated in the pole of the primitive. Now, let AB be that diameter of any circle, STEREOGRAPHIC PROJECTION, 285 whose extremities lie in the circumference, AP,BOH,P,EH,, passing through the poles, E, O, of the primitive H,H,H,H,, whose cen- tre, 0, is the same with the centre of the sphere, and draw AE, BE, intersecting H,H, in a, 6; ab is the stereographic projection of AB. We have, Fig. 127. measure of /EBA = tarcEH,A =+(EH,+ H,A)= meas. of Z Ead ; ZEBA = Ead, and (589) the projection of the circle described on AB is itself a circle situated in the primitive H,H,H,H,, and having the diame- ter, ab. PROPOSITION VII The distance from the centre of the primitive at which (591) any point will be projected, will be equal to the tangent of half the arc intercepted between the point and the pole of the primi- tive, the radius of the sphere being taken for unity. For let A be any point on the surface of the sphere ; we have oa _ sinoKa oF sinoaE or oa = tanzOA, [oE = 1}. (591) Cor. 1. If m be the centre of the circle, ab, we have, dist. of centre om =4(oa-+ 0b) =4(tantOA + tansOB), (592) where it is to be observed that oa, od, or their equivalents, tantOA, tantOB, change signs on A, B, passing O. ‘ Cor. 2. The radius ma = 4(tantOA — tandOB). (593) Cor. 3. The projection of a circle parallel to the primi- (594) tive will be concentric with the primitive. For, if A,B, be parallel to H,H,, we have OA,=OB,, .. om, =+4(tantOA, — tantOA,) = 0, Mod, = +(tantOA, + tantOA,) = tantOA,, Cor, 4. If the pole of the circle be in the primitive, the (595) distance of its projected centre will be the secant of its radius, and the radius of projection the tangent of the same arc. For, if we put H,A, = H,B, = wu, then 0A, = 45° + 4u, JOB, = 45° — 4u; and .%. (592), (593), reduced by (356), become = tanoEKa, 286 STEREOGRAPHIC PROJECTION. PA: om, = —— = Secu, Ccosu sinu M343 = COSU == Ani, Cor. 5. A great circle perpendicular to the primitive (596) will be projected in a diameter of the primitive. For we have (595), OM, — 9 = Sec90° = infinity, MA, 9° = tan90° = infinity. The same is obvious from the figure. Cor. 6, Any great circle will have for the distance of its (597) projected centre, the tangent of the arc measuring its inclination to the primitive, and the radius of projection will be the secant of the same arc, Let P,P, be the diameter of any great circle, P,H,P,H,, and imitate the reasoning in (595.) Scholium J. Any point situated in the primitive, is its (598) own projection. Scholium I. Uf a great circle and the primitive intersect (599) each other in the diameter, H,H,, and a third circle passing through the axis of the primitive in the diameters, P,,P,, H,H,, the points, H,, H,, may be found by spherical trigonometry, and the point, p,, the projection of P,, having been determined by (591), there will be three points given through which to pass the circumference, H,p,H,, the projection of H,P,,H). H,,H, will be found (586), if we put HH) =, BLA 4,0, =. PROPOSITION VIII. To make the Stereographic Projection of the Sphere. I. Let the eye be situated at the south pole in order to project the northern hemisphere. De- scribe the primitive and graduate its circumfer- ence according to the number of meridians which it is intended to lay down; these will be diame- ters passing through the several points of division ‘ (596). | Fig. 128. For the parallels of latitude, we have only to count off from the STEREOGRAPHIC PROJECTION, 287 north point the corresponding degrees, and from the points of division to draw lines to the south point, and the intersections thus formed with the east and west diameter will be in the circumfer- ences of the circles of latitude whose centres will be that of the primitive, (591), (594). II. Let the eye be in the equator. primitive will pass through the poles, N, 8, and the central meridian, NS, and equator, EQ, will be north and south, and east and west lines intersecting in the centre of the Graduate the circumfer- ence as above, and, laying -the corner of a wooden square upon a point of division, di- rect the edge of one arm through the centre, primitive, (596). The Fig. 129. The intersection of the corresponding edge of the other arm with the production of the line, NS, will be the centre, and the distance of this point to the point of division the radius, for the description of a parallel of latitude, (595). The radii just employed set off from the centre upon EQ will give the centres for describing the meridians, which will pass through the points, N, S, (597), (598). III. When the eye is sit- uated otherwise than as above, consult (598), (599), and figure 130. Scholium. It will be ob- served that the middle of the map is comparatively con- tracted in the stereographic and enlarged in the ortho- graphic projection. To avoid this, the lines, NS, EQ, are sometimes divided into equal parts, and the map thus constructed is im- SO Were "Z IN Sy CoOImapwrrHo| WAIHI Wee o} 100 0000000 0043214 0086002 0128372 0170333 0211893 0253059 0293838 0334238 0374265 170 110 0413927 0453230 0492180 0530784 0569049 0606978 06445890 0681859 0718820 0755470 180 2304489 2329961 2355284 2380461 2405492 2430380 2455127 2479733 2504200 2528530 240 3802112 3820170 3838154 3856063 3873898 3891661 3909351 3926970 3944517 3961993 2552725 2576786 2600714 2624511 2648178 2671717 2695129 2718416 2741578 2764618 250 3979400 3996737 4014005 4031205 4048337 4065402 4082400 4099331 4116197 4132998 120 0791812 0827854 0863593 0899051 0934217 0969100 1003705 1038037 1072100 1105897 1172713 1205739 1238516 1271048 1303338 1335389 1367206 139879 1 1430148 Pe 4 (2 —1)(2—2) + D; 1 cQaei3” 130 1139434 | 1461280 140 1492191 1522883 1553360 1583625 1613680 1643529 1673173 1702617 1731863 190 200 2787536 2810334 2833012 2855573 2878017 2900346 2922561 2944662 2966652 2988531 260 4149733 4166405 4183013 4199557 4216039 4232459 4248816 4265113 4281348 4297523 3010300 3031961 3053514 3074960 3096302 3117539 3138672 3159703 3180633 3201463 270 4313638 4329693 4345689 4361626 4377506 4393327 4409091 4424798 4440448 4456042 3222193 3242825 3263359 3283796 3304138 3324385 3344538 3364597 3384565 3404441 280 4471580 4487063 4502491 4517864 4533183 4548449 4563660 4578819 4593925 4608978 210. | 150 1760913 1789769 1818436 1846914 1875207 1903317 1931246 1958997 1986571 2013971 220 3424227 3443923 3463530 3483049 3502480 3521825 3541084 3560259 3579348 3598355 290 4623980 4638930 4653829 4668676 4683473 4698220 4712917 4727564 4742163 4756712 160 2041200 2068259 2095150 2121876 2148438 2174839 22101081 2227165 2253093 2278867 230 3617278 3636120 3654880 3673559 3692159 3710679 3729120 3747483 3765770 3783979 300 4771213 4785665 4800069 4814426 4828736 4842998 4857214 4871384 4885507 4899585 LOGARITHMS OF NUMBERS, 31 7 [CDI FAB wWOR S| |oDIMABEVH S| [ODIMMABYH S| [CDYRMURwWHHO| [CODIMTAWDH Oo 310 4913617 4927604 4941546 4955443 4969296 4983106 4996871 5010593 5024271 5037907 320 5797836 5809250 5820634 5831988 5843312 5854607 5865873 5877110 5888317 5899496 450 6532125 6541765 6551384 6560982 6570559 6580114 6589648 6599162 6608655 6618127 520 7160033 7168377 7176705 7185017 7193313 7201593 7209857 7218106 7226339 7234557 590 “708520 1715875 77123217 717130547 1737864 7745170 7752463 77159743 7767012 77714268 320 5051500 5065050 5078559 5092025 5105450 5118834 5132176 5145478 5158738 5171959 ~~ 390 400 340 5314789 5327544 5340261 535294 | 5365584 5378191 5390761 5403295 5415792 5428254 330 5185159 5198280 5211381 5224442 5237465 5250448 5263393 5276299 5289167 5301997 5910646 5921768 5932861 59435926 5954962 5965971 5976952 5987905 5998831 6009729 460 6627578 6637009 6646420 6655810 6665180 6674530 6683859 6693169 6702459 6711728 530 6020600 6031444 604226 | 6053050 6063814 6074550 6085260 6095944 6106602 | 6211763 6117233 | 6222140 470 480 6720979 | 6812412 6730209 | 6821451 6739420 | 6830470 6748611 | 6839471 6757783 | 6848454 6766936 | 6857417 6776070 | 6866363 6785184 | 6875290 6794279 | 6884198 6803355 | 6893089 ~ 540 550 6127839 6138413 | 6148972 | 6159501 6170003 6180481 6190933 6291361 7242759 7250945 7259116 7267272 7275413 7283533 7291648 7299743 7307823 7315888 7323938 | 7403627 7331973 | 7411516 7339993 | 7419391 7347998 | 7427251 7355989 | 7435098 7363965 | 7442930 7371926 | 7450748 7379873 | 7458552 7387806 | 7466342 7395723 | 7474118 ay 410° tor 350 5440680 5453071 5465427 5477747 5490033 5502284 5514500 5526682 5538830 5550944 420 6232493 6242821 6253125 6263404 6273659 6283889 6294096 6304279 6314438 6324573 360 5563025 5575072 5587086 5599966 5611014 5622929 5634811 5646661 5658478 5670264 370 5682017 5693739 5705429 5717088 5728716 5740313 5751878 5763414 5774918 5786392 |! 430 6334685 6344773 6354837 6364879 6374897 6384893 6394865 6404814 6414741 6424645 440 6434527 6444386 6454223 6464037 6473830 6483600 6493349 6503075 6512780 6522463 | 490 6901961 6910815 6919651 6928469 6937269 6946052 6954817 6963564 6972293 6981005 560 7481880 7489629 7497363 7505084 7512791 7520484 7528164 7535831 7543483 7551123 600 7781513 T7188745 7795965 7803173 7810369 7817554 7824726 7831887 7839036 7846173 610 620 630. 500 6989700 6998377 7007037 7015680 7024305 7032914 7041505 7050080 7058637 7067178 570 7558749 7566061 7573960 7981546 7589119 7596678 7604225 7611758 7619278 510 7075702 7084209 7092700 TLO11L74 7109631 7118072 7126497 7134905 7143298 7151674 580 7634280 7641761 7649230 7656686 7664128 7671559 7678976 768638 L 7693773 7701153 650 7853298 | 7923917 7860412 | 7930916 7867514 | 7937904 7874605 | 7944880 7881684 | 7951846 7888751 | 7958800 7895807 | 7965743 7902852 | 7972675 7909885 | 7979596 7916906 | 7986506 7993405 8000294 8007171 8014037 8020893 8027737 8034571 8041394 8048207 8055009 8061800 8068580 8075350 8082110 8088859 8095597 8102325 8109043 8115750 8122447 8129134 8135810 8142476 8149132 8155777 8162413 8169038 8175654 8182259 8188854 318 LOGARITHMS OF NUMBERS, 660 670 680 690 700 710 70 || 8195439 | 8260748 | 8325089 | 8388491 | 8450980 | 8512583 | 8573325 8202015 | 8267225 | 8331471 | 8394780 | 8457180 | 8518696 | 8579353 8208580 | 8273693 | 8337844 | 8401061 | 8463371 | 8524800 | 8585372 8215135 | 8280151 | 8344207 | 8407332 | 8469553 | 8530895 | 8591383 8221681 | 8286599 | 8350561 | 8413595 | 8475727 | 8536982 | 8597386 8228216 | 8293038 | 8356906 | 8419848 | 8481891 | 8543060 | 8603380 8234742 | 8299467 | 8363241 | 8426092 | 8488047 | 8549130 | 8609366 8241258 | 8305887 | 8369567 | 8432328 | 8494194 | 8555192 | 8615344 8247765 | 8312297 | 8375884 | 8438554 | 8500333 | 8561244 | 8621314 8254261 | 8318698 | 8382192 | 8444772 | 8506462 | 8567289 | 8627275 730 740 750 760 77 780 790 8633229 | 8692317 | 8750613 | 8808136 | 8864907 | 8920946 | 6976271 8639174 | 8698182 | 8756399 | 8813847 | 8870544 | 8926510 | 8981765 8645111 | 8704039 | 8762178 | 8819550 | 8876173 | 8932068 | 8987252 || 8651040 | 8709888 | 8767950 | 8825245 | 8881795 | 8937618 | 8992732 |. B656961 | 8715729 | 8773713 | 8530934 | 8987410 | 8943161 | 8998205 8662873 | 8721563 | 8779470 | 8836614 | 8893017 | 8948697 | 9003671 || 8668778 | 8727388 | 8785218 | 8842288 | 8898617 | 8954225 | 9009131 8674675 | 8733206 | 8790959 | 8817954 | 8904210 | 8959747 | 9014583 8680564 | 8739016 | 8796692 | 8853612 | 8909796 | 8965262 | 9020029 8686444 | 8744818 | 8802418 | 8859263 | 8915375 | 8970770 | 9025468 800 810 820 830 840 850 860 9030900 | 9084850 | 9138139 | 9190781 | 9242793 | 9294189 | 9344985 9036325 | 9090209 | 9143432 | 9196010 | 9247960 | 9299296 | 9350032 9041744 | 9095560 | 9148718 | 9201233 | 9253121 | 9304396 | 9355073 9047155 | 9100905 | 9153998 | 9206450 | 9258276 | 9309490 | 9360108 9052561 | 9106244 | 9159272 | 9211661 | 9263424 | 9314579 | 9365137 9057959 | 9111576 | 9164539 | 9216865 | 9268567 | 9319661 | 9370161 9063350 | 9116902 | 9169800 | 9222063 | 9273704 | 9324738 | 9375179 9068735 | 9122221 | 9175055 | 9227255 | 9278834 | 9329808 | 9380191 9074114 | 9127533 | 9180303 | 9232440 | 9283959 | 9334873 | 9385197 |: 9079485 | 9132839 | 9185545 | 9237620 | 9289077 | 9339932 | 9390198 || 870 880 890 900 910 920 930 9395193 | 9444827 | 9493900 | 9512425 | 9590414 | 9637878 | 9684829 9400182 | 9449759 | 9498777 | 9547248 | 9595184 | 9642596 | 9689497 9405165 | 9454686 | 9503649 | 9552065 | 9599948 | 9647309 | 9694159 9410142 | 9459607 | 9508515 | 9556878 | 9304708 | 9652017 | 9698816 9415114 | 9464523 | 9513375 | 9561684 | 9609462 | 9656720 | 9703469 9420081 | 9469433 | 9518230 | 9566486 | 9614211 | 9661417 | 9708116 9425041 | 9474337 | 9523080 | 9571282 | 9618955 | 9666110 | 9712758 9429996 | 9479236 | 9527924 | 9576073 | 9623693 | 9670797 | 9717396 9434945 | 9484130 | 9532763 | 9580858 | 9628427 | 9675480 | 9722028 9439889 | 9489018 | 9537597 | 9585639 | 9633155 | 9680157 | 9726656 940 950 960 970 980 990 1000 9731279 | 9777236 | 9822712 | 9867717 | 9912261 | 9956352 | 0000000 9735896 | 9781805 | 9827234 | 9872192 | 9916690 | 9960737 9740509 | 9786369 | 9831751 | 9876663 | 9921115 | 9965117 9745117 | 9790929 | 9836263 | 9881128 | 9925535 | 9969492 9749720 | 9795484 | 9840770 | 9885590 | 9929951 | 9973864 9754318 | 9200034 | 9845273 | 9390046 | 9934362 | 9978231 9758911 | 9804579 | 9849771 | 9894498 | 9938769 | 9982593 9763500 | 9809119 | 9854265 | 9898946 | 9943172 | 9986952 9768083 | 9813655 | 9858754 | 9903389 | 9947569 | 9991305 9772662 | 9818186 | 9863238 | 9907827 | 9951963 | 9995655 | CHIMUR WWHO| ] Pau mMIRwWWR S| | CDRA TR WOH S| | OLIATRWWR S| —— [oma ms aRwwrs| LOGARITHMIC SINES. D, D, D, Yo=Yotss LER Aare Nr 5 1 (rt — 1)(t — 2) vias art From 0*1° to 0‘5° inclusive the characteristic is 3, from 0‘6° to 5‘7° it is 2, and after that I. | ge joe if go a 3° 49° 5e° 6° Hh ‘(0} —infin.| 2418553} 5428192; 7188002) 8435845] 9402960| 0192346 | ‘1'| 2418771 | 2832434] 5639994] 7330272! 8542905] 9488739! 0263865 | 21} 5429065} 3210269| 5841933} 7468015} 8647376] 9572843! 0334212 $8} 7189966! 3557835!] 6034886! 7601512] 8749381] 9655337] 0403424 ‘4/ 8439338] 3879622| 6219616/| 7731014} 8849031 | 9736280 | 0471538 ‘| 9408419| 4179190} 6396796| 7856753} 8946433 | 9815729] 0538588 ‘6 | 0200207} 4459409] 6567017! ‘79738941 | 9041685! 9893737] 0604604 ‘7| 0869646! 4722626} 6730804; 8097772} 9134881 | 9970356; 0669619 18] 1449532] 4970784| 6888625; 8213425] 9226105| 0045634 733663 |}. ‘9 | 1961020) 5205514} 7040899} 8326066) 9315439} 0119616| 0796762 | 7° xo go 10° 11° it ee eB (6) | 08589145} 1435553} 1943324) 2396702; 2805988} 3178789} 3520880 1] 0920237] 1489148] 1990913] 2439472] 2844803] 3214297] 3553582 2] 0980662 | 1542076 | 2037974) 2481811 | 2883260] 3249505! 3586027 3} 1040246 | 1594354] 2084516 | 2523729) 2921367] 3284416 | 3618217 | | 141 1099010] 1645998} 2130552] 2565233 | 2959129} 3319035) 3650158 (5) | 1156977) 1697021 | 2176092) 2606330} 2996553} 3353368] 3681853 |; 6 | 1214167) 1747439 | 2221147} 2647030] 3033644] 3387418) 3713304 ie7 | 1270600] 1797265} 2265725 | 2687338 | 3070407) 3421190} 3744517 8} 1326297} 1846512} 2309838] 2727263} 3106849} 3454688) 3775493 9 | 1381275] 1895195 | 2353494 | 2766811 | 3142975} 3487917] 3806237 . 14° io? ed oy. is Re ie 20° 0 | 3836752) 4129962} 4403381) 4659353 | 4599824] 5126419] 5340517 ‘1 | 3867040] 4158152] 4429728} 4681069 | 4923083] 5148371] 5361286 12! 3897106} 4186148] 4455994; 4708631 | 4946205) 5170198} 5381943 3 | 3926952 | 4213950} 4481909) 4733043 | 4969192] 5191904] 5402489 ‘4| 3956581 | 4241563} 4507747) 4757304 | 4992045| 5213488] 5422926 ‘| 3985996 | 4268988} 4533418) 4781418) 5014764| 5234953] 5443253 «6 | 4015201 | 4296228} 4558926] 4805385 | 5037353} 5256298] 5463472 7 | 4044196 | 4323285] 4584271) 4829208} 5059818 | 5277526] 5483585 '|-8| 4072987| 4350161 | 4609456) 4852888} 5082141] 5298638) 5503592 9 | 4101575 | 4376859 | 4634483| 4876426 | 5104343 | 5319635] 5523494 a, nn aaa i te 22° 5543292 5562987 5582579 5602071 5621462 5640754 56059948 5679014 5698043 5716946 5735754 5754468 5773088 5TIL616 5810052 5828397 5346651 5864816 5482892 5900830 LOGARITHMIC SINES. 23° 5918780 5936594 5954322 5971965 5989523 6006997 6024388 6041696 6058923 6076063 28° 29° 30° 24° 6093133 6110118 6127023 6143850 6160599 6177270 61933864 6210382 6226824 6243190 Hie. 6716093 6730319 6744485 6758592 6772640 6786629 6800560 6814434 6328250 6842010 35° 6855712 6869359 6832949 6896484 6909964 6923388 6930758 6959074 6963336 6976545 36° 6989700 7002802 7015852 7028249 7041795 7054689 7067531 7080323 7093063 7105753 37° 75859 13 7596718 7607483 7618208 7628894 7639540 7650147 7660715 7671244 7681735 42° 8255109 8263512 8271887 8280231 8288547 8296833 830509 L 8313320 8321519 832969 | 7092187 7792601 7712976 7723314 7733614 7743876 7754101 7164289 7774439 7784553 43° 8337833 8345948 8354033 8362091 8370121 8378122 8386096 8394041 8401959 8409850 49° 8777799 8784376 8790930 8797462 8803970 BR10455 8816918 8823357 8829774 8836168 50° 8842540 8845889 8855215 8861519 8867801 8374061 8880298 8886513 8892706 8898877 7794630 7804671 7814675 7824643 7834575 7344471 7854332 7864157 7873946 788370 1 44° 8417713 8425548 8433356 8441137 844889 | 8456618 8464318 8471991 8479637 8487257 7118393 7130983 7143524 7156015 7168458 7180851 7193196 7205493 7217742 7229943 38° 7893420 7903104 7912754 7922369 7931949 7941496 7951008 7960436 7969930 7979341 bi itip ra 8494850 8502417 8509957 8517471 8524959 8532421 8539856 8547266 8554650 8562008 25° 6259483 6275701 6291845 6307917 6323916 6339844 6355699 6371484 6387199 6402844 26° 6418420 6433926 6449365 6464735 6480038 6495274 6510444 6525548 6540586 6555559 a7° || 6570468 6585312 6600093 6614810, 6629464 6644056 | 6658586 6673054 | 6687461 6701807 g 32° 33° 34° 7242097 7254204 7266264 7278277 7290244 7302165 7314040 7325870 7337054 7349393 39° 7988718 7998062 8007372 5016649 5025894 8035105 8044284 8053430 8062544 8071626 45°h 7361088 7372737 7384343 7395904 7407421 7418895 7430325 7441712 7453056 7464358 49° 8080675 8089692 8098678 8107631 8116554 8125444 8134303 8143131 8151928 8160694 7475017 | 7436833 7498007 7509140 71520231 7531280 |! 7542288 7553256 7564182 7575068 41° 8169429 8178133 8186807 8195450 8204063 | 8212646 8221198 8229721 || 8238213 8246676 47° UE ARO i 8569341 8576648 8583929 8591186 8598416 8605622 8612803 8619958 8627088 8634194 8641275 8648331 8655362 8662369 8669351 8676309 8683242 8690152 8697037 8703898 Dre 8905026 8911153 8917258 8923342 8929404 8935444 8941462 8947459 8953435 8959389 52° 8965321 8971233 8977123 8982992 8988840 8994667 9000472 9006257 9012021 9017764 9074059 | 9128328 | 9180620 53° 9023486 9029188 9034868 9040529 9046168 9051787 9057386 9062964 9068522 54° 9079576 90850738 9090550 9096007 9101444 9106860 9112257 9117634 9122991 ge 8710735 8717548 8724337 8731102 8737844 8744561 8751256 8757927 8764574 8771198, 9133645 9138943 9144221 9149479 | 9154718 9159937 9165137 | 9170317 9175478 LOGARITHMIC SINES. 56° ~ 9185742. 9190845 9195929 9200994 9206039 9211066 9216073 9221062 9226032 9230982 63° 9498809 9502663 9506500 9510320 9514124 9517912 9521683 9525437 9529175 9532897 7° 9729858 9732610 9735346 9738067 9740774 9743466 9746142 9748804 9751451 9754083 one 9887239 9888982 9890711 9892427 9894128 9895815 9897489 9899148 9900794 9902426 84° 9976143. 9976933 9977710 9978473 9979223 9979960 9980683 9981393 9982089 9982772 57° 9235914 9240827 9245721 9250597 9255454 9260292 9265112 9269913 9274695 9279459 64° 9536602 9540291 9543963 9547619 9551259 9554882 9558490 9562081 9565656 9569215 ai? 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