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To renew call Telephone Center, 333-8400 UNIVERSITY OF ILLINOIS LIBRARY AT URBANA-CHAMPAIGN ‘Whe 14 aC L161—O-1096 Digitized by the Internet Archive In 2022 with funding from University of Illinois Urbana-Champaign https://archive.org/details/highergeometrytrOOscho a \ ; HIGHER GHOMETRY TP RgieG ON: OOM Te ie: THIRD PART OF A SERIES ON ELEMENTARY AND HIGHER GEOMETRY, TRIGONOMETRY, AND MENSURATION, CONTAINING MANY VALUABLE DISCOVERIES AND IMPROVEMENTS IN MATHEMATICAL SCIENCE, ESPECIALLY IN RELATION TO THE QUADRATURE OF THE CIRCLE, AND SOME OTHER CURVES, AS WELL AS THE CUBATURE OF CERTAIN CURVILINEAR SOLIDS 5; DESIGNED AS A TEXT-BOOK FOR COLLEGIATE AND ACADEMIC INSTRUCTION, AND AS A PRACTICAL COMPENDIUM OF MENSURATION. BY NATHAN *SGHOLFIELD. a J i ”, Ps ° : i f} ~— ' é . i ° uN + 4 t s \ » 4 : aS 4 “ 5 | Tm, *- a A Ja ae NEW YORK: PUBLISHED BY COLLINS, BROTHER & CO. No..254 Pearl Street. eeoeeon2e 1845, Entered according to Act of Congress, in the year 1845, by NATHAN SCHOLFIELD, In the Clerk’s Office of the District Court of Connecticut. : : my : J , ’ -*? pe G. W. WOOD, PRINTER, 29 GOLD ST., NEW YORK. gy, Fe" he aR ka TE CaCI | ‘ ‘“ x Sch] \. aTHEMaTics LIBRARY, PREFACE. Tuts part of the series consists of spherical geometry, takgn mostly from Brewster’s translation of Legendre’s work. Ana- lytical plane and spherical trigonometry, based on the subject, as found in Rutherford’s edition of Hutton’s Mathematics, being originally abridged from the larger works of Cagnoli, and others ; but, in this work, much improved and enlarged. To which are added many practical exercises on the subject, by way of application. In this treatise will be found many cu- rious and highly useful problems in trigonometrical surveying, and topographical operations, not before published. The pro- perties of the circle are introduced advantageously into trigo- nometrical problems—hence we are enabled, by geometrical construction, and trigonometrical analysis, to determine many otherwise extremely difficult problems, in a manner at once simple, elegant, and satisfactory. The application of alge- bra to geometry, is discussed in such manner as to combine the principles of the two sciences. The properties of the pa- rabolic, elliptical, and hyperbolic curves, being such as are formed by the sections of a cone, and hence are usually de. nominated conic sections, are also discussed. This subject is, with some alterations and additions, taken from Rutherford’s edition of Hutton’s Mathematics. It is the design of the au- thor, to preserve an unbroken connection from pure elemen- tary to the higher Geometry and mensuration; and with this object in view the present volume, being the third part of the series, is prepared. Eee Bis. iV PREFACE. The well established reputation and the high respectability of the authors from whom our selections have been made, renders it unnecessary for us to discuss their merits in order to secure a favorable reception of this. It will only be neces- sary for us, in the following pages, to preserve the same de- gree of accuracy and perspicuity in our digressions as charac- terise those works, and we shall have nothing to fear from the criticisms of scientific amateurs and mathematicians. ’ CONTENTS. SPHERICAL GEOMETRY. | PAGE. Definitions and General Propositions, - - - - -— = 9 ANALYTICAL PLANE TRIGONOMETRY. Cuap. I.—Definitions and Illustration of Principles, - - - 27 Cuap. II.—General Formule, - . . - > - - 45 Cuap. III.—Formule for the Solution of Triangles, - - - 66 Cuar. [V.—Construction of Trigonometrical Tables, - - - 738 Cuap. V.—Logarithms, aay he Ot = SS Lickers helt Heike =). See ae Cuar. VI.—Solution of Right Angled Triangles, aS Ate grag eae OS Cuap. VII.—Solution of Oblique Angled Triangles, - - - 98 Cuap. VIII.—On the use of Subsidiary Angles, - - - = 107 Cuap. [X.—On the Solution of Geometrical Problems by Trigo- nometry, - - - - - - . oo DU Cuap. X.—Problems in Trigonometrical Surveying, &c. - - 113 SPHERICAL TRIGONOMETRY. Cuar. I.—General Principles and Illustrations, - - - - 131 Cuap. II.—Solution of Right Angled Spherical Triangles,- - 144 Culp. Ill.—Solution of Oblique Angled Spherical Triangles, - 150 Cuap. IV.—On the use of Subsidiary Angles, - - - = 154 vi . CONTENTS. APPLICATION OF ALGEBRA TO GEOMETRY. PAGE. Construction of Algebraical Quantities, - - - - - 158 Geometrical Questions, the modes of forming Equations therefrom, and their Solutions, - - - - + - - 166 Determination of Algebraic Expressions for Surfaces and Solids, 180 CONIC SECTIONS. The Parabola and its Properties, oer, ay mad Bias - - 189 The Ellipse and its Properties, - - - - - sue Ars The Hyperbola and its Properties, - - a Ai Fe at - 215 ( SPHERICAL GEOMETRYS: ©. SPHERICAL GEOMETRY. DEFINITIONS. 1. Tue sphere is a solid terminated by a curve surface, all the points of which are equally distant from a point within, called the centre. The sphere may be con- ceived to be generated by the revolution of a semi- circle DAE about its di- ameter DE; forthe surface described in this move- ment, by the curve DAE, will have all its points equally distant from its centre C. 2. The radius of a sphere is a straight line, drawn from the centre to any point of the surface ; the diameter, or axis, is a line passing through this centre, and terminated on both sides by the surface. All the radii of a sphere are equal; all the diameters are equal, and each double of the radius. 3. It will be shown (Prop. I.) that every section of the sphere, made by a plane, is a circle. This granted, a great circle is a section which passes through the centre; a small circle, one which does not pass through the centre. 4. A plane is tangent to a sphere, when their surfaces have but one point in common. a { . hi. 8 SPHERICAL GEOMETRY. 5. The pole of a circle of a sphere is a point’ in the surface equally distant from all the points in the circumference of this circle. ; 6. A spherical triangle isa portion of the surface of asphere, bounded by three arcs of great circles. Those arcs, named the sides of the triangle, are always supposed to be each less than asemi-circumference. The an- gles which their planes form with each other, are the angles of the triangle. 7. A spherical triangle takes the name of right-angled, isosceles, equilateral, in the same cases as a rectilineal triangle. 8. A spherical polygon is a portion of the surface of a sphere, terminated by several arcs of great circles. 9. A lune is that portion of the surface of a sphere which is included between two great semicircles, meeting in a com- mon diameter. ; 10. A spherical wedge, or ungula, is that portion of the solid sphere which is included between the same great semi-circles, and has the lune for its base. 11. A spherical pyramid is a portion of the solid sphere in- cluded between the planes of a solid angle, whose vertex is the centre. The base of the pyramid is the spherical polygon intercepted by the same planes. 12. A zone is the portion of the surface of the sphere in- cluded between two parallel planes, which form its bases. One of those planes may be tangent to the sphere ; in which case, the zone has only a single base. 13. A spherical segment is the portion of the solid sphere included between two parallel planes which form its bases. One of these planes may be tangent to the sphere ; in which case, the segment has only a single base. 14. The altitude of a zone or of a segment is the distance between the two parallel planes, which form the bases of the zone or segment. 15. Whilst the semicircle DAE (Def. 1.) revolving round its diameter DE, describes the sphere, any circular sector, as DCF or FCH, describes a solid, which is named a spherical sector. 16. The symbal °.* which occurs in this volume, is used to denote because ; when applied in algebraic notation. . . ie SPHERICAL GEOMETRY. 9 PROPOSITION I. THEOREM. Every section of a sphere, made by a plane, is a circle. q © Let AMB be a section, made by a D plane, in the sphere, whose centre is C. YEE From the point C, draw CO perpendicu- 4 — B lar to the plane AMB ; and different lines CM, CM, to different points of the curve AMB, ohich terminates the section. The oblique lines CM, CM, CA, are equal, being radii of the sphere; hence (Prop. VI. B. I. El. S. Geom.) they are equally distant from the perpendicular CO; therefore all the lines OM, MO, OB, are equal. Consequently, the section AMB isa circle, whose centre is O. Cor. 1. If the section passes through the centre of the sphere, its radius will be the radius of the agile ; hence, all great circles are equal. Cor. 2. ‘Two great circles always bisect each other ; for their common intersection, passing through the centre, is a diameter. Cor. 3. Every great circle divides the sphere and its sur- face into two equal parts; for, if the two hemispheres were separated, and afterwards placed on the common base, with their convexities turned the same way, the two surfaces would exactly coincide, no point of the one being nearer the centre than any point of the other. Cor. 4. The centre of a small circle, and that of the sphere, are in the same straight line, perpendicular to the plane of the small circle. Cor. 5. Small circles are the less the further they lie from the centre of the sphere ; for, the greater CO is, the less is the chord AB, the diameter of the small circle AMB. Cor. 6. Anarc of a great circle may always be made to pass through any two given points of the surface of the sphere; for the two given points, and the centre of the sphere, make three points, which determine the position of a plane. But if the two given points were at the extremities of a di- ameter, these two points and the centre would then lie in one straight line, and an infinite number of great circles might be made to pass through the two given points. 10 SPHERICAL GEOMETRY. PROPOSITION Il. THEOREM. In every spherical triangle, any side is less than the sum of the ' + other two. Let O be the centre of the sphere, y and ACB the triangle: draw the radii OA, OB, OC. Imagine the planes AOB, AOC, COB, to be | drawn; those planes will form a solid angle at the centre O; and the angles AOB, AOC, COB, wiil be measured by AB, AC, BC, the sides of the spherical triangle. But each of the three plane angles forming a solid angle is less than the sum of the other two (Prop. XXI., B. I. El. S. Geom.): hence any side of the triangle ABC is less than the sum of the other two. oy we] PROPOSITION Ill, THEOREM. The shortest distance from one point to another, on the surface of a sphere, is the arc of the great circle which joins the two given points. “ Let ADB be the arc of the great circle which joins the points A and B; and with- out this line, if possible, let M be a point of the shortest path between A and B. Through the point M, draw MA, MB, arcs of great ay D circles; and take BD=MB. By the last Proposition, the arc ADB is shorter than AM+MB; take BD=BM re- spectively from both; there will remain B AD PROPOSITION X. THEOREM. The same supposition continuing as in the last proposition, each angle in the one of the triangles, will be measured by the semicircumference minus the side lying opposite to wt in the other triangle. Produce the sides (see the preceding diagram) AB, AC, if 16 “#4 SPHERICAL GEOMETRY. ~ necessary, till they meet EF, inG and H. The point A being the pole of the arc GH, the angle A will be measured by that arc. But the arc EH is a quadrant, and likewise GI, E be- ing the pole of AH, and F of AG; hence EH+GF is equal to the semicircumference. Now, EH+GEF is the same as EF+GH; hence the arc GH, which measures the angle A, is equal to a semicircumference minus the side KF’. In like manner, the angle B will be measured by 3 ctre.—DF: the angle C, by 1 circ.—DE. And this property must be reciprocal in the two triangles, since each of them is described in a similar manner by means of the other. Thus we shall find the angles D, E, F, of the triangle DEF to be measured respectively by 4 circ.—BC, 1 circ.—AC. 1 circ—AB. Thus the angle D, for example, is measured by the arc MI; but MI+BC=MC+BI=1} circ. ; hence the arc MI, the measure of D, is equal to 4 circ.—BC ; and so ofall the rest. Scholium. It must further be observed, that besides the tri- angle DEF’, three others might be formed by the intersection of the three arcs DE, EF, DF. But the proposition immediately before us is ap- D plicable only to the central triangle, which is distinguished from the other ) three by the circumstance (see diagram / to Prop. IX.) that the two angles A and D lie on the same side of BC, the two B and FE on the same side of AC, and the d two C and F on the same side of AB. Various names have been given to the triangles ABC, DEF ; we shall call them polar triangles. PROPOSITION XI. THEOREM. If around the vertices of the two angles of a given spherical trt- angle, as poles, the crcumference of two circles be described which shall pass through the third angle of the triangle ; if then, through the other point in which those circumferences intersect, and the two first angles of the triangle, the arcs of great circles be drawn, the triangle thus formed will have all its parts equal to those of the first triangle. Let ABC be the given triangle, CED, DFC the arc de- scribed about A and B as poles; then will the triangle ADB have all its parts equal to those of ABC. SPHERICAL GEOMETRY. 17 For, by construction, the side AD=AC, A DB=BC, and AB is common; hence those two triangles have their sides equal, /\ each to each. We are now to show, that O the angles opposite these equal sides are F also equal. ¥ If the centre of the sphere is supposed © Re to be at O, a solid angle may be conceiv- E ed as formed at O by the three plane an- gles AOB, AOC, BOC ; likewise another 5 solid angle may be conceived as formed by the three plane angles AOB, AOD, BOD. And because the sides of the tri- ongle ABC are equal to those of the triangle ADB, the plane angles forming the one of these solid angles, must be equal to the plane angles forming the other, each to each. But in this case we have shown (Prop. XXIII. Hl. S. Geom.) that the planes, in which the equal angles lie, are equally inclined to each other ; hence, all the angles of the spherical triangle DAB are respectively equal to those of the triangle CAB, namely, DAB=BAC, DBA=ABC, and ADB—=ACB;; hence, the sides and the angles of the triangle ADB are equal to the sides and the angles of the triangle ACB. Scholium. The equality of those triangles is not, however, ~ an absolute equality, or one of superposition ; for it would be impossible to apply them to each other exactly, unless they were isosceles. The equality meant here is what we have al- ready named an equality by symmetry ; therefore, we shall call the triangles ACB, ADB, symmetrical triangles. PROPOSITION XII. THEOREM. Two triangles on the same sphere, or on equal spheres, are equal in all their parts, when they have each an equal angle included between equal sides. Suppose the side AB=EF, the ie side AC=EG, and the angle BAC E =I EG; the triangle EG may be placed on the triangle ABC, or on ABD symmetrical with ABC, just as two rectilineal triangles are placed upon each other, when they p have an equal angle included be- C tween equal sides. Hence all the . parts of the triangle EF'G will be B F equal to all the parts of the trian- 2 ’ “F 18 SPHERICAL GEOMETRY. gle ABC ; that is, besides the three parts equal by hypothesis, we shall have the side BC=FG, the angle ABC=EFG, and the angle ACB=EGF. inks “ PROPOSITION XIII. THEOREM. Two triangles on the same sphere, or on equal spheres, are equal an all their parts, when two angles and the included side of the one are respectively equal to two angles and the included side of the other. For, one of those triangles, or the triangle symmetrical with it, may be placed on the other, as is done in the corresponding case of rectilineal triangles, (Prop. IX. B. Il. El. Geom.) PROPOSITION XIV. THEOREM. If two triangles on the same sphere, or on equal spheres have all their sides respectively equal, their angles will likewise be all respectively equal, the equal angles lying opposite the equal sides. : This truth is evident from Proposition XI, where it is shown that, with three given sides AB, AC, BC, (see the diagram,) there can only be two triangles ACB, ABD, differing as to the position of their parts, and equal as to the magnitude of those parts. Hence, those two triangles, having all their sides re- spectively equal in both, must either be absolutely equal, or at least symmetrically so; in both of which cases their corres- ponding angles must be equal, and lie opposite to equal sides. PROPOSITION XV. THEOREM. In every isosceles spherical triangle, the angles opposite the equal sides are equal; and conversely, if two angles of a spherical triangle are equal, the triangle is isosceles. First. Suppose the side AB=AC; we shall have the angle C=B. For, if the arc be drawn from the vertex A to the middle point D of the base, the two triangles ABD, ACD will have all the sides of the one respectively equal to the corresponding sides of the other, namely, AD common, BD=DC, and AB= AC: hence, by the last Proposition, their an- gles will be equal; therefore B=C. Secondly. Suppose the angle B=C; we a a SPHERICAL GEOMETRY. 19 shall have the side AC=AB. For, if not, let AB be the greater of the two; take BO=AC, and join OC. The two sides BO, BC are equal to the two AC, BC; the angle OBC, contained by the first two is equal to ACB contained by the second two. Hence (Prop. XII.) the two triangles BOC, ACB have all their other parts equal; hence the angle OCB=ABC: but by hy- pothesis, the angle ABC=ACB; hence we have OCB=ACB, which is absurd; hence it is absurd to suppose AB different from AC ; hence the sides AB, AC, opposite to the equal an- gles B and C, are equal. Scholium. The same demonstration proves the angle BAD =DAC, and the angle BDA=ADC. Hence the two last are right angles; hence the arc drawn from the vertex of an isos- celes spherical triangle to the middle of the base, is at right an- gles to the base, and bisects the vertical angle. PROPOSITION XVI. THEOREM. In any spherical triangle, the greater side is opposite the great- erangle; and conversely, the greater angle is opposite the greater side. Let the angle A be greater than the angle B, then will BC be greater than AC; and conversely, if BC is greater than AC, then will the angle A be greater than B. First. Suppose the angle A A>B; make the angle BAD= B: then (Prop. XV.) we shall have AD=DB; but AD+DC is greater than AC; hence, putting DB in place of AD, we shallhave DB+ DC, or BC> AC. Secondly. If we suppose BC>AC, the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC=AC; if BAC were less than ABC, we should then, as has just been shown, find BC =cos. 4, be Draw a circle A’/B’D’E’ equal to the circle ‘2 ABDE, and from C’ the centre, draw C’P’, Aly making with C’A’ the angle P’C’A’ equal to 4: D’ the angle PCB; that is, to the compliment of PCA, or to (90°—4). E Then, since CP is equal to C’P’, and the angles at S and 8’ are right angles, the angle CPS equal to the angle P’C’S’ the two trian- A gles PCS, P’C’S’ are equal in every respect ; fo—C'S’, CS=P's CS P's’ CA CA =sin. P’C’A’ by Def. =sin. (90°—4) by construct. =cos. 4 by Def. We have hitherto considered an angle PCA less than a right angle, but the same definitions are applied, whatever may be the magnitude of the angle. Therefore, + 32 ANALYTICAL PLANE TRIGONOMETRY. Thus, for example, let us take an angle P,CA situated in the second quadrant, that ts, an angle greater than one right angle, and less than two. From P, let fall P,S, perpendicular on AD, from D draw DT, a tangent to the cir- cle at D, meeting CP, produced i in T, ; then, as before, PBs sin, BCA GA CS, GA 2—cos. P,CA DY: “dl tan. Wg CA ad be CA 2—see. PCA ; AS, CA Again, let the angle in question be situated in the third quad- rant, that is, let it be an angle greater than two, and less than three right angles. Making a construction analogous to that in the two former cases, we shall have 3° sin P,CA Hs cos, Ba. = otan, FAGSs: Sey P,AC sae sin. P,CA. Lastly, let the angle be situated in the fourth quadrant; that is, let it be an angle greater than three, and less than four right angles, then as before, few ANALYTICAL PLANE TRIGONOMETRY. 33 CT GA ee P,CA Se sin. P,CA. We shall now proceed to establish some important general relations, between the trigonometrical quantities which are immediately deducible from the above definitions, and from the principles of Geometry. T Resuming the figure of Def. (8): Since CSP is a right-angled triangle, and - CP the hypothenuse, PS’+CS?=CP* Dividing by CP”, PS? CS 1 CPT CP?” that is, sin.%6+cos.d=1 - - - - - ew + hl) The triangles PSC, TAC, are Peeangulse and similar ; hence, PS_AT CS “CA PS CA AT Therefore cs’ CA CA that is, ona Panta er oe oe a) cos. 6 In last case, for 4 substitute (90°—4) ; then sin. (90°—4) _ o 4 cos. (90°—4) ae Or, S08de | ile = Rovdoe HieW i= Thoth SSS) sin. 8 From (2) and (3) we have sin. é oh F os.6 at Hence, tan. 6= a or, tan. 6cot.d=1- + - - - - (4) By similar triangles ‘CTA, CPS. igs cP CA CS os =CS I 4 34 ANALYTICAL PLANE TRIGONOMETRY. i Or, sec. sieaaee 7 OF; Sec. écos.6=1 - - (5) By Definition, cosec. 6=sec. (90°—#) l f 4 = t case & cos. (90°28) by the last case, es ] . ‘_. « “ = ste or, cosec. sin. é=1 - - (6) Since CAT is a right angled triangle, and CT the hypothe- nuse CA°+AT°=CT? — a Dividing by CA’, sale ba bo ad Wi GAry CAS that Is, 1--tan” d=sec.27) = Poker ate ee By (8) we have a ually j sin. 6 Sig's os.” 4 Therefore, cot.” (= Adding 1 to each side of the equation, wp : lcomee ees ¥. sin. r aa _sin.’ 6+cos.* 4 mae ; r Tae aj by (1) = COseC.: O)Dy (0) ae > =: ou ee) By Definition, abana 5" versin. FeV ae ® _CA—CS = Cider ls CS ™? CA =I—cos.6- - - - - - = (9) By Definition, coversin. 6=versin. (90°—4 = 1—cos.(90°—4,) a the last case. <= ]—dgin) dc 12) eoigiesiod ollie 1h The above results, which are of the highest importance in all trigonometrical investigations, are collected and arranged in the following table, which ought to be committed to memo- ya ANALYTICAL PLANE TRIGONOMETRY. 35 TABLE I. 1.- Sin. 6+-cos.? 6=1 sin. 6 Zt ee tania or cos. 4 fa cos. 4 eas 3. Binh =cot. 4 42 *an:6 Coe ei 5. seciidéosre Sti 6. cosec. 4 sin. 6 =1 # 7. %I+tan.? 4 =sec.? 6 8. 1+cot.’ 4 =cosec.” 6 | 9. v. sin. 4 = }—cos, 6 10. coversin. 6 a] sine 8! 12. The chord of an arc is the ratio of the straight line join- ing the two extremities of the arc to the radius of the circle. PROPOSITION. The chord of any arc is equal to twice the sine of half the are. Take any arc AQ, subtending at the centre of the circle the angle ACQ=4. Draw the straight line CP bisecting the angle ACQ. Join A, Q; from P let fall PS perpen- dicular on CA. Since CP bisects ACQ, the vertical an- gle of the isosceles triangle ACQ, it bi- sects the base AQ at right angles. Therefore, AO=OQ, and the angles at O are right angles. Again, since the triangles AOC, PSC, have the angles CSP, COA, right angles, and the angle PCS common to the two triangles, and also the side CP of the one equal the side CA of the other, these triangles are in every respect equal. . PS =AO=0Q ADF, A emtd tthe: “AQ _,PS "* CA CA or, chord 6 =2 sin. PCA 6 ae) SH 5 We shall now proceed to explain the principle by which the signs of the trigonometrical quantities are regulated — * od 36 ANALYTICAL PLANE TRIGONOMETRY. All lines measured from the point C along B CA, that is, to the left, are considered pos- itive, or have the sine +. All lines measured from the point C along CD, that is, in the opposite direction « = D to the right are considered negative, or have the sign —. All lines measured from the point C along CB, that is, upwards, are considered posi- E tive or have sign +. All lines measured from the point C along CE, that is, in the opposite direction downwards, are considered negative, or have the sign —. Let us determine according to this principle, the signs of the sines and cosines of angles in the different quadrants. B In the first quadrant, sin. I=GA CS Here PS=Cc is reckoned from C along CB upwards, and is therefore positive. CS is reckoned from C along CA, to the left and is therefore positive. In the first quadrant, therefore the sine and cosine are both positive. : sy In the second quadrant, sin. 6= GA CS y= 2 cos.f= Here P,S,=Cc, is reckoned from C along CB upwards, and is therefore posi- tive. CS, is reckoned from C along CD to the right and is therefore negative. In the second quadrant, therefore, the sine is positive and the cosine negative. : P95 In the third quadrant, sin. 6= CA CS cos, 6=_—_3 CA Here P,S,=Cc, is reckoned from C along CE, downwards, and is therefore negative. CS, is reckoned from C along CD, to the right, and is therefore negative. . Me ANANLYTICAL PLANE TRIGONOMETRY. 37 In the third quadrant, therefore, the sine ‘and cosine are both negative. BS 4 In the fourth quadrant, sin. ere cos. jun Oa CA A D Here P,S,=Cc, is reckoned from C along \ CA As the angle goes on increasing, PS goes on diminishing ; and when CP coin- cides with CD, that is, when the angle becomes equal to 180°, PS pea al- together and is equal 0. PS Wende since, gencrally, sin. t= Tai and since, when §=180°, PS=0; -, sin. 180°=0. On the other hand, as the angle increases the cosine increases ; for O87 Gg CSPRECS! CA CA : From C let fall the perpendicular CD on AB. Then, since CDB is a plane triangle right-an- D gled at D, by last proposition , a * “4 < ¥. ANALYTICAL PLANE TRIGONOMETRY. A5 CP—CE dip th +5 ape) nice (Pe) hak” iQ) Again, since CDA is a plane triangle right-angle at A, ; A. SHBG een eo ee ah - - (2) Equating these two values of CD, CB sin 7 api sin. A; C sin. A. Therefore, a In like manner, by dropping perpendiculars from B and A upon the side AC, CB we can prove, CB sin. A, CA_ sin. B, BA sin. G, BA” sin. C. ay In treating of plane triangles, it is convenient to designate % _ the three angles by the capital letters A, B, C, and the sides opposite to these angles by the corresponding small letters a,b,c. According to this notion, the last proposition will be a sinw.A a sin.A 0D sin. B [_—i: +! —_--. we =F b sin. B oc sin.C c sin,c ? CHAPTER II. GENERAL FORMUL. % “~ . ° ; ° ° Given the sines and cosines of two angles, to find the sine of their sum. Let ABC be a plane triangle ; from C let fall C C perpendicular on AB, Let angle CAB=4, and angle CBA=8, Then, AB = BD+DA @ = BC cos, 8 + AC cos. 4, any De. 8B because BDC and ADC are right-angled triangles, ___ Dividing each member of the equation by AB, — BC AC ; ] = AB °°: B + AB cos. a Ae = —G Goos B +5 F cos. 8, by last Prop. in Chap. I. “.sin.C = sin.écos. 8+ sin. 8 cos. 4. _ _ But, sinceABC is a plane triangle, 4+8+ C = 180° - C= 180°—(6+8) sin.C = sin. {180°—(6+ i= sin (6+8,) because 180°—(4+/) is the supplement of (6+.) 5 46 ANALYTICAL PLANE TRIGONOMETRY. Hence, sin. (8+4) = sin. 4 cos 6+ sin. 8 cos. 6- ~ - - (a) That is, the sine of the sum of two arcs or angles is equal the sine of the first multiplied by the cosine of the second, plus the sine of the second mulitplied by the cosine of the first. This expression, from its great importance, is called the fundamental formula of Plane Trigonometry, and nearly the whole science may be derived from it. Given the sines and cosines of two angles, to find the sine of their difference. By formula (a). sin. (0+) = sin. 4 cos. 8+ sin. 8 cos. 4, For 4 substitute 180°—4, the above will become sin. §{180°—(8-—-8)} = sin. (180°—4) cos. 8 + sin. 8 cos.(180°—) But sin.§180°—(é—§)} = sin. (6—8) +. 180°—(é—) is the supplement of (8—8.) And, sin, (180°—#) = sin. 4, And, cos.(180°-—4) = — cos. 4 Substitute, therefore, these values in the above expression, it becomes sin. (6—) = sin. 4 cos. 8—sin. 8 cos. 4 - - - - (b) That is, the sine of the difference of two arcs or angles, is equal the sine of the first X cosine of the second, — the sine of the se- cond X cosine of the first. Given the sines and cosines of two angles, to find the cosine of their sum. By formula (a) sin. (6+ 8)=sin. 4 cos. 8+ sin. 6 cos. 8, For 4 substitute 90°+4, the above will become sin. {90°+(6+8)}= sin. (90°+4) cos. 8 + sin. 8 cos. (90°-+-4) But, sin. $90°+(é+8)}=cos. (46+8) by Table II. And, sin. (90°+4) =cos. 4. And, cos. (90°+3) =—-sin. 4, Substituting, therefore, these values in the above expression, it becomes, cos. (+8) =cos. 4 cos. 8—sin. @ sin.8 - - (ce) That is, the cosine of the sum of two arcs or angles, is equal to ihe cosine of the first multiplied by the cosine of the second, mi- nus the sine of the first multiplied by the sine of the second. . ANALYTICAL PLANE TRIGONOMETRY. 47 Given the sines and cosines of two angles, to find the cosine of their difference. By formula (@:) sin. (6+) =sin. 6 cos. 8+ sin. 8 cos. 4, For 4 substitute 90°—4, the above will become sin. ${90°—(é—8)}= — sin. (90°—4) cos. 8 + sin. 8 cos. (90°—4), But, sin. {90°—(é—£)}=cos, (6—8), By Table II. sin. (90°—4) cose lth oF) 14.2 cos. (90°——4) =a) - BS AO Ala: Substituting, therefore, these values in the above expression, it becomes cos. (6—f) =cos. 4 cos. 6+sin.4sin.8 - - (d) That is, the cosine of the difference of two arcs or angles, is equal to the cosine of the first multiplied by the cosine of the second, plus sine of the first into-the sine of the second. Given the tangents of two angles, to find the tangent of their sum. By Table L.: sin. (+8) tan. (6+ 8) Smeal ROWS) G+8) sin. 6 cos. 8+sin. 8 cos. 4 ~ cos. 6 cos. B—sin. 6 sit. 8 by (@) and (c) Dividing both numerator and denominator of fraction by cos. 6 cos 8: sin. 6 cos. 8 . sin. 8 cos. 4 cos. cos. 8 cos. cos. 4 rr sin. 4 sin. 8 ~ eos. 4 cos. 8 Simplifying, tan. 6+tan. 8 Tt i—tatedisand syode oll rl) se () That is, the tangent of the sum of two ares or angles, is equal to the sum of the tangents of the two arcs, divided by 1 minus the product of the two tangents. Given the tangents of two angles, to find the tangent of their difference. By Table I.: _ sin. (@—8) pga See (6 —8) sin. 6 cos. 8—sin. 8 cos. ~ cos. 4 cos. 8+sin. 4 sin. P. by (b) and (d) 48 ANALYTICAL PLANE TRIGONOMETRY. Dividing both numerator and denominator by cos. 4 cos. 8 : sin. cos. 8 — sin. 8 cos. 4 cos.écos.8 cos. 6 cos. B ste sin. @ sin. 8 ; cos. 4 cos. 8 Simplifying, tan. 6— tan. 8 b+ tan. étane te Ok Lec Oe (7) Hence, the tangent of the difference of two arcs or angles, is equal to the difference of the tangents of the two arcs, divided by 1 plus the product of the two tangents. The student will have no difficulty in deducing the following : cot. 6 cot. P—1 cot. (6+5)= ——_——_— G+8) cot. P+cot. 6 cot. 6 cot. 8+1 (oe 0) ee cot. ()—8)= cot. 8—cot. 6 sec. 4 sec. 8 cosec. 4 cosec. SEC. riley sal adi | cosec. 4 cosec. S—sec. 4 sec. 8 sec. sec. 4 sec. 8 cosec. 4 cosec. B sec. (6—8)= cosec. 4 cosec. 6+sec. 4 sec. 8 sec. § sec. 6 cosec. 4 cosec. 8 sec. 6 cosec. 8+sec. 8 cosec.4 sec. 6 sec. 8 cosec. 6 cosec. B sec. 6 cosec. B—sec. 8 cosec. 6 cosec. (6+8)= cosec. (6—P)= To determine the sine of twice a given angle. By formula (qa) : sin. (6+8)=sin. § cos. 8+sin. 8 cos. 6. Let 6=8, then the above becomes sin. 2 6=sin. 4 cos. é+sin. 4 cos. 4 =2 sin. 6 cus.6- --- -) -. =, (gl) That is, the sine of twice a given angle, is equal to twice the sine of the given angle multiplied by its cosine. 6 In the last formula, for 4 substitute >; then, ‘ 6 sus 6 6 sin. 2X o= Sin. “5 COS. > 6 ) Or, sin. d= 2 sin. —coss—= - - - - = + (g2) 2 2 ANANLYTICAL PLANE TRIGONOMETRY. 49 To determine the cosine of twice a given angle. By formula (c) : cos. (4+8)=cos. 4 cos. 8 —sin. 4 sin. 8 Let 6=§8 then the above becomes cos. 26=cos.*6—sin2 6 - - 2 + - - (hl) By table I. sin.? 6 =1—cos.’ 6; substituting this for sin.* 4: COs. = 2 cos.’ §—— 1 oy OU bh 2 (h2) Again, since cos.” 6=i— sin.’ 4, substitute this for CUS. 2 cos. 2 =I. — 2 sin.’ 4 - ie = = - - = (h8) Hence, the cosine of twice a given arc or angle, is equal to 1 minus twice the square of the sine of the given angle. To determine the tangent of twice a given angle. By formula (e) : tan. 6+tan. 8 ets C18) = —tan.é tan. 8 Let 6=8, the ate becomes 2 tan. 4 2 tan. 2 ace My PSY ekywe bond ke Gtcuetieettiin. Wao (2) The tangent of twice a given arc, is equal to twice the tangent of the given arc, divided by 1 minus the square of the tangent. The student will easily deduce the following : cot.*§—1 cot. 6—tan. 4 cot.2 c= 2cot.4 | 2 sec.” 6 cosec.” 4 5 sec. 2 Oe cogent 6— sec’ 4 sec.” 6 cosec.? 6 sec. 4 cosec.. 4 cosec. 2 = Shrrentne teh 2 sec. 6 cosec. 6 ~ 2 To determine the sine of half a given angle. By formula (23) : r cos. 24=1—2sin26 °” 6 For ¢ substitute >: the above becomes, : 1—2 et cos. 25°= — 2 sin. 5 i 4 Or, cos. 6=1—2 sin.’ > a eS ee 6 ‘2 4 + 50 ANALYTICAL PLANE TRIGONOMETRY. That is, the sine of half a given angle or arc is equal to the square root of 1 minus the cosine of the arc divided by two. Lo determine the cosine of half a given angle. By formal (h2): Cosi Ribil2ae? D cogti=—] . 6 For 64 substitute zi the above becomes, cos, 2 a = 2 cos. ‘a —l 2 2 Or, cos/e) == sz ees, eat os cos." = 1+ cos. 4 4 LA set C08 g, win Mata ee Fo determine the tangent of half a given angle. Divide formula (4) by (A): 6 12g flaca Botta Ice acy il 1+cos. 6. . ball oe 8 1—cos é Or tans =) te oe he ee 7s ; 2 1-++cos. 6 ( ) Multiply both numerator and denominator by Vi—cos. 6; the above becomes, 1—ceos. 6 6 ta at it, = : nt a sae = o ee - - ® i 2 sin. 6 (72) Multiply both numerator and denominator of (71) by V1-+cos. 6; we have, sin. 6 tan. mY = 1chca a ha ee, Cee ( 3) The student will easily deduce the following : 6 Nal 1+cos. 4 cot. oy = mee 1—cos. & Le 1+ cos. 4 a sin. 6 sin. 6 ANALYTICAL PLANE TRIGONOMETRY. wt sec. 4 LR Bice sec. 6 2 sec. 6+1 PER ld ps a / 28002 8 6b 2: sec. 6—] To determine the sine of (n+1) 4, in terms of n 6, (n—1) 4 and 4, By formula (a) and (4): sin. (8+4)= sin. 8 cos. 46+ sin. 6 cos. B sin. (S—4)= sin. § cos. 6—sin. 4 cos. 8 Add these two equations, sin. (8+-4)+sin. (8 —é)=2 sin. 8 cos. 4 Subtract sin. (8 — 4) from each member, sin. (8+6) =2 sin. 8 cos, 6~— sin. (8 — 6) Let 8=n 4, the above becomes sin. (2-+1)§=2 sin. 2 4 cos, é— sin. (n— 1) 4... (m) In the above formula, let n =1; .. n+1=2,2—1=0 *, sin. 2 6=2 sin. 4 cos. 6 —sin. 0 =2 sin. 6 cos. 4, the same result as in (g). Lét n=2 ; .0.n+1=3,n—1=1; sin. 3 6 =2 sin. 2 4 cos. 6— sin. 6 =2X2 sin. 6 cos. 6X cos. 6 — sin. 4 =4 sin. 6 cos.26—sin. 6 « =4 sin. é (1 —sin.’ 6)—sin. 4 =3 sin. é—-4 sin.° 6 - - (x) Let 2==3;..°. xX 1=—4, n——1=2: *. By formula (m): sin. 4 6 =2Xsin. 3 6Xcos. 6— sin. 2 4 =2 (3 sin. 6 —4 sin.* 4) cos. 6—2sin. 6 cos. 4 =(8 cos.* 6 —4 cos. 4) sin. 4 It is manifest that, by continuing the same process, we may find in succession, sin. 5 4, sin. 6 4, > . si -h&c, To determine the cosine of (n+1) 4, in terms of n 6, (n—1) 4, and 6, By formula (c) and (d) : cos, (8 +4)=cos. 8 cos. 6—sin. 8 sin. 4 cos. (8 —4)=cos. Scos. é6+sin. f sin. 6 Add these two equations, cos. (8+4)+cos. (6 — 6)=2 cos. 8 cos. 4 Subtract cos. (8 — 4) from cach member, yn) ver? pe guns 52 ANALYTICAL PLANE TRIGONOMETRY. cos. (8+4)=2 cos. 8 cos. 6 — cos. (8 — 4) Let S=n 4, the above becomes cos. (n+1) 6=2 cos. n 4 cos. 6— cos. (n—1)4- - (0) In the above formula, let n=1;. .. n+1=2, n—1=0; Then, cos. 2 6=2 cos. 6 cos. 6—cos 0 =2 cos.” 6— 1, the same result as in (A2). Let 7n=2, .°. n+1=3, n—-I=I1; “. cos. 3 6=2 cos. 2 4 cos. 6— cos. 4 =2 (2 cos.” 6 — 1) cos. 6— cos. 4 =4 cos..6—3cos.6 - - - - (p) Let n=3; .°. n+1=—4, n—1=2; cos. 4 6=2 cos. 3 4 cos. 6— cos. 2 4 =2 (4 cos.* 6 —8 cos. 4) cos. 6 — (2 cos.” 6 —1) =8 cos.‘ 6— 8 cos.? +1 It is manifest that, by continuing the same process, we may find, in succession, cos. 5 4, cos. 6 8, cities LANES, By adding and subtracting (a) and (b), and by adding and subtracting (c) and (d), we obtain the following formulz, which are of considerable utility. sin. (6+)+sin. (6—f8)= 2 sin. 4 cos. B) sin. (6+8)—sin. (6—f§)= 2 sin. 8 cos. é cos. (6+8)+cos. (6—8)= 2cos.dcos.B( ~~ ~~ (7) cos. (6+8)—cos. (6—@)= —2 sin. 4 sin. 6 Any angle 4 may, by a simple artifice, be put under the form, 6+8 6—f (cae ee eS SRS 2 + 2 And, in like manner, asta h saps 2 2 r ee .. sin, 6=sin. a tai = 6 yt EE os sae ti sin. B cos. = - - -(1) Min Oosaid ae a tO a Be | — s&s 6 =sin. TF ros. =F —sin, ae _ - - - (2) cos. 6=cos. ‘pee aoe “648 —B . 648 . §—8 4 = COS. Tie ada — Sin. my sin. ANALYTICAL PLANE TRIGONOMETRY. 53 cos. 8=cos. alge a ” | : =COB. Raley Rats + sin. es sin, - +. (4) Add together (1) and (2): sin. 6+sin. B=2 sin. TF cos. es - = = = © = (r) Subtract (2) from (1), sin. é— sin. S=2 sin. BE genet Sn oe en fa) Add together (3) and (4), cos. +cos. 8=2 cos. ae a ae $y. dntie- 8, spaade oft) Subtract (4) from (3), cos. 6—cos. B= —2 sin. oe sin. aa - : -/(2) These formule, which are of the greatest importance, might have been immediately deduced from the group (q), by chang- ; ‘ ahh ing 6+8 into 4, 6—8 into 8, 4 into =, into 2 Divide (r) by (s) : 2 sin. cae Cos. smile sin. 4+sin. 8 th 2 2 sin. @—sin.B 4. °—8 6+8 2 sin. 5) COS. 3 tan. ke : i (+48) an See ee Ne et Se (2D) tan. Pa tan. 4 @—f) Multiply (a) by (2); then, _sin. (6+) sin. (8 —8)=sin.’ é cos.” 8 — sin.’ B cos.” 6 =zhit, MAeetsih,® ite are tier Ske Multiply (c) by (d); then, cos. (6+) cos. (3 —8)=cos.? 4 cos.? 8 —sin.? 6 sin.’ 8 =cos”2é@—sin? 6B - - - + (y) bee wae e aA ANALYTICAL PLANE’ TRIGGNOMETRY. We will now investigate a few properties where more than two arcs or angles are concerned, and which may be of use in. the subsequent part of this work. Let 4; 8, y,.be any three arcs or angles. Then, in. 6 sin. y-tsin. 6 sin. (6+8+y) RTE ia sin 7s B sin. (6+8+7) For by formula (a) sin. (6+-8+y)=sin. 4 cos. (8-+y)+cos. 4 sin. (0+y) which [putting cos. 8 cos. y—sin. f sin. y, fur cos. (8+y)], is=sin. 4 cos. 8 cos. y-—sin. 4 sin, 8 sin. y+-cos. 4 sin. (8+y) ; and, mul- tiplying by sin. 8, and adding sin. 6 sin y, there results sin. 4 sin. y+sin. 8 sin. (6+8+y)=sin. 6 cos. 8 cos. y sin. 8+ sin. 6 sin. y cos.? 8+cos. 8 sin. 8 sin. (8+y7)=sin. 4 cos. 6 (sin. 8 cos. y-+cos. 8 sin. y)+cos. 4 sin. 8 sin. (8+y)=(sin. 6 cos. 8 +cos. é sin. 8) sin. (8+y)=sin. (6+) sin. (8+7). Hence, dividing by sin. (6+), we have, ’ in. 6 sin. in. in.(4 es eu aes sin 7+sin 6 sin.(6+8+y7) In a similar manner it may be shown, that __sin. 6 sin. y——sin. 8 sin. (6—8+7) 7 sin. (6 —B) If 6, 2, y, 6, represents any four arcs or angles, then writing y+6 for y in the preceding investigation, there will result ae (6+. Hee é sin. (y+0)+sin. 8 sin. 6+8+y+0) ; Yi sin. (6+ 8) A like process for five arcs or angles will give in. 4 sin. (y-+é in.Bsin.(8 a sin (8 eyo eye tee Sin are) Sin Pein Er Bt eke +2) sin. (6+/) And for any number 4, 8, y, &e. - - - % sn (Gee te..p) Sind sin. (y+9+...A) tsin.f sin.(@+8 ty +...) aD i. Taking again the three 4, 8, y, we have sin. (8 —y)=sin. 8 cos. y— sin. y cos. 6 sin. (y— 46) =sin. y cos. 6—sin. 4 cos. y sin. (6 — 8)=sin. 4 cos, 8 —sin. 8 cos. 4 Multiplying the first of these equations by sin. 4, second by sin. 8, third by sin. y; then adding together the equations thus transformed ; there will result, sin. 6 sin(S—y)-+sin. § sin. (vy — 4)-+sin. y sin. (6—8)=0 sin. 6 sin. (8— y)-++cos. 6 sin. (y—4)+cos. y sin. (6--8)=0 These two equations resulting from any three angles what- ever may evidently be applied. to the three angles of any triangles. | ANALYTICAL PLANE TRIGONOMETRY. 535 Let the series‘of arcs or angles 4,8,y,56 - - - - - 2, be contemplated, then we have formula (x) sin. (6+) sin. (6—8)=sin.’ 6 — sin.’ 8 sin. (8+y) sin. (8 —y)= sin.’ 8 — sin.’ Y- sin. (y+8) sin. (y — 4)=sin.” y — sin.’ 6 Odds es =< ce. . sin. (A-+4) sin. ‘(4 — 4) sin. *}—sin.? 4 Adding these equations together, we have sin. (2+) sin. (8 —8)-+sin. (P+y) sin. (B—y)+sin. (y+) Sin. (Y—O) +. -e ewe ees sin. (A-+4) sin. (A— 6)=0 Proceeding in a similar manner with the sin. (6—§8), cos. ()+f), sin. (8—y), cos. (8+y7), &c., there will at length be obtained cos.(8+8) sin.( — 8)-+cos. (3-+7) sin. (S— ae Fe cos. (A+) sin. (A—4)=0 If the arcs 4, 8, y- - - - % form an arithmetical progression of which the first term is 0 the ratio e and the last term A, any number n of circumferences, then will S—4=2, y —8=s, &c., 6+f8=2, 8-+y=32, &c.; dividing the whole by the sin. ¢, the preceding equations will become sin. g-+sin. 3e+sin. 5¢+é&c.= cos. ¢-+cos. 3g-+cos. 5e-+&c. ai If were equal 22, these equations would become sin. e+sin. (e-+%)+sin. (¢+2%)+sin. (¢+32)+d&c.=0 cos. e-+cos. (e+£)+cos. (e+ 2%)+cos. (g+3%)+&c.=0 The last equations, however, only show the sums of the sines and cosines of arcs or angles in arithmetical progression when the common difference is to the first term in the ratio of 2tol. To find a general expression for an infinite series of this kind, let S+sin. 6+sin. (6+) -+sin. (64+28)+sin. (6+38)+ - - - - &c. Then since this series is a recurring series whose scale of re- lations is 2 cos. S—1, it will arise from the oeve Duman of a fraction whose denominator 1—2y cos. 8+ x? making y=1 Now this fraction will be, _sin, 4+ /sin.(6+8) — 2 sin. 4 cos, & 1 — 2xcos. 8+ x? Therefore, when x=1, we have, g_ sin. 6+sin. (6+8)—#% sin. 4 cos. 8 2—2 cos. 8 : And this because, 2 sin. 4 cos, 8=sin. (4+8) +sin. (§—f) __sin, é— sin. (8 —8) 2 (1—cos. 8) Now putting 4 for (8+) and 8’ for (6--8) we have from formula (s) : by formula (q). 56 ANALYTICAL PLANE TRIGONOMETRY. sin. 4’— sin. 8’=2 cos.1 (4’+8’) sin.2 (6/— 8’) Hence, it follows that, sin. 6—- sin. (6-~ 8) =2 cos. (6 — 18) sin.38 Besides which we have, 1 — cos. S=2 sin.718 Consequently the preceding expression becomes, S=sin. 6+sin. 6+8)+sin. (6+28)+sin. (6+38) &c. ad infinitum, cos. (9 — 18) QsiniB - - - - - = = (22) To find the sum of n+1 terms of this series, we have simply to consider that the sum of the terms past the (n+1)th, that is the sum of | sin. (6+(n+1)8) +sin. (6+(n+2)8) +sin.(8+(n+8)8)+ &c. ad infinitum, is by the preceeding theorem, 2008 Chit He) ba 2 sin.t 8 Deducting this from the former expression, there will remain sin. 6+sin. (6+8)-+sin. (+28)+sin. (0+38)+ - - - - - - - - sin. (6+n8)=cos. (6—38)—cos. (6+(n+1)8) 2 sin. ifs __ sin. (+3 28) sin.t (n+1)8 iP sin. 18 (23) By like means it may be found, that the sum of the cosines of arcs or angles in arithmetical progression, is cos. 6+cos. (+8)+cos. (0+28)+cos. (6+38)+ &c. ad infinitum, _ _ sin.(8 +438) 2 sin. 38 - - = (24) Also, cos. 6+cos. (+8)-+cos. (0+28)+cos. (4+388)+-- . - - - - (cos. +n) cos. (6+18(sin.2 (n+1)8 cok panic, 12 ER SC, To find the numerical value of the sine, cosine, &c., of 45°. In the circle ABD, draw CA, CB, radii at B right angles ; join AB. _ Then by Definition (12) 1 AB A D Chord ABC (90 ey sis | AEB? Chord’ 90 = AGH E AC?-+- BC? _ weAC ‘” “ANALYTICAL PLANE TRIGONOMETRY, 57 PEND ooo, ‘ ca AC? > CSAC a PE an eect ve @) Now, the chord of an arc is equal te twice the sine of half the arc; therefore, 2 sin. 45°=chord 90° 4 sin.” 45°=chord? 90° =2, by Equation (1) ; “ve sin. a Vs ” J 2 Again, by. table I. : sin.? 6+cos.? 6=1 cos.? 45°—=1 — sin.” 45° 1 ay Gata) 1 —- —— 2 cos. 45°= oye sin. 45°. Also, in. 45° tan. 459 ee mn cos. 45° =]=cot. 457. To find the numerical value of the sine, cosine, &:c., of 30°. In the circle ABD, draw CP, making with Pies CA the angle ACP=60° ; join A, | Now, 2 sin. 30°=chord 60° 2 " py3s AC z soe .- AP=AC, + the triangle APC is equiangular, and therefore equilateral. oe sin. spose. z Again, cos. 80°= /] — sin,? 30° =V] iene bs V3 2 2 58 ANALYTICAL PLANE TRIGONOMETRY. Also, sin. 30° tan. 30°= ew cos. 30° — 1 e oo. | ] O07 = ae ee ‘ tan. 30° wae To find the numerical value of the sine, cosine, &c., of 60°. sin. 60°=cos. (90°— 60°) —conrad., = ae by last art. 2 Again, cos. 60°==sin. (90°— 60°) =sin. 30° Also, tan. 60°= V3 } cot. 60°= V3 Jt is required to find the sum of all the natural sines to every minute in the quadrant, radius =1. In this problem, the actual addition of all the terms would be a very tiresome labor, but the solution by means of formula (z8), is rendered very easy. sin. 1( w+ 1)S=sin. 45°, 0’, 30” and sin. To sifi. 30", sin. 45° sin. 45° 0’ 30” ————————————— sin. 39” Applying that formula we have sin. (6+34n€)=sin. 45°, =3438.2467465, the same sum required. Let it be required to find the sum of the sines to every mi- nute of the are of 60°. Here the numerical expression in the equation would become 30° 30° 0/30" Sip. oie sine FOR" = 6.5001 Bee Rdvo1Ne4 5958 sin. 30” = 1719.123373.25 equal the sum of all the natural sines to every minute of the arc of 60°. It may be useful to exhibit the most useful results in this chapter, in the following table. ae ANALYTICAL PLANE TRIGONOMETRY, 59 TABLE III. (1.) sin. @+8) =sin. 6 cos. Psin. B cos. 4 (2.) cos. (8) =cos. dcos. P#sin. 4 sin. 8 ~/-etan..6—tan. (3.) tan. (648) ones 1=¢tan. é tan. 8 (4.) sin. 2 4 =o win © Gos a (5.) cos. 2 =cos.7d—sin.28=2 cos.? 6—1=1I—2 sin.” 9 2 tan. 4 6.) tan. 2 4 —=———_ — ©) 1— tan.’ 4 : 6b L—— cos: 4 de = Nae : (7.) sin 5 5 (8.) cos. a — a / Leos: 8 2 2 (9.) tan if ER A Lae. 6 1—cos.6 sind 2 1+cos.6 sin. 1+cos.d 6 6 10.) sin. 4 = i a (10.) si 2 sin g C8 5 (11.) sin. 3 4 =3 sin. 6—4 sin.* 4 (12.) cos. 3 4 =4 cos.* —3 cos. 4 (13.) sin. (n+1) 6 =2 sin. né cos. 6 —sin. (n—1) 9 (14.) cos. (n+1) 4 =2 cos. nd cos. 6—cos. (n—1) 9 : ; eee § — (15.) sin. 6+sin. 8 =2 sin. hs COR os , . b— 8-++f (16.) sin. é—sin. 8=2 sin. COs. tak y (ale S| (17.) cos.é+cos. S=2 cos. cos. —5 Piety . b— (18) cos. é6—cos. 8= — 2 sin. sin = (19.) sin. é+sin. 8 “" 2 sith 6-—sini 8 §—p tan. 2 (20.) sin. (6+)+sin. (6 —§)=2 sin. 4 cos. B (21.) sin. (6+8)—sin. (—§)=2 sin. 8 cos. 4 (22.) cos. (6+)-+cos. (46 — 8)=2 cos. 4 cos. B tug ."Y ee a 4 60 ANALYTICAL PLANE TRIGONOMETRY. (23.) cos. (6+()—cos. (4 —)= — 2 sin. 4 sin. 8 (24.) sin. (6+ 8) cos. (6—)=sin.’ é—sin.? B=cos.? B— cos.? 6 (25.) cos. (8+) cos. (6—8)=cos.? 6—sin.?8 =cos.?4+cos.28—¥ (26.) sin. 6+-sin. (0+8)+sin. (6+28)+sin. (6+388)+ ---- - Ween (be) = sin. (6-+27f8) sin. 1 (w+1)8° sin. 1 8 (27.) cos. 6+cos. (6+8)-+cos. (6+28)+cos. (64+88)+ - - - - - - cos. (+78) _cos. (6+38) sin. 3 (n+1) B a sin. 2 8 : | 1 (28.) sin. 45° =COS. cab eg (29.) tan. 45° =cot. 45°=1 : 1 (30.) sin. 30° ==COS. 60° =—- 3 (31.) cos. 30° =COSs. 60°= 1 o ee 5 eee See (32.) van. 30 =cot. 60 73 (33.) cot. 80° =tan. 60°= V3 The formule of Trigonometry may be multiplied to almost any extent, and tl same quantity may be expressed in a vast number of different ways. An intimate acquaintance with those given in the above table is essential to the progress of the student. The following, although of less frequent occurrence, may occasionally be found useful, and ean be readily deduced from the above. (34 sin. (45°-E6) cos. =Esin. 6 Dy cbs. ee a) a eae eS (35.) tan. (45°10) _< itetans? 1=tan. é 8 1+sin. 6 36.) tan.? jt = a. 8 Cages). ~ sages 6 1+sin. 6 oo 37.) tan. ode = ‘id ; (37.) tan. (4504. 5) See “ee (38.) sin. (6+) tan, é-+ tan. 8 cot. S=ecot. 6 sin. (6—() ~ tan.é—tan. 8 cot. B—cot. 6~ _— or "a neil : a i at a @. ro yt . ANALYTICAL PLANE TRIGONOMETRY. 61 ' ae (0+-8) cot. B—tan. 4 cot. é—tan. 8 ” Cos. (6—) = cot. B-+tan. § cot. sot. d-+tan. B (40.) Sin. @+sin. 8 tan, SFP cos. 4+cos. 8 2 (4y,) Sine ttsin. BL oot SP cot. d—cos. . 2 (49, sin tosin. Bs tan, SB cos. 6-+cos. 8 2 (43. ) a ae = —cot. +8 cos. 6— cos. 8 2 (44.) se met 2 =- + cot. secre 2 cos. §—cos. 8 2 2 (45.) tan. 6+ tan. 8 sin. (8 + °) cos. 4 cos. 8 (46.) cot. 6+ cot. B = sina @ +8) sin. 4 sin. 8 (47.) tan. — tan. out: 8) , cos. 6 cos. 8 (48.) cot. 6—cot. 6 se! ype ioe od, sin. 6 sin. 8 (49.) tan” 6—tan* @ = sin. (448) sin. (6@—8) cos.? 4 cos.’ 8 (50.) cot.” 6—cot.’ 8 Paya eG +8) sin. (° ash fe sin.” 6 sin? 8 In order to become familiar with the various combinations, and dexterous in the application of these expressions, the stu- dent will do well to exercise himself by verifying the follow- ing values of Sin. 4, Cos 6, Tan. 4, which are extracted from the large work of Cagnoli. 6* 62 — tt v2 1. cos. 6 tan. 4 | 16. 9 cos. 4 ’ cot. 4 17 3. /1—cos.’ 6 18, a 4. /1+cot.? 4 19. ; . 6 a 20 V1-+tan.? é : Pea oe . 2sin. 908-5 91. "7. Vv 1— cos. 24 ao 2 ue t 6 pa ST. 1+tan.’ ; an. 5 94 2 9. 6 F cot. 5 +tan. 5) sek 10, 22: (30° +4)—(sin. 30°——#) : WE in.*(45° + : )\—1 11.22:sin: 5 si A 12. 1—2 sin.’ (45°— 5) : 7 1—tan.’ (45°— >) 27. 13. 14. 15. ‘3 ANALYTICAL PLANE TRIGONOMETRY. TABLE OF THE MOST USEFUL ANALYTICAL VALUES OF SIN. 6, COS. 6, TAN. 6. VALUES OF SIN. 4. | 6 1+tan.’ (45°— 5) 28. Y oO a ase i ACh mae tan.(45 +5) tan.(45 S 8 é tan.(45° +5) + tan.(45°—5) 29. sin. (60°+0)—sin. (60°—#) | 30. VALUES OF cos. 4, sin. 4 tan. 4 . sin. 4 cot. 6 v¥1—sin.? 4 cate: ae Vi1-+tan.’ 6 cot. 6 Vi+cot.d 8 b cos.’ ~—=sin.*— 2 2 1—2s) ae ory 6 . 20s." wen 1—tan.? 1+tan. 5 oe: cot. Y —~fan. 6 cot. 5) + tan. 9 1 eee et 1 + tan. 6 tan. — 2 2 ——_—____, tan.(45°-+5)+cot. (45°+5) 6 ) 2 cos.(45° +5)cos.(45°— 5) Sa cos. (60°+4) +cos. (60°—8) i tie . - “ We . %, » ~~ ANALYTICAL PLANE TRIGONOMETRY. 63 4A, VALUES OF TAN. 4, sin. 4 31. cos. 4 2 cot. oY . ote j 1 cot.2- —1 ._-—-- a 32 cot. 4 2 ry Dia on Rt et cot. — — tan. = 33. cos.” he 2 2 39. cot. 6—2 cot. 2 sin. 4 1—cos. 24 84. [aa ee ee | aE. Waid ais Vv 1—sin.? 6 of: sin. 24 sin. 24 35. ee a: 1+cos. 2 4 pee St cos. 26 ah b 42. l-+cos. 246 an. 3 ‘ 4 Ms s Rotate At 43. tan.(45 +5)— tan. (45 ry, |— tan.’ 2 eA Le 4h ee From certain properties of the circles to be discussed in an- other volume, other important trigonometrical formule, may be deduced, furnishing us with more expeditious means of deter- mining, numerically, the values of some of the trigonometri- cal lines, and ratios, all of which will occur in their order. To develop sin. x and cos. x ina series ascending by the powers of x. The series for sin. z must vanish when z=0, and therefore no term in the series can be independent of z, nor can the even powers of x occur in the series; for if we suppose sin. t= a,t+a,0°+a,0°+a,¢'+a,0°+ 2... } = 2 8 4 5 then sin. (— x) =—a,x+a,2°— a,2°+a,0'—a,0+ ... but sin. (—2)=— sin. « as 2 8 4 5 oe as, aot Cas Og mm ig seus 5 hence a,—0, a,=0.... “SIM. t= @,0-+a,0'+a,2°+a,0°+ - - - «= - (1) * 64 ANALYTICAL PLANE TRIGONOMETRY. Again, the series for cos. x must = 1 when 2 = 0, and there- fore the series must contain a term independent of 2, and it must be 1; also the series can contain no odd powers of a, for if we suppose 608. @— lato. tba ta,z-t 2. es then cos. (—r)=1— a,%#+a,2°—a,a*+a,v'— .... but cos. (—2)=cos. x == 1a aba t+O,t +-¢,0 +" Sr. ae @,=—— 4,, dj=—d5,... .*. dy=0, 2520 3. . Cos t—=)-Paxtt+a, a +aje+ - - -- - = (Q) Hence cos. ¢+sin. =1+a,2+a,0°+a,°+a,0'+a,2° - - (3) cos. & — sin. t=1— a,t&+a,2°—a,2°+a,c'—a,a°+ - - (4) Now in equation (3) write « +h for x, and we have cos.(e+h)+sin(x+h)=1+a, (e«+h) +a, (e+th)’+a,(«+h)'+ (5) but cos. (e«+h)+sin. (w+h)=cos. x cos. h—sin. x sin. h +sin. x cos. h+cos. x sin. h =cos. h (cos. +sin. x)+sin. h (cos. e—sin. 2) rel Lt-c, to hit. }).(1 4-0 t+, 2° +40 ey ) +. (a,h+a,h’+a,h'+.. ee ve estat Diva) Fa ede +a ek 4+, | +a,h—a,*th+a,a,vh+ ... = TR Maat ECP ee - - - (6) | +a,h’ + a ett -aisal Comparing the tee (5) and (6) we have " Ita,e+a,2" +a,2° Te l+a,t+a,27 +a,7 + +a, h+2a,ch+38a,0°h+ +a,h—a,a,th+a,a,x°h— + a,’ +3a, th’-+- -= +a, h? +a,a,ch'+ = a,h’ an ‘e + a,h* car | +J a J and equating the coefficients of the terms involving the same powers of z and h, we have ———q,a.; therefore a, =——— = — 2a, 1443 2 2 1.2 a,a, a,* — a a ° s ° e a = a EN ha 3a 12 3 3 1.2.3 Bag cee Sogo Gan Akg eed + - ANALYTICAL PLANE TRIGONOMETRY. 65 Fie : ay” e ag hence sin, e=a,2 er 23° t T2345? 71.2.3.4.5.6.7% * r a,* 2 a 4 ay cos. t= geo 2 i 9347—19.3.45.6° 7 To effeet and we have only to determine the value of a,. this, we have in. =a, z $y, ae Bucs aif ea ede he Leg, ae Ae e e ° =a,2(1— mec 3g rp Now the value of z may be assumed so small that the se- ries in the parenthesis, and sin. 2, shall differ from 1 and z re- spectively, by less than any assignable quantities ; hence ulti- mately x=a,z, and therefore a,=1; whence : i 7a bok SIMs ho og. TaaAb TBAB GT me te 2. a" x Ey os pon th gt hao ea oe Tae To develop tan. x and cot. x in a series ascending by the powers of X. The development may be obtained from those of sin. x and cos. z, already found. ps 4b & nm 2 t— 159 10.5 99 23 tan. r= = 2 x oes &e cos. x : Las DATUypes ee ae and the series will therefore be of the form PR ee oe a , + ~ i pod he 1.2.3.4.5 Hence, let z+a,27°+a,2°+..= xr Prebrassnn Mir ss Sai A oo ‘ ° 183 TT 284.5 = (1—- -z3t Ory an »-) (ear +a,2'+. ) 66 ANALYTICAL PLANE TRIGONOMETRY. Sago ets ® ae 4 La 1 po es Be + 12.34 Hence, equating the coefficients of the like terms, we have go Pain 1 4s pcx sel Sy OE pos ts! 59.9 Oe 1 1 2° a, = Te ke EE oa a ee = lips “ T1979 345 loads “ts = pega e . 22° OT a tan a+ 1.2.3 -|- 1.2.3.4.5 * eS Mie, a eT Welt e a x oy* * apenas, 2 oO Se eee 1O5e a Sas CHAPTER III. FORMULZ FOR THE SOLUTION OF TRIANGLES. We shall here repeat the enunciations of the two proposi- tions established in Chapter I. PROPOSITION I. In any right-angled plane triangle, Ie. The ratio hich the side opposite to one of the acute angles has to the hypothenuse, is the sine of that angle. 2°. The ratio which the side adjacent to one of the acute angles has to the hypothenuse, ts the cosine of that angle. 3°. The ratio which the side opposite to one of the acute angles has to the side adjacent to that angle, is the tangent of that angle. Thus, in any right-angled triangle ABC, « a . Fae 2 ANALYTICAL PLANE TRIGONOMETRY, 67 Cc ee = sin. A, AG = 098 f.® —_ tan. A => cos. C, =) Siow, = 60t..C Or, CB= ACsin. A) ) = C cos. | | BA = AC cos.A) ! B A SAGO kin’ ok ws isla mak a OR a? CB= BA tan. A = BA cot. C PROPOSITION II. In any plane triangle, the sides are to each other as the sines of the angles opposite to them. We shall, frequently in treating of triangles, make use of the following notation ; denoting the angles of the triangle by the large letters at the angular points, and the sides of the triangle opposite to these angles, by the corrcs- ponding small letters. | Thus, in the triangle ABC, we shall denote the angles, BAC, CBA, BCA, G by the letters, A, B,C, respectively, , and the sides BC, AC, AB, by the letters a, b, c, respectively. 7 According to this, we shall have, by the proposition, Loa sin. A) bas.” sina B a sin. A ve gine Ra aa whitgs TG b x sin. B o- 1 sips CO ae : 68 ANALYTICAL PLANE TRIGONOMETRY. PROPOSITION IH. In any plane triangle, the sum of any two sides, is to their difference, as the tangent of half the sum of the angles opposite to them, is to the tangent of half their difference. Let ABC be any plane triangle, then, by Proposition {I £ Mal sin. A 6b” sin. B a+b sin. A+sin. B A B ———— (ae A—sin..B But, by Trigonometry, Chap. Il. (r) in Ath Era A—B gin. A+sin. b= 2 sin. 5) COS. 5) : oy A——ts sin. A—sin. B = 2 cos. sin. —>— 2 sj A+B cos. A = “ae Proceeding in the same man- ner for the other angles, we shall find, sebfoab nates: oo OF gs ie ac e+h—ec cos. O =. “basin ame PROPOSITION V. To express the sine of an angle of a plane triangle in terms of the sides of the triangle. Let A be the proposed angle ; then by last prop., Y+ce—@a@ cos... A = spent Adding unity to each member of the equation, A= ieee ee 1 + cos. A = ari ee b?+2bce+c’—a’? x Bee Orb cy aE aa 2Q2bhe (b+c+a) (b+c—-a) ; Pe age pce Pot. aaa rt ee ; b? +c? — az Again, cos. A = Te pes Substracting each member of the equation from unity, te b?+-c’—a? ? 1—cos. A = 1— bie Qbc +o x 2he _ &—(b’°—2be+c*) 2be fd Shes or 2Qhe (a+b—c) (a+c—b zilashdr th (ck onnd) gos i eeosbaluis ANALYTICAL PLANE TRIGONOMETRY, 71 Multiplying together equations (1) and (2,)~ (1+c08A).(1—~cosA) = +) OF) ape b) (abe) But (1+cos. A) (l—cos. A)= 1—cos.’? A rs = sin.” Av #(Table 1.) wh _(atb+e) (b+c—a) (frrob) (a+b—c) 4b’¢ Extracting the root on both sides, 1 sin. A= Qhe / (a+b+c)(b+c—a) (a+c—b)(a+b—c)..(3 The above expresison, for the sine ofan angle of a triangle in terms of the sides, is sometimes exhibited under a form somewhat different. Let s denote the semiperimeter, that is to say, half the sum of the sides of the triangle ; then at+b+c rai WK and, 2s = atb+e s—a = ee . 2(s— a) = b+c—a ti Gs ne Ha bd) 2c atec—d s—c = —— . 2 (s —c) = at+b—c Substituting 2 s,2 (s—a),...... naan ei b-+-c—a,..... in the expression for sin.’ A, it becomes 16 s (s—a) (s—b) (s—c int A = 168 G=@) (82) (s=¢) 4 b’c And extracting the root on both sides, ey ee ed ate BT tee DeS Mas (s—a) (s——b) (s—c) Proceeding in the same manner for the other | angles, we shall find E p> (8) sin. B= g's (s—a) (8—b) (S— | ae sin. = ab’ V's (s—a) (s——b) (s—c) 72 ANALYTICAL PLANE TRIGONOMETRY. By equation (1) we have iS 7 1+ cos. _ = (at+b+c) (b+c—a) 2be _ 48 (s—a) 2be But, by Chap. II, ; A 1+cos. A=2 cos. A 4s(s—a) 2 _ —— 2 COS. 3 2 be Extracting the root on boti sides, ref oe oe re And in like manner, | B ee cos. —= s(s—b) eee aN 2 ie oh 2 ac | ae ae 5 (s —s)-| 2 ab ) By equation (2) we have 1—cos. Aa late—b) (a+b —c) 2bc _4(s—b) (-—0) ti 2bc But, by Chap. II, I1—cos. A=2 sin? : .,A 4(s—d) (s—c) gs ARN. ge Extracting the root on both sides, sin =A4/ ee ? Cc And in like manner, sin — \/6=9 69 - sin =A/ 9D) io a? « - we have | ~ e a * _ ANALYTICAL PLANE TRIGONOMETRY. —_73 Dividing the formule marked (%) by those marked (c) tone = 4/ OS b) (s—9) e—9 |. tan = af, Seo) (0) ant ae >) s(s—b) , tan<-= J (s— 4 (s—?) | $s (s—c) RAE THR IV. ON THE CONSTRUCTION OF TRIGONOMETRICAL TABLES. Berore proceeding to apply the formule deduced in the last chapter to the solution of triangles, we shall make a few re- marks upon the construction of those tables, by means of which we are enabled to reduce our trigonometrical calculations to numerical results. It is manifest, from definitions 1°, 2°, 3°, &c. that the various trigonometrical quantities, the sine, the cosine, the tangent, &c. are abstract numbers representing the comparative length of certain lines. We have already obtained the numerical value of these quantities in a few particular cases, and we shall now show how the numbers, corresponding to angles of every de- gree of magnitude, may be obtained by the application of the most simple principles. , | The numbers corresponding to the sine, cosine, &c. of all angles from 1” up to 90°, when arranged in a table, form what is called the Trigonometrical canon. The first operation to be performed is * To compute the numerical value of the sine and cosine of 1’. » We have seen, Chap. II. formula (j) that b ws Se oll 1 — cos. 4 5 zo2aV1—sin. 6 By which formula the sine of any angle is given in terms of the sine of twice that angle. 7* a, * ~ * g® 44 74 ANALYTICAL PLANE TRIGONOMETRY. 6 i Now substitute PY for 4 and it becomes oy 6 ‘ be a/ b te 8 in. 7 or sin. 53= 5 ta sat 3 ; ne / ; é In like’ manner, sin.5a— WE jet 1 1—sin. a att ny 4 W pi ra ye i / 1 — sin.’ 5F &c. = &c. ig 8 a/ ad 6 And generally, sin. gat 4—-FV/ 1—sin.? re 6 Now let 6=30° Wg = 15° and applying the above formula, we have sin, 15°= a/3 —2 1V1—sin. 2 4 Now if we suppose =12° 380! sin. 12° 30’/=1V1+sin. 25° +4V1— sin. 25° cos. 12° 380’=1V1-+sin. 25° = V1— sin. 25° Hence, if the values of the sine and cosine of 12° 30’, and of the sine of 25° obtained by the method already explained, when substituted in these equations, render the two members identical, we conclude that our operations are correct. The values of the sine and cosine of 30°, 45°, 60°, &c. which were obtained in Chap. II, may be employed as for- mule of verification. We can obtain finite expressions, although under an incom- mensurable form, for the sines of arcs of 3°, and all the mul- tiples of 3°, 2. e. for Br, G2, 9°, 19? PS" 1G, 2h 2e , lag GU ce tied OU Brel os ie» 45°, 48°, 51°, 54°, 57°, 60°, 63°, 66°, 69°, 72°, 75°, 78°, 81°, 84°, 87°, 90°. 7 We first obtain the values of the sines 30°, 45°, 60°, 18°, and from these we obtain all the others, by means of the formule, for Sin. (6+/), sin. (6— £8), &c. The numerical value of the trigonometrical functions have been calculated by some to ten places of figures, by others as far as twelve. We must have tables calculated to ten places to have the seconds and tenths of a second with precision, when we make use of the sines of angles which differ but little from 90°, or of the cosines of angles of a few seconds only. Tables in general, however, are calculated as far as seven places only, and these give results sufficiently accurate ior all ordinary purposes, ANALYTICAL PLANE TRIGONOMETRY. 77 Such is the formation of the trigonometrical canon. The labor of the application of this canon, may be much facilita- ted by the application of a system of artificial numbers called logarithms, a description of which, forms the subject of the next chapter. CHAPTER V. LOGARITHMS. DEFINITIONS AND ILLUSTRATIONS. 1. Logarithms are certain functions of natural numbers, by the use of which the tedious operations of multiplication and division are performed by the addition and subtraction of those functions; which consist of artificial numbers having such relations to certain natural numbers, that the sum of any two of those artificial numbers will be a similar function of the product of the natural numbers to which they have such rela- tion; and the difference of any two will be a similar func- tion of the quotient arising from the division of such natural numbers. Or more definitely, logarithms are the numerical exponents of ratios, being a series of numbers in arithmetical progression corresponding to another series in geometrical progression. Thus Me I, 2, 3,4, 5, 6, Indices or logarithms, ?( 1, 2, 4, 8, 16, 32, 64, Geometrical progression. Or 0, 1, 2, 3, 4, 5, 6, Logarithms, ; 3, 9, 27, 81, 243, 729, Geometric progression. Or OS yet2, so; 4, 5 Logarithms, , 1, 10, 100, 1000, 10000, 100000 Geom. progress. Where it is evident that the same indices answer for any geometric series, and therefore there may be an endless variety of systems of logarithms to the same natural numbers by changing the second term 2, 3 or 10, &c., of geometrical se- ries of whole numbers ; and by interpolation the whole system of numbers may be made to enter the geometric series and re- ceive their proportional logarithms, whether intregers or decimals. It also appears from the construction: of these series, that if any two indices be added together, their sum will be the in- dex of that number which is equal to the product of the two terms in the geometric series to which those indices belong. Thus, the indices 2 and 3 being added together, make 5, and the product of 4 and 8, being the terms corresponding to those indices is 82, which is the number corresponding to the index 5. b ee 78 ANALYTICAL PLANE TRIGONOMETRY. In like manner, if any index be subtracted from another, the difference will be the index of that number which is equal to the quotient of the two terms to which those indices belong. Thus the index 6 — the index 4 is=2, and the terms corres- ponding to those indices are 64 and 16, whose quotient is=4, which is the number answering to the index 2. For the same reason, if the logarithm of any number be multiplied by the index of its: power, the product will be equal to the logarithm of that power. Thus, the index or logarithm of 4, in the above series is 2; and if this number be multiplied by 3, the product will be=6, which is the logarithm of 64, or the third power of 4. And, if the logarithms of any number be divided by the in- dex of its root, the quotient will be equal to the logarithm of that root. . Thus, the index or logarithm of 64 is 6, and if this number be divided by 2, the quotient will be 3, which is the logarithm of 8, or the equal root of 64. The logarithms most convenient for practice, are such as are adapted to a series increasing in a tenfold ratio, as in the last of the above forms, and are those which are usually found in most of the common tables on the subject. 2. In a system of logarithms all numbers are considered as the powers of some one number, arbitrarily chosen, which is called the base of the system, and the exponent of that power of the base which is equal to any given number, is called the loga- rithm of that number. Thus, if a be the base of a system of logarithms, N any number, and x such that N ='a* then z is called the logarithm of N in the system whose base is a. The base of the common system of logarithms, (called from their inventor “ Briggs’s Logarithms”), is the number 10. Hence since (loy°= 1 , 0 is the logarithm of 1 __ in this system. (0)'= 10 ,1 ee cae a ee {RO} = vib Onis! 2 Wie elt ce Cee (10)s= 1000 , 3 3008: 6 (10)*= 10000 , 4 10000 ——-———— &e.. =). &e: Sicviny.o; eaeee: with From this it appears, that in the common system the loga- rithms of every number between I and 10,is some number be- ANALYTICAL PLANE TRIGONOMETRY. 79 tween 0 and 1,7. e. is 1 plus a fraction. The logarithm of every number between 10 and 1V0, is some number between 1 and 2,2. e. is 1 plus a fraction. The logarithm of every number between 100 and 1000, is some number between 2 and 3, 7. e. is 2 plus a fraction, and so on. In the common tables the fractional part alone of the loga- rithm is registered, and from what has been said above, the rule usually given for finding the characteristic, or, indea, i. e. the integral part of the logarithm will be readily understood, viz. The index of the logarithm of any number greater than unity is equal to one less than the number of integral figures in the given number. Thus, in searching for the logarithm of such a number as 2970, we find in the tables opposite to 2970 the number 4727564; but since 2970 is a number between 1000 and 10000, its logarithm must be some number between 3 and 4, 2. e. must be 3 plus a fraction; the fractional part is the number 4727564, which we have found in the tables, affix- ing to this the index 3, and interposing a decimal point, we have 3,4727564, the logarithm of 2970. We must not, however, suppose that the number 3.4727564 is the exact logarithm of 2970, or that 9970 = (1 0) 3.4727564 accurately. The above is only an approximate value of the logarithm of 2970; we can obtain the exact logarithm of very few numbers, but taking a sufficient number of decimals, we can approach as nearly as we please to the true logarithm, as will be seen when we come to treat of the construction of tables. It has been shown that in Briggs’ system the logarithm of 1 is 0, consequently, if we wish to extend the application of logarithms of fractions, we must establish a convention by which the logarithms of numbers less than 1 may be repre- sented by numbers less than zero, i.e. by negative numbers. Extending, therefore, the above principles to negative ex- ponents, since = or (10)’"=0.1, —1 isthe logarithm of.1 in this system ! 2 Foo °F (10)"=0.01, —-2 —————_ 01 1 sel 5 ——(. eit —————— 001 5900 OF (10)-*=0.001.—3 1 ——— 10)—"=0.0001,--4 ——-—————__ .0001. 700007 « )-‘=0.00 4 &c. &c. 80 ANALYTICAL PLANE TRIGONOMETRY. It appears, then, from the convention, that the logarithm of every number between 1 and .1l, is some number between 0 and —1; the logarithm of every number between .1 and .01, is some number between — 1 and— 2; the logarithm of every number between .01 and .001, is some number between — 2 and —3; and so on. From this will be understood the rule given in books, of ta- bles, for finding the characteristic or index of the logarithm of a decimal fraction, viz. The index of any decimal fraction is a negative number, equal to unity, added to the number of zeros immediately following the decimal point. 'Thus, in searching for a logarithm of the number such as .00462, we find in the tables opposite to 462 the number 6646420 ; but since .00462 is a number between .001 and .0001. its logarithm must be some number between — 3 and —4, 2. e. must be —3 plus a fraction, the fractional part is the number 6646420, which we have found in the tables, affixing to this the index — 8, and in- terposing a decimal point, we have — 3.6646420, the loga- rithm of .00462. General Properties of Logarithms. Let N and N’ be any two numbers, z and z’ their respec- tive logarithms, a the base of the system. Then, by def. (2), N=a¥- - - - - - (1) Nisa vie) 2) (a—x) @ 82 ANALYTICAL PLANE TRIGONOMETRY. Ex. 7. log. (a*4/a°)=log. a’+ Flog. a’=8 log. a+ log. a 15 = _ log. a Ex, 8. log. V(@—2)"= = log. (a—2)+— log. (@’-+ar-+2') = = slog. (a—zx) +log. (a+x2+z)+log. (a+r—z)} | where z?=az Ex. 9. log. /q?ta? = = flog (a+xz+z)+log. (a+rz—z)}, where z?=2axr V HZ, 4 Ex. 10. log. facie ieee, (a—zx)—3 log. (a+z)} Let us resume the equation, NN == a 1°. If a>1, making z=0, we have N=1; the hypothesis x=1 gives N=a. As z passes from 0 up to 1, and from 1 up infinity, N will increase from 1 up to a, and from a up to in- finity ; so that x being supposed to pass through all interme- diate values, according to the law of continuity, N increases also, but with much greater rapidity. If we attribute negative iB 1 4 values toz. wehave N=a *, or N= —— Here, as x increas- = es, N diminishes, so that z being supposed to increase nega- tively, N will decrease from 1 towards 0, the hypothesis z=o gives N=0. - 1 2°. If a<, put a= he where b>1, and we shall then have 1 clade : \. N= jx GS N=0* , according as we attribute positive or nega- tive values to z. We here arrive at the same conclusion as in the former case, with this difference, that when z is po- sitive N<1, and when z is negative N>1. 3°. If a=1, then N=1. whatever may be the value z. From this it appears, that, 1. In every system of logarithms the logarithm of 1 1s 0, and the logarithm of the base is 1. ANALYTICAL PLANE TRIGONOMETRY. 83 II. If the base be >1, the logarithms of numbers >1 are po- sitive, and the logarithms of numbers<1 are negative. The contrary takes place if the base be< equal to the angles ADB d BDC and BCA; join a,b, and abcd will be similar 5 | to ABCD, and if ab re- y present the side AB, then will de repersent DC. &c. By analysis, In the triangle adc, all the angles and the assumed side, dc are given to find ad and ac. Then in the triangle bcd, all the angles and the assumed side dc, are given to find be, bd. Lastly, in the triangle adb, we have the sides ad, db, and the angle adb, to find the side ab. Then we have, ab: AB:: de: DC:: ad: AD:: bc: BO :: ac: AC:: bd : BD, whence we have the distances, DC, AD, BC, AC and BD. Ml 118 ANALYTICAL PLANE TRIGONOMETRY. PROBLEM V. -_ Given AB, a, b, and the angles c, d, taken at some point Pin the same plane ABC, to find x; and thence PA, PB, PC. Put PAC+PBC=180°—(a+b+e+d)=2s A B PAC] BC =.ch-, ie duets Hen Qe; Then, PAC=s+z, PBC=s—z, and, by lemma 1, sin. @ sin. c sin (s—z)=sin. b sin. d sin. (s+<) _sin,dsin.d sin. (s—) _ tan. s—tan. x "*sin.asin.c sin. (s+x) ~ tan. s+tan. 2 sin. 6 sin. d P Put tan 8.‘ >... cosec a4 Sin 6 sin. a sin c cosec. c sin. d; then we have tan.s— tan. 2 _ tan. 2 I1—tan.f8 tan. 45°—tan. 8 tan. s-+tan. =tan. 6 .-. tan. s_1+tan. 8 ~ tan. 45°+tan. 8. a ew (am 45°—tan. 8 tng sin. (45 mie i 2" F tan. 45°+tan. 6 ‘ sin. (45°+8) °°" Hence z is known, and thence s+z and s—z2 are known. PC sin. (s+xz) AC sin. b PC = AB cosec. (a+) sin. 4 cosec. c sin. (s+2). PROBLEM VI. When the points P and C are on opposite sides of AB. Put PAB+PBA=180°—(c+d)=2s PAB—PBAS®... TUS». G22Qz ; then, PAB=s+2, PBA = s—2x; and, bya lemma 1, sin. a sin. c sin. (s —z) = sin. 6} sin. d sin. (s+2) ; hence, as in the last problem, we have sin. (45°—/) tan? c= sin. (45°-+8) tan. S$; where tan. 8 = cosec. a sin. 6 cosec. c sin. d; and 2s = 180°—(c+d). ANALYTICAL PLANE TRIGONOMETRY. 119 PART II. Many curious and highly useful problems in trigonometrical surveying, may be elegantly solved through the properties of the circle, by geometrical construction, and by analyzing this construction by trigonometrical analysis; a few of which we will give in the continuation of this chapter. PROPOSITION II. LEMMA. If two points be assumed in the circumference of a circle, they will subtend the same angle from any point whatever of the circumference on the same side of the chord joining these two points, which will be half the angle at the centre when the centre is on the same side of that chord as the point of obser- vation, and will be equal to half its complement to 360° when on the opposite side. This proposition is evident from Prop. XIX. Cors. 1 and 8, B. Ill. El. Geom. Hence, if F,G are two points assumed in the circumference ABFG, then will _F, G appear under the same angle from any points A and B situated in the cir- g cumference, and on the same side of the chord FG; which will be half the angle C at the centre; and the points F’,G will also appear under equal angles at every point D, E on the side DE of the chord FG, which will be equal to the angle mea- sured by half the arc FBAG equal the complement of the angle FCG as enunciated. Cor. 1. Hence, any two objects in the circumference of a circle will always appear under the same angle, in any point of the arc of either segment, and in no other point situated out of that circumference, on the same side of the objects, will the angle be the same. Cor. 2. If the angle under which any two objects appear be less than 90 degrees, the place of observation will be some where in the arc of the greater segment; and if the angle be greater than 90 degrees, the place of observation must be some where in the arc of a segment less than a semicircle, and the angles under which the objects appear, will be the same in any point whatever of those arcs. 120 ANALYTICAL PLANE TRIGONOMETRY. Scholium. Hence, having the angles subtended by any two objects from any two given positions, not all in the same cir- cumference, the positions of the objects may be determined by the intersections of the circumferences of two circles, each of which is so described as to pass through the two objects and one of the given positions. PROBLEM VII. Three points in the same plane being given in position, to determine the position of any other point or place of observa- tion in reference to the given points. This problem admits of six cases. The three given points may be the vertices of a triangle, and the required point, or station, may be without the triangle, and opposite one of its sides; it may fall in the same right line with two of the given points, it may fall directly between two of them, it may fall within the triangle or it may fall without the triangle but opposite one of the angles ; and lastly, the given points may be all in the same straight line. Case 1. When the given points are the vertices of a given triangle, and the station regarded, falls without the triangle and opposite one of its sides. Let A, B, C, be the given points C whose positions in reference to each B other are known, and let S be the point required. Having taken, the angles ASC, ASB, describe on AC the segment of a circle that shall contain an angle equal the observed angle ASC; and on CB describe a segment that shall contain an angle equal to the angle CSB, and the point of intersection of the arcs of those segments will determine the position S. | Or, Make the angle EBA = the observed E angle ASC, and the angle BAE = the angle BSE; through A, B, and the in- EZ e tersection at KE, describe the circle A F B AEBS; through E and C draw EC, which produce to meet the circumfer- ence atS. Join AS, BS, and the dis- tances AS, CS, BS will be the required distances of the station S, from the S points A, B, and C. S “ANALYTICAL PLANE TRIGONOMETRY. 121 By trigonometrical analysis. In the triangle ABC, the three sides are given to find the -angle BAC. And in the triangle AEB, we have angle EAB = angle BSE, angle ABE = angle ASE, and therefore the angle AEB, with the side AB, to find AE and BE. Also in - the triangle AEC, we have the sides AC, AF, and the included angle to find the angle AEC. _ Whence the sum of the angles AES, and the observed angles ASE, subtracted from 180° gives the angle SAE. Then in the triangle AES, we have all the angles and the side AE, to find the side AS. And the angle ACS = 180°— angle SAC — angle ASC; then in the triangle ACS we have all the angles and the side AC to find CS. Angle AEB—angle AEC = angle BES, whence we have the side BE of the triangle BES, and the angles E and S, to find BS; then AS, CS, and BS are the station distances required. Scholium. Ist. If the angle BSC, be less than CAB, the point E will be below the point C. 2. When the points E and C fall so near each other that the production of EC toward S is attended with uncertainty, the former method of construction is preferred. Case 2. Let it be required to determine the position of an observer at S in reference to the three objects ABC, when SAC are in the same right line. Having taken the angle at 8, and cal- Ader C culated the angle CAB from the sides of § the triangle ABC which are known by hypothesis, we have the angle ABS=ang. CAB—ang. ASB. Then at B with the side BA, construct the angle ABS. produce the side, BS till meets the production of CA in 8, and SA, SC, SB will be the several distances of S from the points A, C, B, whence hav- ing the angles 8, C and B and the side BC, the sides SB, SC and SA may be obtained. By trigonometrical analysis. Angle SAB = 180°—BAC, angle SBA = CAB—ASB. Hence in the triangle SAB, we have all the angles and the side AB to find the distances AS, BS, &c. Case 3. ~To determine the position of S in reference to three given ob- jects ABC, where the required point is directly between A and B. B it 1]* aad + : » 122 ANALYTICAL PLANE TRIGONOMETRY. Having constructed the triangle ABC which is given by hypothesis, and hav- ing observed the angle BSC, construct from any point A on the line AB an angle BAE=the observed angle and draw CS parallel to EA, and S is the required sta- tion. rs By Analysis. In the triangle ABC all the sides being known, let the an- gle A,B be obtained. Then in the triangle BCS having the angle S and B, and the side CB we may proceed to find the distances BS, CS. Then we shall find AS=AB—SB. Case 4. To determine the position of a station S in reference to three given objects when the required station falls within the tri- angle formed by connecting those given objects. Let the given objects be three vs towns A,B and C which are all visible from a station S, which is included in the triangle formed by lines drawn from A to B, from B to C, and from C to A. 5 A B First take the angles ASB, BSC, CSA, then on‘either side AC describe an arc ASC, which shall contain the observed angle ASC, and one either of the other sides AB, describe an arc which will contain the angle ASB, and the point of in- tersection of those arcs is the station.S, all of which is evident from Lemma II. Otherwise, on AB make an ¢ B angle ABl=the supplement of the angle ASC; and make an angle BAE equal to the supple- ment of the angle BSC, then will ABE=—ASE and BAE = BSE, , since ASE and BSE are respec- tively the supplements of the an- gles ASC and BSC, through the points A, B, and E describe a circle, join EC cutting the circum- ference in the point S which is the station required. ate » * ANALYTICAL PLANE TRIGONOMETRY. 123 By Analysis. Ist. In the triangle AEB the angles Band A being the sup- plements of the observed angles CSA, CSB, are therefore known, and consequently the angle E and the side AB, to find the sides AK, BE. 2d. In the triangle ECA we have the sides EA, AC and their included angle EAC=BAE+CAB, to find the angle AEC. 3d. And the triangle CEB the sides CB, BE, and the in- cluded angle CBE are given to find the angle CEB. 4th. Therefore in the triangle SAB the angle A, being equal the angle E, since they are angles in the same segment, is also known and also the angle ASB, hence we have all the angles and the side AB to find the sides AS, BS which are two ofthe station distances required. 5th. If from the angle CAB we take the angle SAB, we shall have the angle CAS. Therefore in the triangle ACS we have all the angles with the sides AC and AS to find the other station distance CS. Case 5. Tet it be required to find the distance of any station S from each of three objects A, B, C, when one of the angles C of the tri- angle formed by connecting the three objects falls toward the station NS. Make the angle DAB=the observed angle CSB and the angle DBA, equal to the observed angle CSA. On AB de- scribe a circle that shall contain in its greater segment the observed angle ASB, through D and C draw the line DC, till it intersects the circle at S, which intersec- tion determines the position S. By Analysis. Ist. In the triangle DAB, all the angles and side AB are known to find AD, DB. 2d. In the triangle ADC are given the sides AD, AC and then included angle, to find the angle ACD. 3d. The angle CAB=ACD—ASD, since ACD, the out- ward angle is equal to the sum of two inward opposite angles CSA, CAS. Therefore in the triangle ACS we have all the angles and the side AC to find the distances SA, SC. Lastly, in the triangle BSC the sides CB CS, and the angle BSC are given to find the distance CB. 124 ANALYTICAL PLANE TRIGONOMETRY. Case 6. Lo determine the position of any station Sin reference to three objects ABC all in the same straight line. On AB describe an arc of a seg- 4B ‘ ment to contain an angle equal to the angle ASB, and on BC de- scribe an arc containing an angle CSB= the observed angle, subtend- ed by BC, and the point of inter- section of those arcs will determine the position of S ; whence if we S draw the lines SA, SB, SC, those lines will the several dis- tances of S from the objects A, B, C. For (Lemma II) the points A, B appear under the same angle in every point in the arc BSA, and in no point out of that arc, and B,C, appears under the same angle in every part of the arc CSB and no point out of the arc. Hence the point of in- tersection of those arcs is the only point where both of those conditions are united, or where both of the objects appear under the observed angles. Otherwise, at A and on the line AB make BAK=the observed angle CSB and at C an angle=the observed angle ASB; on AC describe a circle that shall contain an angle=the sum of the observed angles at 8S or which is the same, describe a circle which shall pass through the three angles A, E and C, from E through the point B, drawn EBS to cut the circle inS, andSA,SB, SC S determine the relative position of the station S in reference to the three A, B and C. By Trigonometrical analysis. Ist. In the triangle CAE all the angles are given and the side AC to find AE. 2d. In the triangle AEB the sides AE, AB, and their in- cluded angle are given to find the angles AEB and ABE. 3rd. Inthe triangle BSC we have the angle CSB and the angle SBC=ABE, and the angle SCB=angle AEB since they are both angles in the same segment ACS, hence all the angles and the sides BC are given to find the side, SC, SB. and SA. ANALYTICAL PLANE TRIGONOMETRY. 125 PROBLEM VIII. The distances of three objects A, B and C being given and con- sequently the angles which they form with each other. There are also two stations D, I, such that at D the objects A,C and K may | © be seen but not B; at E the object, B, C and D may te seen but not A. , B Hence the anglesCDEK ADC BED © BEC and consequently the angles CDA and CEB are given or known from observation to find the dis- tances DA, DC, DE, EC, and EB. i 4 c ae b Draw cd at pleasure, and atd make @ | an angle cde=the given angle CDE | make also the angle dec=DEC, the angle ceb=CEB, and cda =CDA / a () produced ad, bc, till they meet in s, , and draw sc. Ne wy By Analysis. Assume any value for de; then in the triangle cde all the angles are given and the side de, to find the sides cd, ce, the angle eds=180°—-ade and the angle des=180°—ed ; hence we have in the triangle dse all the angles and side de, to find ds, andes. Inthe triangle cds the angle cds =180°—adc, hence we have two sides cd ds, and their included angle to find the angle dsc=csa and side cs, then from the angle dsc take the angle dsc and we have the angle cse=csb. C Then with the angles csa, csb and the triangle ABC, as data, make the following con- struction by case first Prop. VII, and find the station dis- tances, SA, SB, SC. Then we shall have 126 ANALYTICAL PLANE TRIGONOMETRY.’ cs: CS::de DE:: dc: DCO::ec: EC::ds:DS:: se: SE from SA take SD and there remains DA, also from SB take SE and there remains BE, hence DA, DC, DE, EC and EB are found. Scholium. When AD and BE are pa- rallel, the foregoing method of solution fails. In which case, on AC describe a segment to contain an angle equal to the observed angle CDE, and on CB a segment to contain an vangle equal to CEB, draw the chord CF to cut off the segment CADF containing the an- gle CDE, and another chord CG cutting off a segment CBEG, containing the angle CED, the points F and G will be in the same right line with D and E, join GF which produce both ways till it cuts the circumference in D and E, and the points D and E will be the stations required. PROBLEM IX. The relation of the four points B,C,D,F, to each other are known, or the four sides of a quadrilateral figure and its angles are known, there are also two stations A, EK, such that at A only B, C, E, are visible and at E only the points D, F, A, so that the angles BAC, BAE, AED and DEF and consequently AEF may be known, required the distances AB, AC, ED, EF, EC, and AD. Ist. On BC describe a segment to contain the angle BAC, and draw the chord Cm that shall cut off an an- gle BCm = the supplement of the angle BAE. 2nd. On DFE describe a segment that shall contain an angle DEF, and draw the chord Dn, that shall cut off an angle FDn = the sup- plement of the angle AEF, and the intersections m and n will be in the same right line with the stations A and E. 3d. Through the points of intersection m and n, draw the line mn which produce to E at its intersection with the arc of ANALYTICAL PLANE TRIGONOMETRY. 127 the segment and the points of intersection, A and E will be the points of station. , The analysis of this may be supplied by the student. PROBLEM X. Given the base AB, the perpendicular FD, and vertical angle ADB of a triangle to find the sides AD, DB. On the base AB describe an isosceles triangle ACB whose vertical angle C shall be double the given angle ADB, if that angle is less than 90°, but double its supplement if the angle ADBis greater than 90°, and from this vertice as a centre, with the radius CA or CB describe a D circle, and the vertice D of the triangle will be found somewhere in the circum- ference (LemmalII.) Ata distance FD equal tothe altitude of the given triangle, draw aright line IL parallel to AB, and the point where this line cuts the circum- ference will determine the position of the vertical angle ; hence the sides DA and DB may be drawn. By Analysis. Draw the diameter DCP, and from C draw CH perpen- dicular to AB. Hence having the angle ACB we have also the angles CAB and ABC each equal to ua Rs — there- fore in the triangle ABC having all the angles and the side AB, the two equal sides AC, BC, become known also. And in the right-angled triangle AHC, CH= VAG? AH? And in the similar triangles DFN, CHN we have DF: DN ::CH:CN and by composition :: DF+CH : DN+CN or DF+CH:DC:: DF: DN Also HN= /CN?__CH? ; and BN=i AB—HN FN= VDN*—DF* and BF=BN—NF | Whence we have the right-angled triangled BFD with the sides BF and DF including the right angle, to determine the third side BD which also becomes known, and consequently the side AD as required. ~~ og Yee, ee 128 ANALYTICAL PLANE TRIGONOMETRY. PROBLEM XI. At the distance AB from the bottom of a tower is an object whose length is BD, how far must I ascend the tower that the object may appear under any angle T'? Draw the line AB=the given dis- tance, and produce it to D, then will BD represent the object; draw the in- definite line AE perpendicular to AB, which will represent the side of the tower. Then on DB make an angle BCD=2T, and from the vertex Casa centre, describe an arc of a circle pass- ing through the points D and B, cutting the tower in F,and AF will be the dis- p tance required. For since DCB is an angle in the centre of a circle, and AFBvis an angle in the circumference subtending the same arc, hence (Prop. II.) the angle DFB=4 angle DCB=T. Cor. Since by continuing the are of the circle the perpendi- cular is cut also in G; this point also answers the condition of the question, for angle DGB is evidently=the angle DFB, hence, the question admits of two answers.* By Analysis. First, in the isosceles triangle CBD we have the side BD and the angle C by construction, and since the triangle is isosecles, the angles B and D are each = (180°—2'1')+ 2 =90°—T. Hence, having all the angles and one side, the sides CB or CD are also known. Second, in the triangle ABC we have the sides AB and BC, and the angle B=CBD+2 angle BCD=90°+T, to find the side AC and the angle CAB, which thereby become known. Third, in the triangle ACF we have the sides AC and FC, and the angle CAF=90°— CAB, to find the side AF, the height required. But since the same data given to determine this triangle, apply also to the triangle ACG, the point may be also in G; hence, the problem is ambiguous both by construc- tion and analysis, as explained in case 4th, chap. VII. * This elegant construction was received from Mr. Joseph Gallup, of Norwich, C onnecticut, whose mathematical talent is acknowledged to be of a high order. ~~ pill Pai es cE. ANALYTICAL PLANE TRIGONOMETRY. 129 Scholium. If the circumference which passes through the points D, B should not cut the edge of the tower or perpendi- cular AK, but only touch it, it would admit of only one solution, and that point which would answer the conditions would be the point of contact; but if the circle should not reach the per- pendicular, the question would be impossible. EXAMPLES FOR PRACTICE. Ex. 1. Given the angles of elevation of any distant object, taken at three places on a level plane, no two of which are in the same vertical plane with the object; to find the height of the object, and its distance from either station. Let A, B, C, be the three stations, K the ob- ject, and KH perpendicular to the plane of the triangle ABC. Put BC=a, AC=), AB=c,, HAK=a, sf HBK=, HCK=y, and HK=z; then the angles AHK, BHK, CHK being right angles, we have AH=z cot. «, BH==z cot. 8, CH=z cot. y; whereby from the given data the required may be found. Ex. 2. Given a=30° 40’, 8=40° 33’, y=50° 23’; find 2, when the three stations are in the same straight line, AB be- ing=50° and BC=60 yards. Ans. 77.7175 yards. Ex. 3. Demonstrate that sin. 18°=cos. 72° is =} r (—1+ /5), and sin. 54°=cos. 36° is=1 rk (1+ /5). Ex. 4. Demonstrate that the sum of the sines of two arcs which together make 60°, is equal to the sine of an arc which is greater than 60°, by either of the two arcs: Ex. gr. sin. 3’+sin. 59° 57’ =sin. 60° 30’; and thus that the tables may be continued by addition only. Ex. 5. Show the truth of the following proportion: As the sine of half the difference of two arcs, which together make 60°, or 90°, respectively, is to the difference of their sines ; so is lto /2, or /3, respectively. ua Ex. 6. Demonstrate that the sum of the square of the sine and versed sine of an arc, is equal to the square of double the sine of half the arc. Ex. 7. Demonstrate that the sine of an arc is a mean pro- portional between half the radius and the versed sine of dou- ble the arc. Ex. 8. Show that the secant of an arc is equal to the sum of the tangent and the tangent of half its complement. Ex. 9. Prove that, in any plane triangle, the base is to the difference of the other two sides, as the sine of half the sum of 12 4 4 . . hs oe eae , ae ™ é i830. ANALYTICAL PLANE TRIGONOMETRY. the angles at the base, to the sine of half their difference: also, that the base is to the sum of the other two sides as the cosine of half the sum of the angles at the base, to the cosine of half their difference. Ex. 10. How must three trees, A, B, C, be planted, so that the angle at A may double the angle at B, the angle at B dou- ble that at C; and so that a line “of 400 yards may just go round them? Ex. 11. In a certain triangle, the sines of the three angles are as the numbers 17, 15, and 8, and the perimeter is 160. What are the sides and angles ? ix. 12..The logarithms of two sides of a triangle are 2.2407293 and 2.5378191, and the included angle, is 37° 20’. It is required to determine the other angles, without first find- ing any of the sides ? ‘Ex. 13. The sides of a triangle are to each other as the fractions 1, 1,1: what are the angles? Ex. 14. Show that the secant of 60°, is double the tangent of 45°, and that the secant of 45° is a mean proportional be- tween the tangent of 45° and the secant of 60°. Ix. 15. Demonstrate that four times the rectangle of the sines of two arcs, is equal to the difference of the squares of the chords of the sum and difference of those arcs. Kix. 16. Convert formule ¢, Chap. III, into their equiva- lent logarithmic expressions ; and by means of them and for- mule , Chap. III, find the angles of a triangle whose sides are 5, 6, and 7. Ex. 17. Being on a horizontal plane, and wanting to ascer- ‘tain the height of a tower, standing on the top of an inacces- sible hill, there were measured, the angle of elevation of the top of the hill 40°, and the top of the tower 51°: then measur. ing in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 338° 45’: required the height of the tower. Ans. 83.9983 feet. Ex. 18. From a station P there can be seen three objects, A, B, and C, whose distance from each other are known, viz. AB=s800, AG= 600, and BC=400 yards. There are also measured the horizontal angles APC=38° 45’, BPC—22° 30’. It is required, from these data, to determine the three dis- tances PA, PC, and PB. Ans. PA=710.193, PC=1042.522, PB=934.191 yards. rs *. mee | “. ee SPHERICAL TRIGONOMETRY. Having demonstrated in the treatise on Spherical Geome- try, several important properties of the circle of the sphere, and of spherical triangles, we shall now proceed to deduce various relations which exist between the several parts of a spherical triangle. These constitute what is called Spherical Trigonometry ; and enables us, when a certain number of the parts are given, to determine the rest. The first formula which we shall establish, serves as a key to the rest, and is to spherical trigonometry what the expression for the sine of the sum of two angles is to plane trigonometry. CHAPTER I. ' ra 1. To express the cosine of an angle of a spherical triangle in terms of the sines and cosines of the sides. Let ABC be a spherical triangle, O the centre of the sphere. Let the angles of the triangles be de- noted by the large letters A, B, C, and the sides opposite to them by the corres- ponding small letters, a, 0, c. At the point A, draw AT a tangent to the arc AB, and Ad a tangent to the arc AC. Then the spherical angle A is equal to the angle TAt be- tween the tangents, (Spher. Geom. Prop. VIL.) Join OB, and produce it to meet AT in T. Join OC, and produce it to meet A? in ¢. Join T, t; 132 SPHERICAL TRIGONOMETRY. Then, 4h Gilet, AB=sec. c. OC Or = see. AC=sec. b aaatan. AB=tan. c At OG =tan. AC=tan. d Then in triangle TO¢ T? =OT’?+0? —20T . Ot cos. TOt | Ae ths ong OF OF cos. TO OC OG OG: OC * OC =sec.’ c+sec.” b— 2 sec. c sec, 6 cos. a, - (I) Again, in triangle T At Te =AT’?+Af — 2AT. At cos. TAt TeclAT Aci o¢ wah Ab msriy 9 OG DCE" OG OE HGF A ae =tan.’ c-+tan.* b—2 tan.c tan. bcos. A. - - (2) Equating (1) and (2) tan.” c+tan.? b— 2 tan. c tan. b cos. A _==sec.’? c+sec."b — 2 sec. c sec. b cos. a —1+tan.*? c+1-+tan.? b—2 sec. c sec. b cosa .. —2 tan. c tan. b cos. A=2 —2 sec. c sec. b cos. a sin. c sin. b 1 1 or, pay 2 Rare €O8. A =. iyo ——-., COs: & icos. c ” cos. b cos. c’ cos. b cos. a—cos. b cos.c cos. A— COS: &— ©08. 6 cos. c) sin. & sin. c Similarly we shall have, | cos. b— cos. a cos. c cos, B= ——_______ ———"" P(a) sin. @ sin. ¢ cos. e—ecos.a cos. b cos. C= —-——_—_ sin. @ sin. 0” 2. To express the cosine of a side of a spherical triangle, in terms of the sines and cosines of the angles Let A, B, C, a, b, c, be the angles and sides of a spherical triangle; A’, B’, C’, a’, b’, c’, the corresponding qualities in the Polar triangle, Then by (a), SPHERICAL TRIGONOMETRY. 133 sin. 0’ sin. c’. But (Spherical Geometry, Prop. X.), A’=(180°—a’), a'=(180° — A), b'=(180°— B), c’=(180°—C), cos.(180° — A) — cos.(180° — B)cos.(180°—C) C08. LED" a) ars Sain (180%) .sin(180°=0) Sere, GPE: A+cos. B cos. C) ee sin. B sin. C | Similarly, cos. b= Cisadua rake us a : ‘ (8-) sin. A sin. __ COS. C+cos. A cos. rea Shine ( sin. A sin. B J 3. To express the sine of an angle of a spherical triangle, in terms of the sines of the sides of the triangle. By (a) we have, cos. a—cos. 0, cos. c sin. b sin. c Tittcenk: _cos. a—Ccos. b cos. c-+sin. b sin. c sin. 5 sin. c _©0S. a—(cos. b cos. c—sin. b sin. c) sin. 6 sin. ¢ cos. a—cos. (b+c) =" sin. 0 sin. c atb+c,. b+c—a cos. A = 2 sin. Ji edo wie Chane, Eri. Ch, Ll.) sin. 6 sin. c b+-c Let s Peo eich 22 b+c—a . $—Gi= ari, oatiass. Se &: _a+tb—c 2 sin. s sin. (s—a) woh ch COB: J Sees . iz? 134 SHERICAL TRIGONOMETRY. Again, resuming the expression for cos. A, cos. 4 cos. c+sin. b sin. c—cos. @ 1—cos. A = sin. 6 sin. ¢ __cos. (b—c)—Ccos. a ~ sin. b sin. c atb—c . Str? LO ae sin. b sin. c _2 sin.(s—c)sin.(s—-b) sin. b sin. ¢ 2 sin. (2.) Multiplying equations (1) and (2) 1— cos.? Mecsed STOR IE Par Og Giri lie ie sin.” 6 sin.’ c. -. sin. AS———-2 — Vsin. s sin. (s—a) sin.(s—b)sin.(s—c) } sin.O sin.c Similarly, | 2 . 7 Es / Sin. § Sin. (s-——a) 5i0.($—o) Sit. (S—c sin. B= a aes See eae Po 2 __* __ “ysin. s sin (s—a)sin. (s-6)sin.(sc) sin,a@ sin.b sin. C= Now, by equation (1) we have, 2 sin. s sin.(s—a) ICCB sin. b sin. c or A 2sin.s sin. (s—a) 2 cos. 9 sin. bsin.c A Aes Mata) s sin. (s—a) aie OF sin. 6 sin. ¢ a wae. sin. $ sin. (s—) \ (72) 2, sin. @ sin. ¢ C Ne s sin. (s—c) ~ cos. =e sin. a sin. 6 Next, by equation (2), 2 sin. (s—b)sin. (s—c) cos. A = - - sin. 0 sin: c SPHERICAL TRIGONOMETRY. 135 Or, A ye 2 fink nes 8) a sin. 6 sin. c} epee sin. (s—4)sin. sin. (s—6)sin. (s—c) ) aie pa ae Os) sin. b sin. ¢ Similarly, B sin. (s—a)sin.(s—c) ais Naor aimee f pia sin. (s—a)sin. (s—4) sin ANY aN sin. a sin. 6 J Finally, dividing a expressions (y.3 by those y. 2), we obtain, tan.“ ale! \ J a OS (s—b)sin. (s—c) } sin. $ sin.(s—da) _ Bo. /sin. (s—a)sin.(s—c) PEM dati. s ainoy tan =! / sin. (s—a)sin.(s—b) sin. ssin.(s—c) 4. To express the sine of a side of a spherical triangle in terms of the sines and cosines of the angles. By (8) we have, ‘gp. COS. A-+cos.B cos. C en sin. B sin. C cos. A-+-cos. B cos. C-+sin. B sin. C .. 1+cos. a= sin. B sin. C ane eet ee sib. sm..G pre Nh Tels <0 ao cer SB is oe) oie le 2. (PlaneTrig.Ch.IL.) sin. B sin. C Let Se gee vB celia and, pees A+B—C 136 SPHERICAL TRIGONOMETRY. Hence, ee fe. eee 2 cos. (s’—C) cos. (s B) sin. B sin. C ster hi antes Resuming expression for cos. a, cos. B cos. C —sin. B sin. C-+cos. A sin. B sin C cos. (B+C)+cos. A Rs sin. B sin. C A+B+C B+C—A = 12 cos. ——5-—— cos. sin. B sin. C oe! cos. 8! cos. (8X) Pray Seep A sO) Multiplying Equations (1.) and (2.). . PLS Ass 4 cos. s’ cos. (s! -s) cos. ‘S —B) cos. (s’/——C} sin.” B sin.’ C 1—cos.a@ = — 2 sin. B sin. C X /—cos. s’ cos. (s'/— A) cos. (s’/— B) cos. (s’/—C) ) Similarly, sin. a = sin. 6 = sin. A sin. C 518) x aN Ga ed at 7 ( X /__ cos. s’ cos. (s/— A) cos. (s’/— B) cos. (s‘/—C) By Equation (1) we have, 2 cos. (s’/—— B) cos. (s’-—C) Wye nee sin. B. sin. C ,@ 2-os. (s'—B) cos. (s'— C) "92 COS. > sin. B sin. C ee cos. (s’_B) cos. (s'—C) ) Nid ae La V sin. B sin. C Similarly, P hed cos. (s/— A) cos. (s‘/—C) > (6, 2.) pee =f sin. A sin. C pile cos. (s‘/—— A) cos. (s‘/—— B) i Ne Say / sin. A sin. B SPHERICAL TRIGONOMETRY. 137 By equation (2.) 2 cos. s’ cos. (s’— A) Patna Page 1 or Sky l —'cos))-@ = : : sin. B- sin. C ake 2 cos. s’ cos. (s’/—A) ove = RY a ne bz 2 sin. B sin. C siniiee an A Fal0818' 208. (8 A), 2 sin. B sin. C pe f= s’ cos. (s/— B) | 1. = ae anaes Seen ae Oe. 2 sin. A sin. C " oD sittiom = pape eS SPS EE) | 2 sin. A sin. B ) Finally, dividing the expressions (6. 3.) by the expressions (6. 2.) tan oe — cos. s’ cos. (s’— A) ) 2 cos. (s‘/—B)cos. (s’/—C) 6 , / — cos. s’ cos. (s’/—B) | tan. > = ae al _— 2 cos. (s/——A) cos. (s’/—C) i: c — cos. s’ cos. (s/— C) tg ee! A I re ge es igen cos. (s’/—A) cos. (s’—B) It is to be remarked that although the expressions (6. 1.) (5. 8.), (6. 4.), appear under an impossible form, they are in re- ality always possible. For by Prop. XVIII. of Spherical Geometry, the sum of the angles of a spherical triangle, is always greater than two right angles, and less than six right angles. », AF+B+C > 180° and < 540° A+B+C i —— or SSF and. 210% Hence, cosine s’ is always negative, and ... — cos. s’ is always positive. Again, if a’, b’, c', be the three sides of the polar triangle, since the sum of any two sides of a spherical triangle is greater than the third side: b+. c' > a! 180°— B+-180°— C > 180° — A B+ C —A< 180° B+ - my tae | “a 138 SPHERICAL TRIGONOMETRY. .. cos. (s’/— A) is always positive, and in like manner, cos. (s’ — B), cos. (s'-—-C), are always positive ; hence the above expressions are in every case possible. » 5. The sines of the angles of a spherical triangle are to each other as sines of the two sides opposite to them. Taking the expressions (vy. 1.) and calling the common ra- dical quantity N for the sake of brevity: ) 2N sin. & SS sin. 6 sin. c ; 2N sin. B= sin. @ sin. ¢ Dividing the first of these by the second: sin, A sin. @ sin. c / sin. a ) sin. Bs sin. Osin.c ~~ ssin. D Similarly, sin. A __ sin. @ sin. b “ad sin. a («.) sin. C sin. c sin. 6 sin. Cc sin. B sin. b sin. a sin. b sin. C sin. csin.a@ sin. € J 6. To express the tangent of the sum and difference of two angles of a spherical triangle, in terms of the sides opposite to these angles, and the third angle of the triangle. By (a) we have, cos. a — cos. b cos. c COS. A =. . ° - - - - - = - 1. sin. 0 sin. c ib And, cos. c— cos. a cos: 6 cos. C = sin. a sin. b cos. c = cos. a cos.b+ sin. asin.bcos.C - - - (2) Substituting this value of cos. ein Equation (1.): cos. a— cos. a cos.’ b— cos. D sin. a sin. b cos. C cos. A =. sin. 6 sin. c cos. a (1 —cos.’ 6) —cos. 6 sin. a sin. b cos. C sin. 0 sin. ¢ cos, a sin. b— cos. b sin. a cos. C cats TRELINE ReMi ee ORR cine Eee iy ti (3.) sin. c 2 i ‘ wail ye ‘ 7. . . » pea Pk SPHERICAL TRIGONOMETRY. 139 In like manner, substituting the value of cos. ¢ in Equation (2), in the expression for cos. B, we shall find, cos. 6 sin. a— cos. @ sin. 6 cos. C ea megane © + ee b00 oe) «€ cos. B = Adding equations (8) and (4) : cos. A + cos. B sin. a cos. b-+sin. 6 cos. a—(sin. a cos. b+sin. b cos. a) cos. C ——s —— sin. c __ sin. (a-+b) — sin. (a+0) cos. C an sin. ¢ sin. (a+b) (1 — cos. C) re sinc # BPR we >. * Again, by Equation (s) we have, sin. A + sin. a sin. B .. Speed . sin. A+sin. B __ sin. asin. 6 sin. B a sin. b ; 1 : sin. B *, sin. A=Esin. B* =< (sin. a = hed) — sin. b C sin. = (sin. asin. b) -— ae (sin. a+ sin. 6) cate (6.) Dividing Equation (6) by Equation (5), and taking first the positive sign: sin. A+sin. B sin. a+sin. 0 sin. C cos. A+cos. B ~ sin. (a+b) 1—cos. C erie Ste noe iar 2 sin. Ch Cos. Gi 2 pe 2 C BE Append +o: a uc oe 2 COS. 5) COs. heal 2 sin. 9 COS. 9 a tan. 15 = ee Colm 2 a+ 6b 2 cos. — Again, dividing Equation (6) by Equation (5) and taking the negative sign. sin. A—sin. B sin. a— sin. D sin. C cos. At+cos.B sin. (a+4) 1 — cos. G « - as ay f- . “4 - ct - SPHERICAL TRIGONOMETRY. 2 2 sin. - sin. sin. a-— a—b 2 a+b COs. 2 COS. cot. COs: ===> 2 We have thus obtained the required expression, viz. Similarly, tan. tan. tan. tan. tan. A+B COS. COS. sin. sin. @—) COS. “ae. COs. sin. sin. COs. COS. sin. sin. cot. cot. cot. cot. cot. cot. vo | & 9 | D> vo] > 0} Q a | a /- ‘ oe. & SPHERICAL TRIGONOMETRY. 141 7. To express the tangent of the sum and difference of twe sides of a spherical triangle, in terms of the angles opposite to them and the third side of the triangle. Let A, B, C, a,b, c, be the sides and angles of a spherical triangle, A’, BY, C’, a’, b’, c’, the corresponding parts of the po- lar triangle then by expression (2), ) g'—)p' ee oh aammere” cos vain j (2 Therefore, yg 80° A) —(180°—B) tan, 180° +180°-5 2 “oot, (18O=9) 2 ~_ (180°--A)-+(180°-B) “~" —_2 cos. ———_____—_—— ; 2 z cos.( a a+b ee 8 c s0— = ¥ a - om tan. (180 > = ere cot. (90 5 cos.( 180 eae ; 2 A—B . COS. Gc t at+b_ < fan. —— an Py bs 2 cos. — a’ —— b! sin 7 C’ oa Ala Died “eot. — . 2 = a’+ b! 2 d sin. oY ae Therefore, ._ (180°—A)—(180°—B) (180°—a)-(180°-b) ee isha” + kan. a Le 8) Cee oT ~, (180°—A)+(180°-B) ei j sin. __ aa 2 . A—B Sil. = . a eee Bit 2 | tai PF 2 _ A+B 2 sin. —-—— 4s 142 SPHERICAL TRIGONOMETRY. We shall thus obtain another group of formule analogous to the last. ? (Aves ) Poe: cos res tan, = oe 2 A+B 2 COS. 2 As—B sin. —-—— a—b 2 c | Te ieee) Ai tan. ==> het o ye A Ae ba | sin. _ 2 ) | cos pie | bre 20h gad ee ea OF ae COS. 5 | cage eee C. (¢') ey sin “IST a | tan. 5 = 7 tan. 9 sin. 5 J sedate ) ; oy b sie A+C 2 | COs. 5 _A]Ee oes sin. ss eee ee tan. [=a tan. 5 HAS G 5} sin 5 J 8. To express the cotangent of an angle of a spherical trian- gle, in terms of the side opposite one of the other sides and the angle contained between these two sides. By («) A cos. @a—cos. b cos. c 1 COS. == e . = - - = e ea ae sin. b sin. c (1) and, cos. C— COS. a cos. b cos. C= ——————_ sin. @ sin. 0 & _ a =s ~ . SPHERICAL TRIGONOMETRY. | 143 Hence, ‘cos. c=cos. b+sin. a sin. db cos. C. Substituting this value of cos. ¢ in equation (1), it becomes cos. a—cos. a cos.” b—sin. a sin. b cos. b cos. C cos. A= — sin. 0 sin. c cos. a (1—— cos.” b)——sin. a sin. b cos. b cos. C r* sin. 6 sin. ¢ . cos. a (l—cos.* b)—— sin. a sin. b cos. 6 cos. C fs cos. A= i aS ee er ere e sin. b sin. c *. cos. A sin. c=cos. @ sin. b—sin. a cos. b cos. C But, ‘ Sita 1A sin. c= — sin. a, by (¢ sin. A » by ), sin. C .4 ] ; cos. A rg sin. @=cos. a sin. b—sin. a cos. b cos C cot. A=cot. asin. d cosec. C—cos. d cot. C. Tn rlicn the cotangent of A is expressed 1 in the required manner. If in Equation (1), instead of substituting for cos. c, we had — substituted for cos. b, the value derived from the Equation. ' cos. b— Cos. a COS. C ERs cos. B= sin. @ sin. c we should have found a value for cot. A in terms of a, c, B, or cot. A=cot. a sin. c cosec. B— cos. c cot. B Proceeding in like manner for the other angles, we shall ob- tain similaf results, and presenting them at one view, we have cot. A=cot. a sin. b cosec. C—cos. b cot. C ) ) =cot. a sin. c cosec. B — cos. c cot. B cot. B=cot. 8 sin. a@ cosec. C — cos. a cot. C =cot. b sin. c cosec. A — cos. c cot. A | (1) cot. C=cot. c sin. a cosec. B — cos. a cot. B =cot. c sin. b cosec. A —— cos. 6 cot. A J 9. To express the cotangent of a side of a spherical tr tangle, in terms of the opposite angle, one of the other angles, and the side interjacent to those two angles. Let A, B, C, a, d, c; be the angles and sides ofa spherical triangle, and Al, BY, Cee es the corresponding parts in the polar triangle. 144 SPHERICAL TRIGONOMETRY. Then by (x) cot. A’=cot. a’ ‘sin. b’ cosec. C’ — cos. b’ cot. C’ .°. cot. (180°—a)=cot.(180°—A) sin. (180°—B) cosec. (180°—c} — cos. (180°—B) cot. (180°— c) —cot. a= —cot. A sin. B cosec. e— cos. B cot. c *, cot. a=cot. A sin. B cosee. c+cos. B cot. c. Applying the same process to each of the expressions in (7), we shall obtain analogous results, and thus have a new set ei. - =cot. B sin. C cosec. a+cos. C cot. of formule: | (8) cot. c=cot, C sin. A cosec. b-+-cos. A cot. =cot. C sin. B cosec. a+cos. B cot. a J cot. a=cot. A sin. B cosee. c+cos. B cot. =cot. A sin. C cosec. 6+cos.. C cot. cot. b=cot. B sin. A cosec. c+cos. A cot. waa era By aid of the nine groups of formule marked, («), (9), (7); (6), (2), (2), (25, (n), (4), we shall be enabled to solve all the cases of spherical triangles, whether right-angled, or oblique-angled ; and we shall proceed in the next chapter to apply them. CHAPTER II. ON THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. Spherical triangles, that have one right angle only, are the subject of the investigation of this chapter; those.that have two or three right angles are excluded. A spherical triangle consists of 6 parts, the 3 sides and 3 angles, and any 3 of these being given, the rest may be found. In the present case, one of the angles is by supposition a right angle; if any other two parts be given, the other three may be determined. Now the combination of 5 quantities taken, oD. 4. ‘ 3 and BG tg g 108 therefore ten different cases present themselves in the solution of right-angled triangles. The manner in which each case may be solved individual- ly, by applying the formulze already deduced, will be pointed out at the conclusion of this chapter; but we shall in the first place explain two rules, by aid of which the computist is en- abled to solve every case of right-angled triangles. These are known by the name of Napier’s Rules for Circular Parts ; and it has been well observed by the late Professor Wood- SPHERICAL TRIGONOMETRY. 145 house, that, in the whole compass of mathematical science, there cannot be found rules which more completely attain that which is the proper object of all rules, namely, facility and brevity of computation. The rules and their descriptions are as follow: Description of the Circular parts. The right angle is thrown altogether out of consideration. The two sides, the complements of the two angles, and the complement of the hypothenuse, are called the circular parts. And one of these circular parts may be called a middle part (M), and then the two circular parts immediately adjacent to the right and left of M are called adjacent parts; the other two remaining circular parts, each separated from M the middle part by an adjacent part, are called opposite parts, or opposite extremes. This being premised, we now give Napier’s Rules. 1. The product of sin. M and tabular radius=product of the tangents of the adjacent parts. % 2. The product of sin. M and tabular radius=product of the cosines of the opposite parts. These rules will be clearly understood if we show the man- ner in which they are applied in various cases. Let A, B, C, be a sphericai triangle, right angle at C. Let a be assumed as the middle part. Then (90°—— B) and 0 are the adjacent parts. And (90°— c) and (90°—A) are the opposite parts. Then by rule (1) R Xsin. a=tan. (90° — B) tan. b s=cot.) B tana Do mists f= ping on a; uae = I(T) By Rule (2) R. sin. a=cos. (90°—— A) cos. (90° — c) S=Site A-SI. €C, et UU ap tea ess = (2) 2. Let b be the middle part, Then (90°—A) and a are adjacent parts, Then (90°—c)and (90° — B) are opposite parts. Then by Role J, R. sin. b=tan. (90°—A) tan. a S=COLPAMtAt @ an = mg Swett we GB) 12* 146 _ SPHERICAL TRIGONOMETRY. And Rule II, and R. sin. b=cos. (90° — B) cos. se ggasi =sin. B sin. c 3. Let (90° —c) be the middle part. Then (90° — A), and (90° — B) are adjacent parts, And b and a are opposite parts. Then, R sin. (90° — c)=tan. (90°——A) tan. taal, R. cos. c=cot. A cot. B And, R. sin. (90°— c)=cos. a cos. 8. R. cos. c=cos. a cos, 8 - 4, Let (90° — A) be the middle part. Then (90° — c) and db are adjacent parts, And (90° —B) and a are opposite parts. Then Rule I. R. sin. (90°—— A)=tan. (90° — ce) tan. b. ..., .R..cos, A=cot.‘c fen: 0" And Rule II. R. sin. (90°— A)=cos. (90°— B) cos. a, R. cos. A=sin. B cos. a 5. Met (90°— B) be the middle part. 8a od’) -. + @ Then (90°—c) and a are the adjacent parts, And (90°— Then Rule I. cos. B=tan. (90°—c) tan. a, =(ani-a-cote:c cos. B=cos. (90° A) cos. a =sin. A cos. b Collecting the above results, and making R= sin. a=cot: B tan. b* - sin. asin. A sine?-\ >. sin. b=cot. A tan.a- - sin. b=sin. Bsin.c - - cos.c=cot. Acot. BB - cos. c=cos. a cos.) - - cos. A=tan. 0 cot. c - cos. A=sin. B cos.a_ - cos. B=tan. acot.c - cos. B=sin. Acos. b- - 1 A) and 0 are the opposite parts. =. (9) = We) nsd(L0) , we shall have Sane ne (1) -- + Q) - => (8) - = = (4) == (5) -- + ©) re) -- = @) yo pai) - - = (10) ci 4 SPHERICAL TRIGONOMETRY. 147 | It now remains for us to show that these conclusions are accurate, and in accordance with the formule already deduced. Now by (a). cos. c— cos. a cos. 6 sin. a sin. But when C=90°, then cos. C=0 __ COS. C— COS. @ COS. b ” sin. a sin. b . cos. c=cos. acos. b, which is formula (6) in the above table. cos. C= Again by (8) sin. @_ sin. : sin. c sin. C But when C=90° sin. C=1 sin. a=sin. A sin. c, which is formula (2) above. Sihjerly, sin. 6 sin. B sin. c sin. C * .. sin. b= sin. B sin. c, which is formula (4). . Next since by (a) ‘ cos. a—cos. 6 cos. c cos. A= : . ——; substitute for cos.c its value in (6). sin. 6 sin. ¢ - cos. a—cos. a cos. *b > sim 2 sin. c _cQs. a sin. "gin. € cos. a sin. b = sina sin. A .. sin.b=cot. A tan. a, which is formula (3.) ——,; Be perinie for sin. c, its value as found in (2.) Again, a cos. a—cos. b cos. ¢ ——— S: © substitute for cos. ds its value in (6.) cos. A= sin. 6 sin. c cos. ¢ —— — cos. bcos. ¢ » =cos. 0 sin. 6 sin. c cos. c sin. 8 ~~ gin. c cos. b =tan. b cot. c, which is formula (7.) 148 SPHERICAL TRIGONOMETRY. Again by (a.) cos. B __ cos. b—cos. a cos. ¢ «gin. @ SIN. € cos. )—cos. b cos.? a oF sin. @ sin. c substitute for cos. c its value from (6) cos. 6 sin. a N ; ’ r ee ee substitute for sin. c, its value from (4.) cos. 0 sin. a =~ sin sin. B Sin. a=cot. B tan. b, which is formula (1.) Again, cos. B * ail cos. b — cos. @ COS. € ‘ ; * ia aT a a ee substitute for cos. b, its value in (6.) ‘ * Se cos. aieOe. c | C08. a= J sin. @ sin. c COS. C SIN. & ~ gin. C COS. @ =tan. a cot. ec, which is formula (9.) Next by (8.) cos. A+cos. B cos. C oe ae sin. B sin. C But C=90° .*..cos. C=0, and sin. C=1. y cos. A ane cos. O—— sin ar. sin. B .. cos. A=sin. B cos. a, which is formula (8.) Again, . B+cos. A cos. C COS. WG Sisbdeah So —.——— and when C=90°. sin. A sin. C _ COs. B ; ~ sin. A : +, cos. B=sin. A cos. b, which is formula (10.) Lastly, COS. C-+cos. A cos. B din th; COR ere ne and in this case, SPHERICAL TRIGONOMETRY. 149 __cos. A cos. B sin. A sin. B | =cot. A cot. B, which is formula (5.) We have thus proved the truth of the results derived from the application of Napier’s rules, and may therefore apply these rules without scruple to the solution of various cases of right- angled triangles. | Let us then take each combination of the two data, and de- termine in each case the other three quantities, adapting our _ formule to computation by tables. — 1. Given A, B, required a, 8, ¢. , : cos. A R cos. A=sin. B cég,'a ... cos. dR = - + (i) = an ’ sin. F » _ »R cos. B=sin. A cos. b .. cos. b5=R — tae (2) . Racos. c=tot. A.cot R - = o-- - + = = (3) ¥ 2. Given a, b, required A, B, ¢, R sin. a =cot. B tan. d .. cot. B=R sin. a cot. b - (4) R sin. b =cot. A tan.a .. cot. A=R sin. b cot.a@ - (5) 4 R cos. c=cos.acos.b6 = - = - = + =- = = (6) 3. Given a, c, required A, B, d ie sin. @ R sin. a =sin. A sin. c .*. sin. A=R — - - (7) ba iIC COs.'C R cos. c =cos. a cos. b .*. cos. b5=R ~ @ - (8) Cos. a w © “Roos. B=tan.acot.c - - - apa toga tea (9) . 4, Given 8, c, required A, B, a. : : : sin. 6 R sin. d=sin. B sin. c .. sin. B=R- - (10) sin. ¢ - ~ COS. C ™ Rcos.c =¢os.acos.b .. ads. a=R (11) ' cos. b Rcos. A=tan.bcot.c - = - - = = = - = (12) 5. Given A, c, required B, a, 8, Roos. A=tan. b cot. c .. tan.b=Reos. atan.c - (138) R cos. c =cot, A.cot. B.*. cot. B=Rtan. A cos. c* + (14) R sin. aysin. A sin. ca ee + 6. Given B, c, required A, a, d. R cos. B=cot. c tan. a .*. tan. a=R cos. B tan. ¢ - (16) * i? 150 SPHERICAL TRIGONOMETRY. R. cos. c =cot. A cot. B .:.: cot. A=R fan*B cos. & - (17) Resin. d=sin.Bsin.ec - - -~ = = ~ = = = (18) 7. Given A, b, required B, c¢, a. R cos. A=cot.c tan. b.. cot. c=R cos. A cot. b - (19) R sin. b =cot. A tan.a.'. tan.a=R tan. A sin.b - (20) R cos. B=sin. Acos.b - - = - - - - - = (21) 8. Given B, a, required A, c, b. R-cos. B=cot. c tan. @ .*...cot. c=R cos. B cot. a - (22) R sin. a =cot. B tan. b .. tan. 2=R tan. B sin. a - (23) BR cos. A=sin: B gos.a_ 0 © SF ae ae 24) 9. Given A, a, required B, 8, c. ae es : cos. A & R cos. A=sin. Bcos. a .. sin. B=>R —— = - (25) COS. a R sin. a=sin. A sit.e «. sin c=R—— ~~ (26) sin. A : Rsin.b =cot.Atan.a@ - - - - - - - = ~ (27) 10. Given B, b, required A, a, c. cos. B R cos. B=sin. A cos. b .. sin. A=R - (28) cos. b Risin. @—sin. Bisin: c .°: sini.c =F es - - = (29) sin. B R gin. a:=cot.Btan.G ~- ~. -~ -.-,- - .@= (30) CHAPTER IIL. ON THE SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. The different cases which present themselves are contained in the following enumerations. 1, When two sides and the included angle are given. 2. When two angles and the side between them are given. 3. When two sides and the angle opposite to one of them are given. 4. When two angles and the side opposite to one of them are given. 5. When three sides are given. 6. When three angles are given. SPHERICAL TRIGONOMETRY. 154 I. When two sides and the included angle are given. The remaining angles may be determi mula (¢.) a ate y be determined from the for- Thus, ig a, b, C, be given, A, B, c, required. a—b EB PME 20 C an. — a ak ° 9 2 . a—)d ; A—B sin 9 ae C tan. = paar Se, i A x —B Whence 5 and a are known from the tables. Let ADB, : 2 A+B. 5) Q A=b+9 B=t—o A and B being known, ¢ may be obtained from (:.) sin. to sinwf For ; == sin.@ sin. A sin. C sin. And, in like manner, if any two other sides and the included angle be given, the remaining parts may be determined. ? sin. C=SIN. @ II. When two angles and the side between them are given. The remaining sides may be determined from the formula(Z’.) Thus, let A, B, c, be given; a, 4, C, required. B 7 ; eae COS. 2a i ati are ALB* 9.2 COs. yh. icp s an Bs Px, sin. —> tan. — tan. —- 2 _ _ A+B 2 sin. joi ak Whence or" and = are known from the tables. 152 : SPHERICAL TRIGONOMETRY. . % a+b Let aree’ R a—b Ps 2 “= @ a=s' +o! b=s'—o! a and b being known, C may be obtained by (.) 7 sin.C sin. c ue sin. A sin. @ sin. C=sin. A —< si n. a And, in like manner, if any two other angles and the in- cluded side are given, the remaining parts may be determined. III. When two sides and the angle opposite to one of them are given. The angle opposite to the other side may be found from formula (e.) Thus, let a, b, A be given, B, C, c, required. sin. B sin. d sin, A sin. a sinh eae abe sin. a The angle B being determined, the remaining angle C will be found from (¢.) a—b cos. —— A+B 2 C For tan. Te een cot. =r Hotels Sebacp. co tivrtolt cos.- 5 The angle C being determined, the remaining side c will be found from (é.) For sin. c sin. C sin. a sin. A sin. C sin. c=sin. @ -— sin. A or c may be found from (2’.) And, in like manner, if any other two sides and the angle opposite to one of them be given, the remaining parts may be determined. — . 4 © te a lal SPHERICAL TRIGONOMETRY. 153 IV. When two angles and the side opposite to one of them are given. m The side opposite to the other angle may be found from for- mula (¢.) Thus, let A, B, a, be given; 8, c, C, required. sin. 6 sin. sin. a sin. A Sp eat sin. B i sin, b= sin. a — The side 6 being determined, the remaining side c will be found from (%') cos Sear a+b We ng 2 c : eat Ste agf et, ok aaa as 4; 1 ee Fox tan. 9 ALB 9 COs. 9 A+B ty COS. aT ean tan 5) A-B an. 2 COS. 5) The side c being determined, the remaining angle C will be found from (:.) Sai BIH. C For =a sin. A sin. @ sin. ¢ sin. C=sin. A — sin. @ or c may be found from (c.) And, in like manner, any other two sides being given and the angle opposite to one of them, the remaining parts may be determined. V. When three sides are given. The three angles may be immediately determined from any one of the formule (vy 1,) (7 2,) (7 38,) (vy 4.) The choice of the formula, which it will be advantageous to employ in practice, will depend upon the consideration already noticed in the solution of the analogous case in plane trigo- nometry. VI. When three angles are given. The three sides may be immediately determined from any of the groups of formule (6 1,) (62,) (63,) (0 4.) 14 , 154 SPHERICAL TRIGONOMETRY. CHAPTER IV. ON THE USE OF SUBSIDIARY ANGLES. We have already explained in plane Trigonometry, the meaning of Subsidiary Angles, and the purpose for which they are introduced ; we shall now proceed to point out under what circumstances they may be employed with advantage, in Spherical Trigonometry. In the solution of case I., where two sides and the included angle were given, we first determined the two remaining an- les, and having found these, we were enabled to find the side also. It frequently happens, however, that the side alone is the object of our investigations, and it is therefore convenient to have a method of determining it, independently of the angle. Thus, for example, let b, c, A be given, and let it be required to determine a, independently of the angles B, c. By («,) we have cos. a—cos. b Gos. € es sin. b sin. c Whence cos. a=cos. A sin. d sin. c+cos. 6 cos. c. From which equation a is determined, but the expression is not in a form adapted to the logarithmic computation; we can, however, effect the necessary transformation by the intro- duction of a subsidiary angle. Add and subtract sin. 0 sin. c on the right hand side of the equation. Then cos. a =cos. A sin. b sin. c+cos. b cos. c+sin. b sin. c—sin. b sin. ¢ =cos. b cos. c-+sin. b sin. c+sin. b sin. c cos. A—sin. b sin. c =cos. () —c)— sin. bsin.c vers. A 1— cos. a=1— cos. (b —c)-+sin. b sin. c vers. A vers, a=vers. (b—c)-+sin. } sin. c vers. A sin. 6 sin. c vers. A =vers. (b——c) ) 1+ jae Goo cos. A= (Redan 4—Sin- b sin. c vers. A vers. (b—c) *, vers. a=vers. (b—c){1+tan.? 6} =vers. (b—c) sec.’ 4 from which a may be determined by the tables, 4 being known from the equation sin. b sin. c vers. A i rane vers. (b—c) SPHERICAL TRIGONOMETRY. 155 In like manner in case II, where two angles and the in- cluded side were given, we first determined the remaining sides, and then we were enabled to find the remaining angle. Now, let us suppose that A, B, c, are given, and that we are required to find C independently of a and b. cos. C+-cos. A cos. B Brom: (3) ip: Cur ae sin. A sin. B Ae cos. C=cos. c sin. A sin. B—cos. A cos. B . 1—cos. C=1—-sin. A sin. B(1—vers. c)-++cos. A cos. B. =1-+cos. (A+B)-+sin. A sin. B vers. c. Feat A sin. B vers. c ied or °2 sin.’ 5=2 cos.” A+B sin. A sin. B vers. ¢ =2 cos.” ee ee 2 2°cos 2 A+B “we z -. C A+B .. sin.2— = cos.’ sec.? 6 Se 2 If we assume : sin. A sin. B vers. c fan. 6= 2 2 cos 9 In case IJ, where two sides and the angle opposite to one of them were given, we first determined the angle opposite to the other side, and then the remaining angles and the remain- ing side in succession. Now, let us suppose the side ¢, inde- pendently of the angle B and of each other, under a form adapted for logarithmic computation. To find C, we have (7.) cot. A=cot. a sin. b cosec. C — cos. b cot. C or cot.Asin.C=cot. a sin. b—cos. b cos. C or sin. C=cot. a sin. 6 tan. A —cos. 0 cos. C tan. A .. sin. C+cos. C cos. 6 tan. A=cot. a sin. b tan. A. sin. 4 Let cos. 6 tan. A=tan. = cos. 4 ; sin. 6 : sin. C+ cos. C=cot. a sin. b tan. A cos. 4 *, sin. C cos. 6+cos.C sin. 4=cot. a sin. 6 tan. A cos. 4 . : sin. 4 sin. (C-+é)=cot. a sin. b tan. ae SEP cae| f =cot. a tan. b sin. 4 whence C is known, é being previously determined from equa- tion 156 SPHERICAL TRIGONOMETRY. tan. 6=cos. b tan. A. To find c, we have from («.) cos. a—cos. b cos. c cos. A= : sin. 6 sin. c oe sin. c sin. b cos. A=cos. a— cos. b cos. ¢ ; COs. a sin. c tan. b cos. A= — COs. Cc cos. b ' COs. @ sin. c tan. b cos. Am——= — cos. c cos. } 1 ‘e 6 Let tan. b cos. A=tan. 6= “2 _. cos. 4 ; sin. 4 cos. a sin. c—_—_—_ + COS. c= cos. 4 cos. b cos. a cos. 6 cos. b whence ¢ may be found, é being previously determined from the equation : tan. 6=tan. b cos. A. In like manner, in case IV, when two angles and the side opposite to one of them were given, we first determined the side opposite to the other angle, then the remaining side and the remaining angle in succession. Now, let A, B, a, be ~ given, and let it be required to determine c and C, independ- ently of 6 and of each other, and under a form adapted to lo- garithmic computations. If we take the formula (é.) cot. a=cot. A sin. B cosec. c+ cos. B cot. ¢ or cot.asin. c=cot. A sin. B+cos. B cos. c or sin. c=cot. A sin. B tan. a+cos. B cos. c tan. @ .. sin. c—cos.c cos. B tan. a=cot. A sin. B tan. a. , sin. 4 Let cos. B tan. a=tan. 6— =" cos. 4 ‘ sin. 6 ; sin. c— cos. c=cot. A sin. B tan. @ COS. sin. (c —6)=cot. A sin. B tan. a cos. 4 sin. 6 cos. B tan. a =cot. A tan. B sin. 4 whence c may be determined, 4 being previously known from equation tan. 6=cos. B tan. a. =cot. A sin. B tan. a SPHERICAL TRIGONOMETRY, Peer To find C, we have from (8) Soak ass 20% A-+cos. B cos. C sin. B sin. C. sin. B sin. C cos. a=cos. A+cos. B cos. C cos. A sin. C tan. B cos. a=- +cos. C cos. B sain Ctany Beeosa—cos C8S8 4 cos. B in. 8 Let tan. B cos. a=tan. 6=~" é cos. 4 sin. Cx S12: pees cos. C=°0®: A Os cos. B cos. A cos. 4 — cos, (C +46) =——— ae) cos. B whence C may be found, 4 being known from equation tan. é=tan. B cos. a. In the fifth and sixth cases, any one of the angles or sides required, may be found independently of the rest by the for- mule referred to. EXAMPLES IN SPHERICAL TRIGONOMETRY. Ex. 1. In the right-angled spherical triangle ABC, the hy- pothenuse AB is 65° 5’, and the angle A is 48° 12’; find the sides AC, CB, and the angle B. Ans. AC=55° 7’ 32" BC=42 32 19. angle B=64 46 14 Ex. 2. In the oblique-angled spherical triangle ABC, given AB=76° 20’, BC=119° 17’, and angle B=52° 5’; to find AC and the angles A and C. Ans. AC=66° 5’ 36” angle A=131 10 42 angle C= 56 58 58 Ex. 3. In an oblique spherical triangle the three sides are G=Bioal 7?) pa=-114 foc on lees required the angles A, B, C. Ans. A= 62°39! 42” B=124 50 50 C= 50 34 42. 14* ‘ APPLICATION OF ALGEBRA TO GEOMETRY. . ON THE GEOMETRICAL CONSTRUCTION OF ALGEBRAICAL QUANTITIES. As lines, surfaces, and solids are quantities which admit of increase and decrease, like other quantities, they may, like others, be made the subjects of algebraical operations, either by their numerical representatives, or by symbols expressing such quantities. Itis only necessaryfor this purpose, that their representatives should possess values in relation to each other corresponding to the magnitudes of the quantities which they represent ; and that those values should be expressed accord- ing to the properties or relations of the lines, surfaces, or solids, to each other; subject to geometrical construction and algebraical notation. ‘J We are enabled, also, to express by lines, surfaces, and so- lids, the solutions furnished by Algebra. This is founded on the known properties of geometrical figures, corresponding to similar properties of the quantities algebraically expressed. In this view of the subject, all quantities under algebraic ex- pressions, may be conceived to be susceptible of some kind of geometrical construction. 2. We will proceed to explain the manner of constructing those expressions, and representing, under a geometrical form, the conditions of an equation. This is called constructing the algebraic quantities. Ex. 1. Let it be proposed to construct such a quantity as b the following: — the value of the letters composing the quan- tity being known. From any point A draw two indefinite lines A AM, AN, making any angle with each other ; upon one of these lines AM take AB=c, and AD=a; then upon the line AN take AC=b. Having drawn the line BC, draw also DE par- allel to BC, this will determine AE as the va- lue of ee For the parallels DE, BC, give this proportion AB: AD:: AC : AB, (Prop. XIV. Cor. 2 B. IV. El. Geom.) orc: a::6: AE. M N B Cc . : % ald =" > 2 _ APPLICATION, ETC. 159 Me ab Therefore, AE=— Res MA te oe ey Hence, the line AE being the fourth proportional to the three lines represented by C, a, b, it may be used for the con- struction of the quantity—. 2d. Hence, also if it were proposed to construct the quan- Ges ‘ G tity — it may evidently be done in a similar manner, since in ab this case. the lines and a would be equal, for if — = c then a=b. b 3d. If it were proposed to construct eset it may be ob- served that the quantity may be resolved into the expression (a+d) b ct+d b shall have — to be constructed, which may be referred to the , hence representing a+d by m, and c+d by n, we former case. a’ —) 4th. Let the quantity to be constructed ba — 3 it may be observed that a?— b’ is equivalent to (a+2) x @a); hence, 2p? $e - fe WED, be represented under the form eo a) and we have only to find the fourth proportional to c, a+, a—b. bth. If the quantity to be constructed be —— oe , it may be put under the form Be ek md and having constructed ie in the manner just explained, we call the line given by this con- f ab Cc mc : struction, m; then a becomes —— which may also be constructed as above shown. 7 6th. It will be presumed, therefore, that in order to con- ab . a’ b struct a - H 3 ! a whence if we construct ‘ and represent its value by m, we mb may proceed to construct 7 160 APPLICATION OF Thus the whole art consists in decomposing the quantity : ; : : a a’ into portions, each of which retains the form sereOL rene ts and although this process may appear sometimes difficult, yet we may easily arrive at the object proposed by employing trans- formations. e a’+b? 7th. If, for example, is to be constructed, we may take °+5° a+am .. Le becomes an which may atam (at+m)a ai ‘ aa, Or Gan? * quantity easy to be con- structed after what has been said, when m and n are known. Now to determine m and 2, the equations l°=a’m, c’=an, : bs Cc give, m=—a and n= b'=a'm, and c’=an; then, be reduced to A which may be constructed by the methods already explained. | Thus, while the quantity is rational, that is, without radical expressions, if the dimensions of the numerator do not exceed those of the denominator except by unity, we may always re- duce the construction to the finding of a fourth proportional to three given lines. It sometimes happens, that quantities present themselves under a form, that renders recourse to transformations of no use ; this is when the quantity is not homogeneous, that is, when each of the terms of the numerator and denominator is not composed of the same number of factors ; when the quantity, for example, is such as ao But it should be observed, that we never arrive at a result of this kind, except when, in the course of an investigation, we suppose, with a view of simplifying the calculation, some one of the quantities equal to unity. If, for example, in 3 2 3 os we suppose b equal to 1, we shall have a But, as we never undertake to construct a quantity without know- ing the elements which we are to use for this construction, we always know in each case what is the quantity which is sup- posed equal to unity. We can always therefore restore it, and the above difficulty cannot occur ; because, as the number of dimensions must be the same in each term of the numera- tor, and also of the denominator, although the number of terms may be different in the one from what it is in the other, we restore in each term a power of the line, which is taken for ALGEBRA TO GEOMETRY. 161 unity, sufficiently raised to complete the number of dimen- a LZ b +c 2 a+ be the line which is taken for unity, we may write formula a+bd+ed Saab which may be constructed by making l’?=d m, c= dn, and a’ = d@’ p, which will change it into ip+bd*>+d'n sions ; thus, if we construct ; d being supposed to ad+dm dp+bd+dn (p+b+n)d 7 ean ip Ll yiap or me gie. qyl ee quantity easily con- structed, when we have constructed the value of m, n, and p; b? 2 3 namely, m = pt. Tn P= oe which is readily done after what has been said. Hitherto we have supposed that the number of factors, or the dimensions of each term of the numerator exceeds the number of factors, or the dimensions of the denominator only by unity. It may exceed that number by two or even three, but never by more than three, unless some line has been sup- posed equal to unity, or some of the factors do not represent numbers. 3. When the dimensions of the numerator of the pro- posed quantity exceed by two the dimensions of the deno- minator, the quantity expressed is a surface, the construction of which can always be referred to that of a rhomboid. and consequently to that of a square. If, for example, the quan- : a’ 2 b tity to be constructed be pike a’+ab a+ab, it may be considered as a X relat Now po easily constructed, after what has been laid down, by considering it a+b f asa X Te Let us suppose therefore that m is the value * b of the line thus obtained ; then a X ie will become a xm. Now if we make a the altitude and m the base of a rhomboid, we shall have a@ x m for the surface of this rhom- boid, (Prop. VI, B. IV, El. Geom.) therefore, reciprocally, : ; : *b this surface will represent a X m, or ities eli ate : . @+be+d In like manner, the quantity Pern Ee may bereduced to atc 162 APPLICATION OF a similar construction by making bc =a m, and d?= an; for a+amec+tand (et mar be) GC ais RS eae a+me+nd ae IR 2g refers itself to the preceding constructions, as also the values of m, n. Having found the value of this factor, if we represent it by p, we have only to construct a X p, that is, to make a rhomboid whose altitude is a and base p. 4. Lastly, if the dimensions of the numerator exceed the dimensions of the denominator by three, the quantity expresses a solid, the construction of which may always be reduced to a parallelopiped. If, for example, we were to construct a’ b + a’ 6 ’ Taman ahha might consider this quantity as the same as a+ab a’ : tab, estates and, having constructed thiol in the man- ner already explained, if we represent by m, the line given by this construction, the question will be reduced to this, namely, to constructab xm. Nowab represents, as we have seen, a rhomboid ; if, therefore, we conceive a parallelopiped, having for its base this rhomboid, and for its altitude the line m, the solidity of this parallelopiped will represent a b X m, that is, @b+a@ B? atc 5. What has been said will suffice for constructing any rational quantity ; we proceed now to rational quantities of the second degree. Ist. In order to construct Vad, let us draw ‘p an indefinite line AB, upon which we may take the part CA, equal to a, and the part BC, equal aC B it will then become Now the factor abx to b; upon the whole AB as a diameter, describe a semicircle, cutting in D, the perpendicular CD, raised upon AB at the point C ; then CD will be the value of Jab; that is, the value of Vg % is obtained by finding a mean proportional between the two quantities represented by a, 6. Indeed, we have AC :CD::CD:CB, or a~CD:: CD: 6; whence OD? =ab, or. CD = Jap Qnd. If we were to construct V3 ab + 8°, or which is the same thing, /(3 a + 6) 6, we should find a mean proportional between 3 a + 6 and 6. > ALGEBRA TO GEOMETRY. 163 8d. In like manner, if the quantity to be constructed were va’ — 0’, we might consider this the same as V (a +6) (a—b); and then find a mean proportional between a + 6 and a—b, If the quantity were Va? + bc, and we make bc = am, then we shall have Va’ + am, or (a4 + m) a, which is constructed by find- ing a mean proportional between a + m and a after having i be constructed the value of m = > by the rules already given. 4th. To construct Va? + 6’, we can in like manner make b? = am, and construct Va’ + am, in the manner just explained. But the property of a right angled triangle furnishes a more simple construction. If we draw the line AB, c equal to a, and at its extremity A erect a perpen- dicular AC, equal to b, joining BC, we shall have BC? = AB’ + AC? = a? + 6b’, and, consequently, BC = va’? + Bb’. 5th. We can also, by means of a right angled triangle, construct Va? — 6? in a manner different 4 “ from that above given; draw a line, AB, equal c to a,and having described upon AB, as a diam- ON eter, the semicircle ACB, draw from the point A , a chord AC, equal to 6; then, if we draw BC, sig this line will be the value of Va?— 6’; for the triangle ABC being right angled, we shall have AB? = AC? + BC’; con- sequently, BC? = AB’ — AC’ = a?— 0’; therefore BC = Va? — 6°. 6th. Hence, also, Va? + be admits of a different construc- tion from the above. Make be = m’, and construct Va? + m’?, as just shown, first finding for m a mean proportional between b and ¢, as indicated by the equation be = m’, which gives m= V be: 7th. If there are more than two terms under the radical sign, the construction is to be reduced to one of the preceding methods by means of transformations. — If, for example, we have Va’ + bc + ef we may make be = am, ef =an, and we have Va? + am + an, or V(a + m + n) a, which may be constructed by finding 4 mean proportional between a and a+ m+n, after having constructed the values of m and n, be ef ; namely, m = —,n = —. We might, moreover, make a a be = m?, ef = n3, ’ 164 APPLICATION OF and then we should have to construct Va? +m? + n?. Now, when there are several positive squares contained under the radical sign, as Va? -+m*-++n?+p?+ &c. we may make Va?-+m?=h, | Vh2+n? =i, Vi2?+p? = k, and so on; and, as each of the quantities is determined by the preceding, the last will give the value of Va?+m?+n?+p?+ &e. In order to construct these quantities in the most simple manner, each hypothenuse is to be regarded successively as aside ; having, for example, taken AB = a, and raised the perpendicular AC E = e, we may join BC, which will be 1; then at the point C if we raise upon BC the perpendicular CD = ; and having drawn ¢ BD, which will be i, at the extremity D, we may raise upon BD the perpendicular DE = p, and BE will be 4, and equal to Va2+m?-+n?+p?. If some of the squares are negative, we may combine the method just given with A B that for constructing Va? — b2 8th. Lastly, if the quantity to be constructed be of this form atts b+e Vd+e multiplying by Vd + e, will change it into a V (b+ c) (d+e); dyke then, by finding a mean proportional between 6 + ¢ and d+e, : : am . é : and calling it m, we have 7 pr which is easily constructed. The construction often becomes much more simple by set- ting out always from the same principles; but these simplifi- cations are derived from certain considerations which are pe- culiar to each question, and consequently can be made known only as the occasion presents itself. We will merely remark, in concluding, that although the construction of the radical quantities, which we have been considering, reduces itself to finding fourth proportionals, mean proportionals, and con- structing right-angled triangles, still we can arrive at con- structions more or less simple or elegant by the method em- ployed for finding these mean proportionals; we shall now, therefore, introduce two other methods of finding a mean pro- portional between two given lines. The first consists in describing upon the greater AB (see 3d diagram to Art. 5) of two given lines a semicircle ACB, 2% ALGEBRA TO GEOMETRY. 165 and, having taken a part AD equal to the less, raising a_per- pendicular DC and drawing the chord AC, which will be a mean proportional between AB and AD; for, by drawing CB, the triangle ACB is right angled, (Prop. XIX, Cor. 2, B. III fil. Geom.) and consequently AC is a mean proportional be- tween the hypothenuse AB and the segment AD. (Prop. XVII, Cor. 5, B. LV, Hl. Geom.) f The second method consists in draw- D ing a line AB, equal to the greater given line, and having taken a part AC cqual to the less, describing upon the remain- 5, der BC.a semicircle CDB, to which if H 7 we draw the tangent AD; this tangent is a mean proportional between AB and AC. (Prop. XX VII, B. IV, El. Geom.) It is evident, therefore, that rational quantities may always be constructed by means of straight lines, and radical quanti- ties of the second degree may be constructed by means of the circle and straight line united. As to radical quantities of higher degrees, their construc- tion depends upon the combination of different curved lines. We will now proceed to the consideration of questions, the solution of which depends either upon rational quantities or radical quantities of the second degree. Geometrical questions, and modes of forming equations there-~— from, and their solutions. 6. The precepts usually given in algebra for putting questions into equations, are equally applicable to questions in geometry. Here, also, the thing sought is to be represented by some sym- _ bol ; and the equation is to be constructed in such manner, as to express the relations of the quantity represented by such symbol, in quantities that are known, or in those whose values are attainable ; and the reasoning is to be conducted by the aid of this symbol, and of those which represent the other quantities, algebraically, as if the whole were known, and we were proceeding to verify it; this method of proceeding is called analysis. Although in expressing geometrical questions by algebraic equations, we have more resources and more facilities accord- ing as we are acquainted with a greater number of the pro- perties of lines, surfaces, &c., still, as algebra itself furnishes the means of discovering these properties, the number of pro- positions really necessary is very limited. The two proposi- tions that similar triangles have their homologous sides propor- tional ; and, that, in a right angled triangle the square of the | 15 “ * * PI 166 APPLICATION OF ae hypothenuse is equivalent to the sum of the squares of the twe other sides, are the fundamental propositions and the basis of the application of algebra to geometry. But there are many ways of making use of these proposi- tions, according to the nature of the question, and there is al- ways a discretion to be exercised in the choice of the means, and manner of applying them; and this discertion can only be acquired by practice. When a geometrical question is to be resolved algebraically, it will be necessary to construct a figure that shall represent the several parts or conditions of the problem under considera- tion, and, if possible, get such expressions for the unknown quantities in terms of those that are known, as may easily be determined, according to the known properties of the figure. But if it so happens, that the required quantity can have no ex- pression which will render it available under the present con- struction, we may, frequently, by drawing lines having certain relations to the known parts of the figure and also to the un- known, so connect the known to those that are required, as to get available expressions for their values. Having proposed a figure as above, we may, by means of the proper geometri- cal theorems, proceed to make out as many independent equa- tions as there are unknown quantities ; and the resolutions of these will give the solution. PROBLEM I. The base BC, and the sum of the hypothenuse AB and perpen- dicular AC, of aright angled triangle being given, to determine the triangle. Let BC = 6, and AC = 2, and if AB-+ AC A be represented by s, then will the hypothenuse » AB be represented by s—a. Therefore, by the properties of the right angled triangle (Prop. XXIV, B. 1V, Hl. Geom.) we have ACG?_+ BC? = AB’ Or, Y+B = sv—2 sx + 2’, C B omitting 2? which is common to both sides of the equation, and transposing the other numbers we have, 23x = sx — Or, x= s'—)? which is the value of the perpendicular AC; where s and d may be any numbers whatever, provided s be greater than b. (Prop. X, B. I, £1. Geom.) o'> a4 ' * gh i ALGEBRA TO GEOMETRY. 167 If this quantity is to be constructed, it may be resolved (Art. 2, Hz. 4,) into the form __ (s+) (s—b) ve 2s and constructed as follows. From any point C, draw an in- definite line, CM, and perpendicular thereto another indefinite line, CN, Set off on CM, CB = 2s, and N take also CE = one of the factors of the numerator, as s+, and take CF = the other factor, viz: s—b, draw BEF and also EH parallel thereto, and CH will be the value of x required. In like manner, if the base and the difference of-the hypothenuse M and perpendicular be given, we shall have by putting d for the difference and the other letters b and x as before ; d+ for the hypothenuse. Whence we have, z+ Qde4+- 2 = 8 + 2 /? «o= bd? 2d B E C Which may be constructed as before PROBLEM II. To desribe a square in a given triangle. (By a given triangle is understood a triangle in which the construction is known, viz: one whose sides, angles, altitude, &c., are known.) It will be perceived that this question resolves itself into the determination of some point, G, in the altitude EF, through which a line AB, drawn parallel to HI, shall be equal to GE; we may, therefore, determine in algebraic expression, for AB, and also for GF, and put them equal to each other and we shall have a solution. E A \\. oral \ Be es D Fo I + 168 APPLICATION OF Let us therefore designate the known altitude EF by a, the known base HI by b, and the unknown line, GF by z; then will EG = a—z. Since AB is parallel to HI, we shall have BP: EG :: FP: GB:: El: BB:: HI: AB consequently, EF: EG:: HI: AB Or, a:a—x::b: AB, whence AB= Gesaes But, AB, GF i= .2, therefore, ap as e2 and ab — be = ax Or, ab = ax + ber = (a + Db) g, h | Ug b In order to construct this quantity, it is necessary to find a fourth proportional to a+), b, and a (Art. 2,) which may be done as follows : From F to O apply a line FO-= a+4, that is, = EF + HI, and join EO; then, having taken FM = HI = 6, draw MG parallel to EKO, which, by its meeting with EF, gives the de- termination of GF, or the value of x; for the similar triangles EFO, GFM, give FO: FM:: FE: FG or, a+b6:b::a:FG ab therefore, FG = aig PROBLEM IIl. Given the base BC, and the angles B and C of the triangle ABC, to determine the altitude AD. (Angles are made to enter into an algebraic expression by the aid of lines employed in trigonometry, viz, sines, tangents, &e. Thus when it is said that an angle is given, it may be understood that the value of its sine or tangent is given.) _If we designate BC by a, and AB by y, we shall have CD: AD:: radius : tan. ACD, (Trigonometry) or if we designate the radius by r, au the tangent of the angle C by ¢, we have, syssrit ALGEBRA TO GEOMETRY. 169 whence CD = ao . In like manner designating the tangent Co ae of the angle B by ¢’ we shall have, | nS ieee B . Drags r'ot whence BD = a Oe. But, CD+BD=BC=a therefore, a + 2 = ¢ whence by changing the construction we may present at t! | ; I FU + rt This expression is susceptible of greater simplicity, by in- troducing, instead of the tangents of the angles C, B, their co- tangents ; which, let us designate by q and q’; observing that the tangent is to radius, as radius to the cotangent, (‘Trigono- metry) and we shall have Liters cans and’ $7262 gee 2 Emcee 7? % whence ¢ = —, and : substituting these values for ¢ and ¢’ in the former equations we have, ar. =. qq’ ar qq ar Schohum | From the above it may be perceived that when among quantities that are given those employed do not lead to results so simple as may be desired, it is not always necessary to commence the work anew in order to arrive at a more simple result; but it may be sufficient to express, by equations, the ratios of the quantities first employed to those which we would introduce, as we have expressed ¢ and ?@’ by the equa- 2 2 r ; tions ¢ = — and?! = — by which a solution dependent upon q and q’ is obtained. PROBLEM IV. Given the three sides of a triangle ABC, to find the segments AD, DC, formed by the perpendicular BD, and also to find the perpendicular BD. 15* i70 APPLICATION OF Let BD = y, CD = x, BC = a, AB= 8, AC = ¢, then AD or AG— CD = c—z. Hence, we have | , C+yY=@and?i—-2er+27+y= 0 Let the second equation be sub- tracted from the first, and we have 9 cre C = a’ — 2D’; whence, we have © | Ga Is a—P+c a—b’ —>= OD C Se = 2c 2c —-+te A which may be resolved into 2= fe b) fae By reference to article 2 it will be perceived that to obtain the value of 2, we have to find a fourth proportional to c, a+b and a— 6, to take one half of this and add it to 4e, or one half the side AC. Scholium. Several important conclusions may be drawn from this solution, some of which we will notice; showing at once, different modes of putting geometrical questions into equations, and how, by varying the propositions of these equations, new propositions may be discovered. | Ist. The equation 2 cz —c? = a? —}? is resolveable into ce (2a-——c) = (a + db) (a—D). Now since the product of the first two factors is equal to the product of the last two, we may consider the first two as the extremes, and the last two as the means of a proportion; hence we have c:a+b::a—b: 2xr—c, or e—(c—x); or,. AC: BC+ AB:: BC—AB:CD—AD 2nd. If from the point C as a centre, and with a radius BC; we describe the arc BO, and draw the chord BO we have BD? + DO? = BO? ; now DO = CO — CD = BC— CD = a—z, therefore BO? = y? + a? —2axr + 2?; but we have found above y? + 22 =aq?; consequently BO? = 2 a? —2 ax = 2a (a—2). ‘ a q? eS fe + c2 Putting for x its value iggy wise since 2 ac — a? — ¢? = — (a? — 2ac +c?) =—(c —a)!, we shall have $2 ~— q?-—— p2 eines dy et 8 2 BO: =2a(a+————) =2a(@e— ee +t ZC 2c Sauer” ALGEBRA TO GEOMETRY. «171 = < (0 — (c — ay), Now, by considering c— a as a single quantity, we find bY — (c—a)! = (b+ c—a) (b—c +a), hence BO’ = = 6+ e—a) @—c +), which may be put under this form, BO = — (a+b +e—2a) (a+b +c—20). If, therefore, we designate the sum of the three sides by 2s, we shall have BO? = — (2s—2a) (2s—2ce)= 4— (s— a) (s—c). Letting fall from the point C upon OB the perpendicular Cl, we obtain from the right angled triangle CIO this proportion, CO: aes oR: sin. ed pat peonmne ey that is, 1 : + BO: -R:.sin. OCI; whence 141 BO = dows ol, or BO = iensiny OEE : R R 4 a (sin. OCI)? consequently BO? = iat ese Putting these two valves of BO* equal to each other we have 4a’ (sin. OCI 4a = 8) 0), or, dividing by 4 a, and fan the denominators to disappear, ac Gin. OCI)’ = R*(s—a) (s—e): that is, dividing by ac, putting R equal to 1, and extracting the square root, sin. OCL = / (s—@ S—s) ac which agrees with a formula, in Trig. 3d. We may, from the equation y? + 2? = a?, deduce the following: y? = a? —x? = (4+ 2) (a—2), putting for x its value, as found in the problem we have a—b+ec b— qa?’ y= (ct Qe ) (a+ Qe ) = (Soe eS) eee aes) oe Qc oy Sey Rl SE (a + c) ver) é oa) ={ 2c 2c 172 APPLICATION OF r & tie 30) (ai+ saat) (” +c—a) (b—c+ 2) : i 2c 2c ; consequently, 4 c2y? = (a+¢+4+ 6) (a+c—b) (6+c¢—a) (b6—c+a) =(a+6+c) (a+b +ce— 2b) (atb+ce— 2a (a+b+e — 2c) ; or, designating the sum of the three sides a + 6 + ¢ by 2s, 4 c?y2=2 s (2s —2b) (2s —2a) (2s—2e) = 16 s (s—b) (s—c) (s—a), or, dividing by 16 and taking the square root, cy § qe Vy @ an.) et ee 3 = Vs (s—8) (s—e) s—a) But ~, or ees is the surface of the triangle ABC. Hence, to find the surface of a triangle by means of the three sides, we must subtract each side successively from the half sum, multiply the half sum and the three remainders continually to- gether, and take the square root of this product; which agrees with Prop. XL, B. IV, El. Geom. 4th. The equations 2 cx —c? = a? —b? may be resolved as follows, b? = a? +c? —2 cx but if the perpendicular fall without the tri- 8 angle as in the present diagram, AD will © then be c + 2 instead of c — x, hence, desig- nating the sides as before, we have y? + x? = a’, and y? + c? + 2.cx+a?=0?, the first 5 subtracted from the second gives c? + 2ca = - = 6? —a?, or c (c + 2x) = (b+ a) (b—a); whence, c:b+a::b—a:ct 2 Now, c+2z,ore+c+a=CD+AD; consequently, AC: AB+ BC:: AB—BC:CD+ AD 5th. The same equation c? + 2 cx = 6? —a?, may also be put under the following form b? = a? + c? + 2 cx, which answers to the last figure. comparing this with the equation, b? = a? + 2c —2 cx, which answers to the former fig- ure, we observe that b?the square of the side AB opposite to the acute angle C, is less than the sum of the squares of the other two sides a? +c? by 2 cx; on the contrary, the square of the side AB opposite the obtuse angle, (see last figure,) is equal to a? + c? + 2 ca, that is, greater than the sum of the squares of the other two sides, by 2cx, which agrees with propositions XX VI and XX VI, B. IV, El. Geom; by these propositions we may determine when the angles of a triangle ALGEBRA TO GEOMETRY. 173 are to be calculated by means of the sides, whether the angle sought, be acute, or obtuse. 6th. The two equations 62 = a? + c? —2 cx and 63 = a§ +07 +2 cx confirms the theory of positive and negative quantities, for it is plain that the segment CD takes different directions, accord- ing as the perpendicular BD falls within the triangle or with- out it. In these two equations the term 2 cz has, in fact, con- trary signs. Hence, whatever result we obtain with regard to one of these triangles, we obtain that which belongs to the analogous case of the other by merely changing its sign of that part which takes a different direction on the same line. Now, since in the above theorem, respecting the surface of a triangle the segment CD does not come into consideration ; therefore, the proposition is equally applicable to all kinds of plane triangles. PROBLEM. V. Having the lengths of the three perpendiculars, EF, EI, EH, drawn from a certain point EK, within an equilateral trian- gle ABC, to its three sides, to determine the sides. Draw the perpendicular AD, and having ri joined EA, EB, and EC, put EF=a EI=4, EH=c, and BD (which is 3BC)=z. Then, since AB, BC, or CA, are each=2z, we shall have, Prop. XXIV. B. lV. El. Geom, K AD= vy (AB’ — BD’)= vy (42° — 2’) = 32? rma af che And because the area of any plane triangle 8 a ial is equal to half the rectangle of its base and perpendicular, it follows that triangle ABC=1BC x AD=2xzy3=2' y3, BEC=iBCXEF=rxa. =aa, AEC=3ACxEI=«xb =r, - ALB=$ AB Mig X given, we draw through the centre O the straight line AOC, the line AB is L to be considered as known, and consequently the line AC. In order to know how the line AE is to be drawn, we have only to determine what ought to be the magnitude of AD, that, when produced, the part DE should be equal to the given line. We will designate AD by z, AB by a, AC by 0, and the given line, to which DE is to be made equal, by c. Since the figure BDEC is a circle, the secants AC, AE, must be reciprocally proportional to the parts without the cir- cle ; that is, AC: AE:: AD: AB (Prop. XXXVI. B. IV. El. Geom.),or b:a+ce:: @© : a; whence 4 cxr = 2b; an equation of the second degree, which, being resolved, gives r= — 1 c£EV1 c+ab. of which the first value only, —ic+ Vici +ab, satisfies the question under consideration. In order to finish the solution, it is necessary to construct this quantity, which can be done without employing the trans- formations made known, art. 2. For this purpose, we draw from the point A the tangent AT, which, being a mean pro- portional between AB and AC, gives AT’=ab; the value of x therefore becomes | g=—1e+ViF+AT, | The radius TO being drawn, becomes a perpendicular at AT; ALGEBRA TO GEOMETRY. 175 if then we take TI equal to 1c, by drawing AI, we shall have Al=Vv? c’+AT"; therefore, in order to obtain x, we have only to apply TI from I to R, and to describe from the point A. as a centre, and with the radius AR, the are RD, which will determine D, the peint sought; for Kutt AD, or AR=AI—IR=AI — TI=V1 7+AT?— ! cmz In order now to know what the second value of = signifies, namely, 3a g=—1c— v1 ¢?+ab, it must be observed that, as it is wholly negative, it can only fall in the direction opposite to that toward which AD tends. Let us see, then, if there be a question depending upon the same quantities and the same reasoning, which fulfils this con- dition. If now we suppose a and 0) negative, the equation x°+cxr=ab, undergoes no change; since, therefore, when the circle BDEC becomes B’D’E’C’, situated toward the left in the same manner that BDEC is toward the right, it follows that the solution of this case is contained in the same equation; the second value of z, or—}c~—VvV1ic’+ ab, belongs to the same case, and satisfies the same conditions; if, therefore, in the preceding construction, we apply IT from I to R’ on AI produced, and from the point A, as a centre, and with a ra- dius equal to AR’, we describe an arc cutting the circumfer- ence B’D’'E’C’ in E’, the point E’ will be such that the part intercepted, E’D’, will be equal to c. Indeed, AK/=AR’'=AI+IR’= V1 77+AT’+2 6, that is, AE’ is equal to the second value of z, the signs being changed. Now, since we apply this quantity in a direction opposite to that in which z extends, it follows that AE’ is in reality the second value of z. Hence, as the two circles are equal and situated in the same manner, the two solutions may both belong to the same circle, so that if we describe from the point A, as a centre, and with a radius AR’, the arc R’E, the line AE will also resolve the question ; indeed, it is evident that the point E, determined in this manner, is in the line AD, (obtained by the first construc- tion,) produced. But of the two solutions, furnished by alge- bra, the first falls on the right of the point A, and appertains to the point D of the convex circumference, while the second falls on the left, and appertains to the point E’ of the concave part of the circumference. a > 4 Fe, “ 176 APPLICATION OF PROBLEM VII. Let it be required to find the direction of a given line AB from a point C, such, that its distance from the point A, shall be a mean proportional between its distance from the point B and the whole line. Let the given line AB be designated by a, and the dis- tance AC required by z; then BC will be a—zx; and, since the proportion required G ae is AB+s AC.:: AG.: CB, or EEG Me segaee sete eae we shall have m=O ON) OV LOLA an equation of the second degree, which, being resolved, gives r= —Latviadt+a’ In order to construct the first value of z, we must, accord- ing to what has been said, (Art. 5,) raise the point B the per- pendicular BD=} a; and, having drawn AD, we shall have AD=VBD?+AB?= V2 a’+a’; we have then only to subtract from this line the quantity 1 a, which is done by applying BD from D to O; then we shall have AO=V1 a?+a?—} a, that is, it will be equal toz. We then apply AO from A to C toward B, and C will be the point sought. As to the second value of z, namely, r= —1La—V1a@ +a’, ¥ we apply BD from D to O' on AD produced, then we shall ave AO’'=1 a+ V1 a’+a? ; and, as the value of z is this quantity taken negatively, we ap- ply AO’ from A to C’ on AB produced in a direction opposite to that toward which z is supposed in the solution to extend ; and we shall have a second point C’, which will also be such, that its distance from the point A, will be a mean proportional between its distance from the point B and the whole line AB. Scholium. 1. We may observe that this question contains that of dividing a line in extreme and mean ratio; also the - construction which we have obtained, is the same as that given in the Elements of Geometry, (Prop. IV. B. IV.) But it will be perceived, that we are made acquainted with this construction by algebra, whereas in the Elements of Geometry we supposed the construction, and only demonstrated its truth. wg, “ » - 5 ALGEBRA TO GEOMETRY. 177 2. With a little attention to the course pursued in the pre- ceding questions, it will be evident that we have always taken for the unknown quantity a line, which being once known, serves, by observing the conditions of the question, to deter- mine all the others. This is the course to be pursued in all cases, but there is a choice with regard to the line to be used; there are often several, each of which has the property of de- termining all the others, if once known. Among these some would lead to more simple equations than others. The follow- ing rule is given to aid in such cases. 3. If among the lines or quantities, which would, when taken each for the unknown quantity, serve to determine all the other quantities, there are two which would in the same way answer this purpose, and it would be foreseen that such would lead to the same equation, (the signs + and — excepted); then we ought to employ neither of these, but take for the unknown quantity one which depends equally on both ; that is, their half sum, or their half difference, or a mean proportional between them, or &c., and we shall always arrive at an equation more simple than by employing either the one or the other. 4. The question we have resolved, (Prob. VI,) may be used to illustrate what is here said. In this question there is no reason for taking AD rather than AE, for the unknown quanti- ty; by taking AD for the unknown quantity z, we have z-+c for AE; and, by taking AE for the unknown quantity z, we - should have «—c for AD; and, as to the rest, the mode of proceeding is the same for each case; so that the equations differ only in the signs. If, therefore, instead of taking either for the unknown quantity, we take their half sum, and desig- nate it by 2, since their half difference DE=c is given, we shall have | AE=2+} c, and AD=z—} ca, whence, according to the proposition adopted in the first so- lution, (c+3 6) @—2¢)=ab or x — Fea, a more simple equation than the former, and which gives z=V1 c+ab; and, since AE=z+}-c, we have immediately , AE=1c+v! c?+ab, and AD=—ict yi @+ab, as before found. 178 APPLICATION OF PROBLEM VIII. Let it be required to draw a right line BFE from one of the an- gles B of a given square BC, so that the part FE intercepted by DE and DC, shall be of a given length. Draw EG perpendicular A D to BE to meet BC produced in G, and from the angle E draw EH perpendicular to BG. Let BC or DC=a,FE=8, BF=y, and CG=z. Since the triangle EHG is similar to the triangle BCF 8 4 Mi and the side EH=the side BC, hence the hypothenuse EG= the hypothenuse BF. But BE’?+EG’=BG,, or 2y°+2bytVH=a+2ae+2° and because the triangle BCF and BEG are similar, BF: BC:: BG: BE or yah: aol: Fb hence, y+by=a+axr multiplying this equation by 2, we have 2y?+2by=2a’+ 2ax. Subtracting the last from the former equation, we have E | | b= —a’?+a", or Y+a’=2z’, hence, r= V+? having the value of z, y may be found in the equation y?+by=a+ar completing the square y’+by+jb=a’+azr+ jb hence y= Va +ax+ib—hb Let DE=z, then ieee | We a “and yz=ab ab hence, | is y Scholium. This problem is susceptible of several modes of solution, but perhaps none more simple than the one here given, for most of the modes of which it is susceptible, involve powers and equations higher than quadratics. ig ALGEBRA TO GEOMETRY. 179 EXAMPLES FOR PRACTICE. Ex. 1. To find the side of a square, inscribed in a given ici hose diameter is d. 1 semicircle, w Fees lays Ex. 2. To find the side of an equilateral triangle inscribed in a circle whose diameter is d; and that of another circum- scribed about the same circle. Ans. id./3, and dV3 Ex. 3. To find the sides of a rectangle, the perimeter of which shall be equal to that of a square, whose side is a, and its area half that of a square. Ans. a+ia/2 and a—tla/2 Ex. 4. Having given the perimeter (12) of a rhombus, and the sum (8) of its two diagonals, to find the diagonals. Ans.1/2 + /54+/2 and 4—-v2 Ex, 5. Required the area of a right angled triangle, whose hypothenuse is z** and the base and perpendicular x?” and 2’, Ans. 1.029085 Ex. 6. Having given the two contiguous sides (a, 6) of a pa- rallelogram, and one of its diagonals (d), to find the other di- agonal. Ans. /(2a°+2b? — d’) Ex. 7. Given the base (194) of a plane triangle, the line that bisects the vertical angle (66), and the diameter (200) of the circumscribing circle, to find the other two sides. Ans. 81.86587 and 157, 48865 Ex. 8. The lengths of two lines that bisect the acute angles of a right angled plane triangle, being 40 and 50 respectively, it is required to determine the three sides of the triangle. Ans. 35.80737, 47.40728, and 59.41143 Ex. 9. Given the hypothenuse (10) of a right angled trian- gle, and the difference of two lines drawn from its extremities to the centre of the inscribed circle (2), to determine the base and perpendicular. Ans. 8.08004 and 5.87447 Ex. 10. Having given the lengths (a, 6) of two chords, cut- ting each other at right angles, in a circle, and the distance (c) of their point of intersection from the centre, to determine the diameter of a circle. Ane 1(a?-L°) + 2c? Ex. 11. Two trees, standing on a horizontal plane, are 120 feet asunder ; the height of the highest of which is 100 feet, and that of the shortest 80; where in the plane must a per- son place himself, so that his distance from the top of each tree, and the distance of the tops themselves, shall be all equal to each other ? Ans. 20./21 feet from the bottom of the shortest and 40,/3 feet from the bottom of the other 180 APPLICATION OF Ex. 12, Having given the sides of a trapezium, inscribed in a circle, equal to 6, 4, 5, and 8, respectively, to determine the diameter of the circle. Ans. 35 / (130X138) or 6.574572 Ex. 13. Supposing the town 4 to be 30 miles from 8, 8 25 miles from c, and c 20 miles from a; where must a,house be erected that it shall be at an equal distance from each of them? Ans. 15.118578 miles from each, viz :- in the centre of a circle whose circumference passes through each of the three towns. | eu Ex. 14. In a plane triangle, having given the perpendicular — (p), and the radii (7 R) of its inscribed and circumscribing cir- cles, to determine the triangle. Qr / (2pr—4rR—7" Ans. The base rab ed me p—2r DETERMINATION. OF ALGEBRAICAL EXPRESSIONS FOR SUR- FACES AND SOLIDS. 7. We have seen in the Hlements of Geometry, that surfaces depend upon the product of two dimensions, and solids upon the product of three dimensions; so that, if the several di- mensions of one or two solids, or two surfaces, which we would compare, have to the several dimensions of the other, each the same ratio, the two surfaces will be to each other as the squares, and the two solids as the cubes, of the homolo- gous dimensions ; and more generally still, if any two quanti- ties of the same nature are expressed each by the same num- ber of factors, and if the several factors of the one have to the several factors of the other, each the same ratio, the two quantities will be to each other as their homologous factors, raised to a power whose exponent is equal to the number of factors. If, for example, the two quantities were a 0 c d, a’ b’ c' d', and we had Gamt 0 = Ps aCe £ oe then we should have a'b a'c a’ d y= — cc = —, d' = —, a a a and consequently, a’*becd abed:dbeid::abcd: ban? a’ 4 a : ae a’ ss gi What is here said is true not only of simple quantities; the ——~ |=. . s 7 g~s = “s - ’ J ALGEBRA TO GEOMETRY. 161 same may be shown with respect to compound quantities. Let the quantities whose dimensions are proportional be ab +cd,a’bl' -c'.d'; since, by supposition, ) esa ost: ose B:dzd, we shall have t / ee ee / a a and consequently mh 12 ab+cd:iad bl te'd::ab+ed:~— + ves 42 b 12 1p acy ee ::@(ab+cd):a"(ab+ cd), ° a : a’, It follows, from what is here proved, that the surfaces of similar figures are as the squares of their homologous dimen- ~ sions, and that the solidities of similar solids are as the cubes of their homologous dimensions ; for, whatever these figures and these solids may be, the former may always be considered as composed of similar triangles, having their altitudes and bases proportional, (Prop. XXIII. B. IV. Hl. Geom.,) and the latter as composed of similar pyramids, having their three di- ~ mensions also proportional. (Prop. XXXII, Cor. 5, B. Il, £2. Sol. Geom.) It will hence be perceived, that quantities may be readily compared, when they are expressed algebraically ; and this may be done, whether the quantities be of the same or of a different species, as a cone and a sphere, a prism and a cylin- der, provided only that they are of the same nature, that is, both solids, or both surfaces. Let it be required to investigate the properties of a pyramid and and also of a frustum of a pyramid, Let h = the altitude, s = the greater base and s’ the smaller base, and h’ = the altitude of the vertical pyramid taken from the top of the frustum, then we shall have /s’: /s::h':h + h' or the altitude of the whole pyramid, and consequently, A+R) vi =h Vs=h vs! + hi vs' and hV/s—h' ~/se=hys dividing by Ys— vs’ 16* » i82 APPLICATION OF we have h'= é le, J/Si— JSS whence h’ becomes known. Let now h + h’ be represented by k, and we may have SD hus from the first equation ie a then we shall have for the solidity of the whole pyramid-=*(1) 1 hi the solidity of the small pyramid “ substituting for k its value, we have for the solidity of the hol 7 bal whole pyramid =~ asi, oigeti Bd sta ae h' 78/8 h' sfs—s' Js! NED J s! s') or 3 ( Js’ ) mpl SF thay ae putting for h’ its value found above we have, kif} y 8s J/s—s'V/s! : 3 (/s— Vs’) vs which being reduced gives, ; j hence the solidity of the frustum will be t 3 (s + Vss' +s’) - - - += - - = (3) that is, the solidity of the frustum is equal to the sum of the greater base, the smaller base, and a mean proportional be- tween the two bases, multiplied by the altitude of the frustum ; which agrees with the proposition in geometry. And if the two bases are equal, viz: if s = s’ then the so- lid becomes a prism, and the expression will become Sis +s +s) or 4h (88) =hs - =i saga) that is, the solidity of the prism is equal to its base multiplied by its altitude. Let the lateral surface of the pyramid be used as an ele- ment in its investigation. To find the lateral surface of a frustum of a regular pyra- mid, having the two bases and slant height given, as well as the radius of the circle inscribed in the larger base. Let the larger base be called s, the smaller s’, the slant height h, and the radius of the circle inscribed in the larger base, r. The perimeter of the larger base will be = and the peri- eT ALGEBRA TO GEOMETRY. 183 meter of the smaller base may be found from the following tee ne Neritacine patna, anailae VS: VS 3: tr ip vs e€ perimeter oO € smaller base : whence (= + : ) h = the lateral surface. aged ates Or we may investigate the surface of the frustum in con- nection with the whole pyramid of which it is a part. ssvs' hs : Thus ep es h: TF the slant height of the whole pyramid, which make = f, and the vertical pyramid cut from the frustum will be k —h. k Hence, we have 7's. Bes cases — the lateral surface of the whole pyramid. - - - - - - - - + = = - (5) And since the lateral surfaces of similar pyramids are pro- sk stk portional to their bases, we may make s: s’:: ah — the sur- face of the vertical pyramid cut from the frustum. Ske Site Hence aes te dee Coc os pans 9; or (s— s’) the difference of the two bases, multiplied by . the ratio of the slant height of the pyramid to the radius of the base, is equal to the lateral surface of the frustum. Mee It may be observed that the ratio ~ 1s constant whether ap- plied to the whole pyramid, to the pyramid cut off, or to the frustum ; and is such as would be represented by the sine of the angle formed by its slant side with the plane of the base. Cor. Whence we have for the lateral surface of the frustum of a pyramid, this rule: Multiply the difference of the bases by the sine of the angle which the slant side makes with the base, or by the ratio of the whole slant height of a perfect pyramid on the same base to the radius of the base, which will give the lateral surface. 8. If x represent the ratio of the circumference of a circle to the diameter, a ratio whichis known with sufficient accu- racy for practical purposes (Prop. XIX. B. V. El. Geom.,) the circumference of any circle whose radius is 7, will be 277 - (1,) and its surface rr’ - - - - - - (2.) Hence it is evident that the areas of circles increase as the squares of their radii, r being always of the same value, the quantity *r’ depends on, and js proportional tor’ - (3.) 8 184 APPLICATION OF If h be the altitude of a cylinder the, radius of whose base is r, for its convex surface we shall have trh_ - - (4) for its solidity 77r*h - - - - - . and for the same reason we shall have mr”h’ for the solidity of another cylinder, whose altidude is h’, and the radius of whose base is 7’, hence the solidities of the two _eylinders are to each other as the altitudes multiplied by the squares of the radii of the bases. If their altitudes and radii of their bases are proportional, in which case the cylinders will be similar, we shall neve) hte hiieun Sa" he consequently =h’ < and the ratio PAT h rh:rh becomes oe or multiplying by x and dividing by 4, r* - - 6,) that is the solidities are as the cubes of the vail of their bases, as before shown in geometry. Also if the altitude of a cone is A, andr the radius of the base, its convex surface may be expressed by Cae gene AT [Vr + hig KG SPE OK BS ss ss (7) Or let k = the slant height of the cone, then will its lateral surface be rr Xk = rkr - - - (8 which result will be also rb ee Tee we take © ar the area of the base, and increase it in the ratio ofr: k, viz. rhe or rkr rsksiaqr’: The solidity of the cone will be h_ «rh r cr X 3 3 - : - - - - (9) or if we multiply the convex surface of the cone by one-third of its distance from the centre of the base, (Prop. IX. B. IIL. El. S. Geum.) we shall obtain the same result. The distance from the centre of the base to the surface may h be expressed k:h:: r “the distance, - - (10.) hence the solidity will also be rh «rh rkrX raha as before. If h' = the altitude of a frustum of the cone, then may the h! slant height of the frustum be h : k : h’ ae the slant height ALGEBRA TO GEOMETRY. 185 required, which call k’ ; let the radius of the pei base of re+2r! the frustum be 7’, then will the lateral surface he ax k! =rhiatrkig 0. Z ¥ - (Er 2” = the greater base, and rr” may stelitoe base » (Prop. a B. Ill. Hi. S. Geom.) the solidity == (rr ar? + J gpp)-(12) 2h! iad ( And since _ and ~ : cone on one or the other of the bases, and whose altitude is equal to that of the frustum, hence if one of those expressions is taken from that of the frustum, the remainder will express a conesected frustum. *h. Thus 1h (#r?-+ar?+ J apy") cae 1 aa = har? + Jerr?) - - (13) which i is a conesected frustum, eee a conical cavity on its larger base and 3 h (t1°+ Vay?) - - - > (14) expresses a eer da frustum the cavity of which is formed on the smaller base. Or if we multiply the convex surface by } its distance from the centre of either base, we shall have the solidity of a cone- sected frustum, whose cavity is taken from the opposite base (Prop. XI, Cor. B. II, El. Sol. Geom.) Thus rkr + r'k'a, formula (11) the expression of the lateral each expresses the solidity of a h surface, multiplied formula (10,) the distance of the surface from the centre of the larger base gives rhe + rvhkix (15) When the two bases of the frustum are equal, the coinsected frustum becomes a conesected cylinder, and r and 7’, and h’ becomes identical. Hence the uae es becomes = 29 ke | - - - - - (16) where & represents the altitude. 9. Applying the same notation to express the sphere, we have for the surface of any sphere whose radius is r, 4rr° ; and 4nr’X1r=—4ar'’, will be its Bona GErop: SL Bal Damiee Sol. Geom.) - - - - “pe apd) If the surface of a saonet zone is required, it may be ex- pressed by the product of the altitude of the zone multiplied by the circumference of the sphere; let h=the altitude, and we have 2arh for the spherical surface of the zone, - - (2.) The solidity of the sector of which this is the jodie base, is QerhXir=20crh, - - (3) Now, since ‘the sector CBAD Ges the diagt am to the Prob- lem on the 187 page,) may be considered to be made 186 APPLICATION OF up of a cone CBD and a segment BDA; in order to express these portions, it will be necessary to find the area of the cir- cular section BD; for this purpose, since CP=CA — AP =r—h, and CB=r, we have in the right angled-triangle BPC, BP= V CB? — PC?= V7? — r’°+2rh—h?= V 2rh —h?, - (4.) The radius of the circular section BPD, hence the area of the circle will be 2rrh — ch’, - - n Yae S GR And the solidity of the cone vill te (2erh — ah) Ae -h _ 2r *h ab h+ch - (6) Hence the solidity of the segment will be enth __Qer Be wh: 2, pa _ (0) Which may be resolved into rh?(r —h,) - - (8.) Hence, the solidity of the segment is equal to the es otek of a circle, whose radius is the altitude of the segment multiplied by the radius of the sphere, minus a third of this altitude. 10. Let 7’ be the radius of a sphere inscribed in a vertical polyedroid, and r the radius of the circumscribed sphere. Then (Prop. XVIII. B. III, Ei. Sol. Geom.) the surface of the polyedroid will be 2r’"X2r=4r'rr, -- - - - - - = (1) And since the surface of the circumscribed sphere is=4r’x, formula (1, Art. 9,) it follows that the surface of the polyedroid and that of its circumscribed sphere, are to each other as 7’: r, since those are the only variable quantities which enter into their expressions. If h be the height of any zone of the sphere, its surface, formula (2, Art. 9,) will be 27’7xh=2rhr, - - - - (2) The surface of the corresponding zone of the polved old. will be 27/7 x h=2r'ha, (P. XVII. Cor. B. Il. HLS. G) - (3.) And hence the surface of any zone or segments of the po- lyedroid and sphere made by the same perpendicular to their common axis, will be in the ratio of their whole surface, viz. 7: TSP AMRE phils ty RS Ht TA ae oh HoeziqarR MA) The base of any segment of the polyedroid made by a plane passing through a circle common to the polyedroid and sphere, may be found from formula (5, Art. 9.) Let 2rrh — xh’, be the base of acommon segment of the sphere and polyedroid, and h the common altitude of the polyedroidal and spherical sector. The solidity of the ee ea tat sector will be Qrlhax— = = gr he a) oR! sts es (5.) And since the corresponding Ae sector is 17°hr, for- mula (3, Art. 9) it follows that the solidities of the corresponding ALGEBRA TO GEOMETRY. 187 sector of these solids are as the ratio of: r’, as has been shown with regard to the convex surface. ‘The cone whose base is the base of the segment, and whose vertice is the centre of the polyedroid or sphere,formula (6, Art. 9,) may be expressed Q077*h — 38erh?+rh* — hence the segment will be “ar! "he —2arh+rri?—ch* | e=aah(ar er’ trh— hy ~~ - - = (&) If the polyedroidal segments consists only of a vertical cone, its solidity will be 2%rh — rh’) xX th=acrh’? —} rh’, subtract this _ from the spherical segment on the same base, formula (7, Art. 9,) arh?— nh’, and we have }rrh’?. - - - - - - = - (7%) ~ which is the value of that portion of the segment of the sphere not included in that of the inscribed polyedroid, which is such a portion of the sphere as would be generated by the re- ~ volutions of a circular segment as BD, about the axis B DC, passing through the centre of curvature, and per- 0f/ pendicular to the arc of the segment at the point of contact D 2 =! PROBLEM. It is required to find when the spherical segment and the cone composing @ spherical sector are equal to each other. Let ABED represent a sphere gene- rated by the revolution of a semicircle ABE about its diameter AE. The sec- - tor ABC, by this revolution, generates a spherical sector, which is composed of a spherical segment generated by the revolution of a semisegment ABP, and of a cone generated by the revolution of the right-angled triangle BPC. The solidity of the sector, formula (3, Art. 9) will be 247*h. The solidity of the cone, 277°h — arb, formula (6.) Now, in order that the cone may be eh to the segment, the sector, which is the sum of both, must be double the cone: hence, 27r°h=4arh=2rrh*?+2ch', dividing by 2, transposing, &c., 188 APPLICATION OF rrh?—lerh-+idch’ dividing by xh rh=17r741h* transposing 1h'—rh=17r h?—3rh=—r* from which we obtain ds h=2re Vv Er? Of these two solutions it is evident that only h=2r— Vv 47% can satisfy the conditions of the question, since 3r+V4r* is more than 27, or more than the diameter of the sphere. EXAMPLES FOR EXERCISE. 1. What is the solidity of the spherical segments of which the frigid zones are the convex surfaces, the altitude of each segment being 327 miles, and the radius of the base 15'75,28 miles ? Ans. 1282921583 solid miles nearly. 2. What is the solidity of the spherical segments of which the temperate zones are the convex surface, the radius of the superior base being 1575,28 miles, that of the inferior 83628,86 miles, and the altitude 2053,7 miles ? Ans. 55021192817 solid miles nearly. 3. What is the solidity of the spherical segment of which the torrid zone is the convex surface, the radii of the bases be- ing 3628,86 miles, and the altitude 3150,6 ? Ans. 146715018499 solid miles nearly. 4. Having two vats or two tubs in the form of conical frusta, whose dimensions are as follows, viz.: the first has a base whose diameter is 3 feet, its altitude is 31 feet, and the slant height of its side is 4 feet: the diameter of the base of the se- cond is 31 feet, its altitude is 5 feet, and the curve surface is 60 square feet, what must be the dimensions of one capable of containing as much as the other two, if*the diameter of the bottom and top, and the altitude are in the proportion of 2 21 and 3. 5. What is the difference in surface of a vertical hexedroid circumscribing a sphere whose diameter is 10, and the whole surface of a conesected frustum of a cone inscribed in the same sphere, and whose wanting base is 6, and perfect base 4? CONIC SECTIONS. There are three curves, whose properties are extensively applied in mathematical investigations, which, being the sec- tions of a cone made by a plane in different positions, as will be shown in another place, are called the Conic Sections. These are, 1. The Parabola. 2. The Ellipse. 3. The Hypetboth. PARABOLA. DEFINITIONS. j. A Parabola is a plane curve, such, that if from any point in the curve two straight lines be drawn; one to a given fixed point, the other perpendicular to a straight line given in po- sition: these two straight lines will always be equal to one another. 2. The given fixed point is called the focus of the parabola. 3. The straight line given in position, is called the directrix of the parabola. Thus, let QAq be a parabola, S the fo- cus, Nv the directrix ; Take any number of points, P,, P., P;, cae in the curve; Join 8, P,;8,P,;8, P, 3... and draw PN SP NEL PS ING peteoperpendiéuiar’s of) “\ to the directrix; then ia SPUN! (Pls PNY, SPs PE SE 19, £ 4. A straight line drawn perpendicular to the directrix. and cutting the curve, is called a diameter ; and the point in which it cuts the curve is called the vertex of the diumeler. 5. The diameter which passes through the focus is called the axis, and the point in which it cuts the curve is called the principal vertex. 17 190 CONIC SECTION. Thus: draw N,P,W,, N,P,W,, N,P,W,, KASX, through the points n, P,, P,, P,, 8, perpendicular to the di- rectrix ; each of these lines is a dia- meter; P,,.P;, P,, A, are: the- ver-. K tices of these diameters; ASX is the axis of the parabola, A the principal vertex. N; 6. A straight line which meets the 2 curve in any point, but which, when produced both ways, does not cut it, is called a tangent to the curve at that point. 7. A straight line drawn from any point in the curve, par- allel to the tangent at the vertex of any diameter, and termi- nated both ways by the curve, is called an ordinaie to that diameter. 8. The ordinate which passes through the focus, is called the parameter of that diameter. 9. The part of a diameter intercepted between its vertex and the point in which it is intersected by one of its own ordinates, is called the abscissa of the diameter. 10. The part of a diameter intercepted between one of its own ordinates and its intersection with a tangent, at the extre- mity of the ordinate, is called the sub-tangent of the diameter. Thus: let TP¢ be the tangent at P, the vertex of the diameter PW. From any point Q in the curve, draw Qgq parallel to Té and cutting PW in v.. Through § draw RSr parallel to Té. Let QZ, a tangent at Q, cut WP, produced in Z. Then Qq is an ordinate to the di- ameter PW; Rr is the parameter of PW. Pv is the abscissa of PW, corresponding to the point Q. vZ is the sub-tangent of PW, corresponding to the point Q. 11. A stright line drawn from any point in the curve, per- pendicular to the axis, and terminated both ways by the curve, is called an ordinate to the axis. 12. The ordinate to the axis which passes through the focus is called the principal parameter, or latus rectum of the para- bola. 13. The part of the axis intercepted between its vertex and the point in which it is intersected by one of its ordinates, is called the abscissa of the axis. 14. The part of the axis intercepted between one of its own = » all we i ie ‘ ~ PARABOLA. 191 ordinates, and its intersection with a tangent at the extremity of the ordinate, is called the sub-tangent of the azis. Thus: from any point P in the curve draw Pp perpendicular to AX and cutting AX in M. Through $ draw LS/ perpendicular to AX. Let PT, a tangent at P, cut XA ' produced in T. Then, Pp is an ordinate to the axis ; L/ is the latus rectum of the curve. AM is the abscissa of the axis cor- responding to the point P. “ea the subtangent of the axis corresponding to the oint ©. : It will be proved in Prop. III, that the tangent at the princi- pal vertix is perpendicular to the axis; hence, the four last definitions are in reality included in the four which immedi- ately precede them. Cor. It is manifest from Def. 1, that the parts of the curve on each side of the axis are similar and equal, and that every ordinate Pp is bisected by the axis. 15. If a tangent be drawn at any point, and a straight line be drawn from the point of contact perpendicular to it, and terminated by the curve, that straight line is called a normal. 16. The part of the axis intercepted between the intersec- tions of the normal and the ordinate, is called the sub-normal. Thus: Let TP be a tangent at any point P. From P draw PG perpendicular to the tangent, and PM perpendicular to the axis. T Then PG is the normal correspond- ing to the point P; MG is the sub-nor- mal corresponding to the point P. 192 CONIC SECTIONS. PROPOSITION I. THEOREM. The distance of the focus from any point in the curve, is equal to the sum of the abscissa of the axis corresponding to that point, and the distance from the focus to the vertex. That is, SP=AM-+AS. For SP=PN by Def. (1,) =KM *.* NM is a parallelogram. =AM+AK =AM+AS -.: AK=AS, by Def. (1.) PROPOSITION Il. THEOREM. The latus rectum is equal to four times the distance from the focus to the vertex. That is Li=4 AS. For, | j Li=2 LS, Def. (14.) cor =2 LN oe sath ane” AS—AR. PROPOSITION IH. PROBLEM. To draw a tangent to the parabola at any point. Let P be the given point. t Join S, P; draw PN perpendicular to ” the directrix. ) Bisect the angle SPN by the straight % line Tt. Tt is a tangent at the point P. For if Tt be not a tangent, let Tt cut the curve in some other point p. Join 8S, p; draw pn perpendicular to the directrix ; join S, N. CONIC SECTIONS. 193 Since SP=PN, PO common to the triangles SPO, NPO, and angle SPO=angle NPO by construction, “. SO=NO, and angle SOP=angle NOP. Again, since SO=NO, Op common to the triangles SOp, NOp, and angle SOp=angle NOp. ”, Sp=Np. But since p is a point in the curve, and pz is drawn perpen- dicular to the directrix, oe pN=pn. That is, the hypothenuse of a right-angled triangle equal to one of the sides, which is impossible, .*. p is not a point in the curve; and in the same manner it may be proved that no point in the straight line T¢ can be in the curve, except P. .. Tt is a tangent to the curve at P. «Cor. 1. A tangent at the vertex A, is a perpendicular to the axis. Cor. 2. SP=ST, For, since NW is parallel to TX “. angle STP= angle NPT = angle SPT by construction, SP=ST Cor. 3. Let Qq be an ordinate to the diameter PW, cutting SP in z. Then, Pz=Pv t Q For, since Qq is parallel to Té *, angle Prv=angle «PT = NPT by construc- tion, = Pvz interior oppo- T site angle, Pz=Pv Cor. 4. Draw the normal PG, (see diagram Prop. V.) Then, SP=SG, For since angle GPT is a right angle, angle GPT=PGT+PTG=PGT+S8SPT Take away the common angle SPT and there remains angle SPG=angle SGP ROSE E=SG. 17* 194 , PARABOLA. ily PROPOSITION IV. THEOREM. n ee The subtangent to the axis is equal to twice the abscissa. ’ \ % " RP That is, — MT=2 AM For, MT=MS+ST —MS+SP. Prop. IIL cor, 2... << =MS+SA+AM. Prop. I. A\F 2 =2 AM. Cor. MT is bisected in A. PROPOSITION V. THEOREM. The subnormal is equal to one half of the latus rectum. That is, MG = Z 5 we denote the latus rectum by L. For, MG sc. —SM a SP—SM. Prop. III, cor. 4. AS + AM—SM. Prop. I. S+ AS + SM—SM AS el tl i Wl A 2 L 2 Pr op. Il. PROPOSITION VI. THEOREM. If a straight line be drawn from the focus perpendicular to the tangent at any point, it will be a mean proportional between the distance from the focus to that point, and the distance from the focus to the vertex. That is, if SY be a perpendicular let fall from S upon Tt the tangent at any point P SP.; oh Ae fos boo. Join A, Y. Pt Since SP= ST, and SP is drawn per- yu pendicular to the line PT, le Yr. Also by per A ay A\S ™M A=AM -- Since AY an the sides of a tri- angle TPM proportionally ; AY is paral- le] to MP, ~" > se > we ee ee PARABOLA. ‘ 198 AY i is perpendicular to AM. * he the triangles SYA, SYT, are similar, BST : SY 3 RP SSA . or, SP:SY::SY:SA .. SP=ST by Prop. III, cor. 2. Cor. 1. Multiplying extremes and means, SY*=SP SA Ol rt Cor: 20,6 Pa SA tee 2 SY" Cor. 3. By Cor. 1, SA’ And since SA is constant for the same parabola, SP proportional to SY’. Cor. 4. By Cor. 1., AS .SP SY? - 48Y?=4AS.SP = L.SP. Prop. IL lI ll PROPOSITION VII. THEOREM. The square of any semi-ordinate to the axis is equal to the rectangle under the latus rectum and the abscissa. That is, if P be any point in the curve = L. AM. For, PM? = SP?-——-SM? (Prop. XXIV. B. IV. Ei. by Geom. = (AM=AS)*— (AM — AS)’ *SP=AM+AS(Prop. 1,) & SM=AM—AS , = 4 AS. AM. (Prop. X. and XI, B. IV. °G Mm El. Geum.) aR Prop. I. Cor. 1. Since L is constant for the same parabola PM? proportional to AM, That is, The abscisse are propotional to the squares of the ordinates. PROPOSITION VIII THEOREM. If Qq be an ordinate to the diam- eter PW and Pv, the corresponding abscissa, then, Qu = 45P x. Bo. Draw PM an ordinate to the axis. 'r Join S,Q; and through Q draw DQWN perpendicular to the axis. ” ¥ . f 196 CONIC SECTIONS. From § let fall SY pepibacienlat on the tangent at P. The triangles SPAY, QD», are similar. Qu: QD’: : SP? :S¥? Bcti se : SA, Prop, VI. Cor. 2. The triangles PTM, QD», are also similar ; a Ds SD : MT 23 PM : PM .MT :: 4AS. AM:2PM.AM > 4AS : 2PM 2PM. QD = 4AS. Dv But, — QN’= 4AS . AM—4AS. AN=4AS(AM~— AN) = 4AS.MN And, PM’— QN?= PM+QN) (PM—QN) = (PM+QN) . QD - (PM+QN) .. QD = 4AS. MN = 4AS. DP But, 2PM .QD = 4AS. Dv * (PM—QN). QD = 4AS. Po Or, QD? = 4AS. Pv 4s Qv’? :4AS.Pv::SP:SA. Qv’?.= 48P . Pe: Cor. 1. In like manner it may be proved, that qv = 48P x Pv. Hence, Qu = qv; and since the same may be proved for any ordinate, it follows, that A diameter bisects all its own ordinates. Cor. 2. Let Rr be the parameter to the diameter PW. Then, by Prop. HI. Cor. 3. Pa abe. Ceol bys Now, by the Proposition, RVs 400 wry = 4SP’ ., 4RV’ or Rr’? = 16SP? Rr= 4SP. Hence the proposition may be thus enunciated : The square of the semi-ordinate to any diameter is equal to the rectangle under the parameter and abscissa It will be seen, that Prop. VII. is a particular case of the present proposition. ELLIPSE. DEFINITIONS. 1. Aw Evuirse is a plane curve, such that, if from any point in the curve two straight lines be drawn to two given fixed points, the sum of these straight lines will: always be the same. 2. The two given fixed points are called the foci. Thus, let ABa be an ellipse, S and H the foci. Birks oft Take any number of points in the curve P,, PL, P,,---- Oi” Sie esl ees bay Dek es OA S.P,, H.P, ; - - - - then, a a SP, + HP, = SP, +.HP, = SP, + HP, =---- 3. If a straight line be drawn join- ing the foci and bisected, the point Ps of bisection is called the centre. 4. The distance from the centre to either focus is called the eccentricity. . 5. Any straight line drawn through the centre, and termi- nated both ways by the curve, is called a diameter. 6. The points in which any diameter meets the curve are called the vertices of that diameter. 7. The diameter which passes through the foci is called the axis major, and the points in which it meets the curve are called the principal vertices. 8. The diameter at right angles to the axis major is called the axis minor. | Thus, let ABa be an ellipse, S_and Hi the foci. Join S,H; bisect the straight line SH in C, and produce it to meet at the curve in A and a. Through C draw any straight line Pp, terminated by the curve in the oints P, p. Through C draw Bod at right angles to Aa. Then, C is the centre, CS or CH the eccentricity. Pp is a diameter, P and p its vertices, Aa is the major axis, Bd is the minor axis. 198 CONIC SECTIONS. 9. A straight line which meets the curve in any point, but which, being produced both ways, does not cut it, is called a tangent to the curve at that point. . 10. A. diameter drawn parallel to the tangent at the vertex of any diameter, is called the conjugate diameter to the latter, and the twodiameters are called a pair of conjugate diameters. 11. Any straight line drawn parallel to the tangent at the vertex of any diameter and terminated both ways by the-curve, is called an ordinate to that diameter. 12. The segments into which any diameter is divided by one of its own ordinates are called the abscisse@ of the dia- meter. 13. The ordinate to any diameter, which passes through the focus, is called the parameter of that diameter. Thus, let Pp be any diameter, and Té a tangent at P. Draw the diameter Dd parallel to Tt. Take any point Q in the curve, draw Qq parallel to Tt, cutting Pp in v. Through S draw Rr parallel to Tt Then, Dd is the conjugate diameter to Pp. Qq is the ordinate to the diameter Pp, corresponding to the point Q. Pv, vp are the abscissee of the diameter Pp, corresponding to the point Q. Rr is the parameter of the diameter Pp. 14. Any straight line drawn at right angles to the major axis, and terminated both ways by the curve, is called an ordinate to the axis. 15. The segments into which the major axis is divided by one of its own ordinates are called: the abscisse to the azis. 16. The ordinate to the axis which passes through either focus is called the latus rectum. (It will be proved in Prop. IV., that the tangents at the prin- cipal vertices are perpendicular to the major axis ; hence, de- finitions 14, 15,16, are in reality included in the three which immediately precede them.) 17. Ifa tangent be drawn at the extremity of the latus rec- tum and produced to meet the major axis, and if a straight line be drawn through the point of intersection at right angles to the major axis, the tangent is called the focal tangent, and the straight line the directriz. , ELLIPSE. 199 Thus, from P any point in the curve, draw PMp Risle pas ue to Aa, cutting Aa in M. Through S draw Li perpendicu- a lar to Aa. | Let LT, a tangent at L, cut Aa produced in T. Through T draw Nz perpendicular to Aa. Then, Pp is the ordinate to the axis, corresponding to the oint P. : ! AM, Ma are the abscissx of the axis, corresponding to the point P. L1 is the latus rectum. LT is the focal tangent. Nv is the directrix. 18. A straight line drawn at right angles to a tangent from the point of contact, and terminated by the major axis, is called a normal. The part of the major axis intercepted between the inter- sections of the normal and the ordinate, is called the subnor- mal. Let Tt be a tangent at any point r ‘From P draw PG perpendicular N to Tt meeting Aain G. JIN t From P draw PM perpendicular , i to Aa. ; 4 Then PG is the normal corres- ponding to the point P. ? MG is the subnormal correspond- ing to the point P. PROPOSITION 1. THEOREM. The sum of two straight lines drawn from the faci to any point in the curve is equal to the major axis. That is, if P be any point in the curve. SP + HP = Aa. For SP + HP = AS +AH Ls = AS + SH, And, > Def. 1. SP + HP = aS + aH : = 2aH + SH, > o .. 2(SP+HP) = 2 (AS + SH + Ha) ie 200 CONIC SECTIONS. Or, SP + HP = Aa. Cor. The centre bisects the axis 5 Tn for 2AS+SH=2aH+SH. AS = aH - And, SC = CH by definition 3. AC»ra¥rau, Cor. 2. SP + HP =2 AC; "SP = 2 ACHP mP = 2 ACG—SP hence SP — HP = 2 AC .— 2 HF PROPOSITION Il. THEOREM. The centre bisects all diameters. Take any point P in the curve. Jon 8,P ;-H,P38,H4 Complete the parallelogram SPHp Join C.p 3 CP; Then, since the opposite sides of a parallelogram are equal, SP = Hp, HP =Sp..SP+ PH i Sp pu . p isa point in the curve. Again, since the diagonals of paral- lelogram bisect each other, and since SH is bisected in C, *. Pp is a straight line, and a diameter, and is bisected in C. And in like manner, it may be proved that every other di- ameter is bisected in C. PROPOSITION Ill THEOREM. The distance of either focus from the extremily of the axis mi- nor is equal to the semi-axis major. That is, | SB or HB = AC. Since SC = HC, ard CB is com- mon to the two right-angle triangles SCB, HCB, R --SB = HB. But. SB + HB = 2 AC. Prop. I. - SB =\HB =. AC. A a Cor. 1. BC? = AS. Sa. For, 6 BC? = SB* — SC’ . s ie wl. é ELLIPSE: ©, 201 AB? a SCs = (AC + SC). (AS—SC) = ANSdt Scat Cor. 2. The square ‘sor i eccentricity is equal to the differ- ence of the squares of the semi-axes ; For, SC? = SB — BC? = AC’— BC’. PROPOSITION IV. PROBLEM. To draw a tangent to the ellipse at any point. Let P be the given point. Join 8.P ; H,P produce SP. Bisect the exterior angle HPK by the straight line T¢. T?é is a tangent to the curve at P. For, if Tt be not a tangent, let Tt cut the curve in some other point p. Jon S,p; Hyp; make PR= PH, join p, K; H,1S cutting T¢ in Z. Since HP =PK, PZ common to the triangles HPZ, KPZ, and the angie HP Z=angle KPZ by construction, HIZ=K-Z, and the angle HZP =angle KZP. Pit since HZ=-KZ, Zp common to the triangles ees KZp, and angle HZp=angle KZp, pK sspit. But, since any two sides of a triangle are greater than the — third side, | Sp + pK > SK SSP PKR > SP + PH:.: PK=PH by construction: > Sp + pH, by definition 1, ve PAN or pide But we have just proved that pX=pH, which is absurd, .. p is not a point in the curve, and in the same manner it may be proved that no point in the straight line T¢ can be in the curve ees r. . Tt is a tangent to the curve at P. Cor. 1. Hence, tangents at A and a, are perpendicular to the major axis, and tangents at B and bd are perpendicular to the minor axis. 18 202 CONIC SECTIONS. Cor. 2. SP and HP make equal angles with every tangent. Cor. 3. Since HPK, the exterior angle of the triangle SPH, is bisected by the straight line Té, cutting the base SH pro- duced in T * ST2 HT 22 SP :-HP: t K P EL T re Oe a PROPOSITION V. THEOREM. Tangents drawn at the vertices of any diameter are parallel. Let Tt, Ww, be tangents at P,p, the vertices of the diame- ter'P’ Cp. JomuS Pas Hs 25.9% op, Hi: » Then, by Prop. Il, SH isa paral- a lelogram, and since the opposite | 7 angles of parallelograms are equal, “. ang. SPH= angle SpH supplement of ang. SPH=supple- | * S ment of ang. SpH 8 Rae P or, ang. SP T+ang. HPt=ang. SpW+ mck sag, SPT HP ut ang. = ang. t Rs - alae Spans pw | by Prop. IV. Cor. 2. Hence, these four angles are all equal, ss .. ang. SPT=ang. Hpw. And since SP is parallel to Hp, ang. SPp = ang. PpH, -. whole ang. TPp = whole ang. wpP, and they are alternate angles, —f Wem, .. Tt is parallel to Ww. Cor. Hence, if tangents be drawn at the vertices of any ‘wo diameters, they will form a parallelogram circumscribing he ellipse. ee PY a ee are 7 a ELLIPSE. ~ 203 PROPOSITION VI. THEOREM. If straight lines be drawn from the foci to a vertex of any di ameter, the distance from the vertex to the insertion of the con- jugate diameter, with either focal distance, is equal to the semi axis, major. + al That is, if Dd be a diameter conjugate to Pp, cutting SP in E, and HP ine, PT or Pes AC. Draw PF perpendicular to Dd, and HI parallel to Dd or Tt, cutting PF in O, Then, since the angles at 0 are rightangles, the ang. [PO=ang.HPO, 4 and PO common to the two triangles pe sTO MANA ME Celt =etir: Also, since SC = HC, and CE is parallel HI, the base of a triangle SHI, ory kere Pak, Hence, 2PKH=2EKI+2 IP ; = SE + EI 4. IP + HP = SP + HP = 2AC a ACs Also, ang. PEe = ang. Pek. .”. PE = Pe, and Pe= AG, PROPOSITION VII. THEOREM. Perpendiculars, from the foci upon the tangent at any point intersect the tangent in the circumference of a circle, whose diameter is the major axis. From § let fall SY perpendicular on Té a tangent. Join 8, P: H, P; produce HP to meet SY produced in K. Join CY ; Then, since angle SPY =angle KPY (Prop. IV.) and the angles at Y are right angles, and PY common to the a-—< two triangles, SPY, KPY. ° asd SP=PK i 204 CONIC SECTIONS. And S¥¥k, x And, since SY=YK, and HC=CS, CY cuts the sides of the triangle HSK a aw ea ™ . CY is parallel to HK. Also, since CY is parallel to HK, SY=YK, HC=C8, . CY=LHK Hence, a circle inscribed with Centre C and radius CA will pass through Y. And in like manner, if HZ be drawn perpendicular to T%, it may be proved that the same circle will pass through Z also. PROPOSITION VIII. THEOREM. Lhe rectangle, contained by the perpendiculars, from the foci upon the tangent at any point, is equal to the square of the semi-axis, minor. That is, Se 2 HZ=EC’. Let T¢ be a tangent at any point P. On Aa describe a circle cutting Tt € in Y and Z. Join S, Y; H, Z; Then, by the last Prop., SY, HZ are / perpendicular to Tv. i Produce YS to meet the circumfer- ence in y. Jon C, y; C,Z; a Since yYZ isa right angle, the seg- 9 ¥™ ment in which it lies is a semicircle, and 7. =. are stirs ex- tremities of a diameter. “. yCZ is a straight line and a diameter. - Hence the triangles CSy, HCZ, are in every respect equal. 40 BY =HZ CA ee. i= AS. SA=BC’ Prop. III. Cor. 1. ‘a fet ae ~. - sisal ae ey Be PROPOSITION IX. THEOREM. 205 Perpendiculars let fall from the foci, upon the tangent at any point are to each other as the focal distance of the point of contact. aad That is, mSY: HZ::SP: HP. For the triangles SPY, HPZ, are TY nianifestly similar, SP SY.y HA SP AP. Cor. Hence, op end w WF ee i? SY aS Ve. A ap =BC?. as last Prop. =BC 550-80 . So also, oar, is Aga 9 aly Sp eer d wste Sas PROPOSITION X. THEOREM. i} a tangent be applied at any point, and from the same point an ordinate to the axis be drawn, the semi-azxis major is a mean proportional between the distance from the centre to the intersection of the ordinate with the axis, and the distance ,com the centre to the intersection of the tangent with the axis. ‘That is, CT:CA::CA:CM. a Since the exterior angle HPK is bisected by Tt, Prop. IV. ST : HT :: SP: HP. (B. IV. Prop. XVI. Hi. Geom.) ST+HT : ST—HT : : SP+HP: SP—HP 18* ¥ 206 CONIC SECTIONS. or, 2CT :. SH. :: 2AC :SP—HP aT PEED ACK Mes OSH > SPtaps. . (1) But since PM is drawn from the vertex of the triangle SPH perpendicular on base SH, ~, SM+tHM : SP—HP: : SP+HP : SM—HM or, © SH : SPHP, 2s: 2 ee 2 CM oe (2. Comparing this with the proportion marked (1,) we have, Spe 28! cM RE er my Ogre id tars Cae 4} or, CT eos, AG es Se Ms PROPOSITION XI. THEOREM. Lf a circle be described on the major axis of an ellipse, and if any ordinate to this axis be produced to meet the circle, tan- gents drawn to the ellipse and circle, at points in which they are intersected by the ordinate, will cut the major axis in the same point. Let AQa, be a circle described on Aa. Take any point P in the ellipse, draw PM perpendicular to Aa, and produce MP to meet the cir- cle in Q, join C, Q. Draw PT a tangent to theel- \ lipse at P cutting CA produced in T. 9 “~~~ Jon TQ. Then QT is a tangent to the circle at Q. For if TQ be not a tangent, draw QT’ a tangent to Q cut- ting CA in ‘T’. Then CQT” is a right angle. .. Since QM is drawn from the right angle CQT’ perpen- dicular on the hypothenuse. “, CT’: CQ:: CQ: CM. (Prop. XVII, Cor. 2.B. IV. E. G.) or ,CTHeCa 2: CAs CM as27CQ=CA; But, by the last proposition, CT’: €A ::CA SCM, ab Col =OTy, | which is absurd, therefore QT’ is not a tangent at Q; in the same manner it may be proved that no line but QT can bea tangent atQ, .. &e. oe as - > it i i i. : ELLIPSE. 207 Cor. 1. Describe a circle on the minor axis. Draw Pm an ordinate to the mi- nor axis cutting the circle in q. Let a tangent at P cut the mi- * nor axis produced in @. Then, since Pm is parallel to AC, and PM to BC, Cia Gin? CT: aT =A CS eet ::Cq’ :Cm? .-. the triangles CQT, Cmg are similar. Ps, BEE ao peed Ore CB::CB:CM. Which j is analogous to the property proved in the last pro- position for the major axis. Cor. 2. Join tg. We can prove as above, that ¢q is a tangent to the circle Bqb. PROPOSITION XIf. THEOREM. The square of any semt-ordinate to the axis, is to the rectangle under the absciss@, as the square of the semi-axis minor is to the square of the semt-axis major. That is, if P be any point in the curve, PM*SAM.: Mats 2 Berner Describe a circle on Aa, : and produce MP to meet it in Q. / At the point P and Q draw the tangent PT, QT, which« will intersect the axis in the \ same point T, (Prop. XI.) Let the tangent to the el- lipse intersect the circle inY,Z. Joie 3 HA; S¥ and HZ are perpendicular to Tt, (Prop. VIL.) Hence the triangles PMT, SYT, HZT, are similar to each other. oo RM Ss SB Y & t-PA PY and: PAs tar Zee: 2° WO sae . PM Se Zee MOP ave ALZ, or PM's (BO pe :: MIP eo TQta(Prop. XI.andProp. XVL B. IV. El. Geom.) >: QM’? : CQ’? -- MQT, MQC are si- milar triangles. We ee a ee oS *. ¢ ’ * s ) * i 208 ' * CONIC. SECTIONS. . ~-::AM.Ma: AC? *- PM?: AM. Ma : me BOT? AG: Cor. 1 Let P,M ,P,M.,... be ordinates | to the axis from any points P,, P, Then by Prop. Bo Mo2 s AM, Mass. BC Sten. POM,’ : AM, . M,a:: BC? ws Oks (RM? .PM;? : AM, .M,a: AM, . Mga. That is, the square of the or ‘dinates to the axis are to each other as the rectangles of their abscisse. Cor. 2. By the fifth proportion in Frop. PM:QM::BC: AC. Cor. 3. By Prop. PM?: AM. Ma:: BC’: AC’. But as aa Ma=AC—CM, ; (AC+CM) (AC — CM) : : BC’ AC? Pars AC? — CM? Cor. 4. Describe a circle on Bb, draw Pm, an ordinate to the minor axis cutting the circle in g. Then, Pm=CM , PM=Cm. Then by Cor. 3. Te ae ACG? Pmt: AC. 2 Cm : BC? Pm? le TA Gist: BCYCm : BC: crue +Cm) (BC —Cm) : BC? Bm . mb : BC? or, Pm > Bm. Hho AC? : BC? Which is analogous to the Storer proved in the proposition © for the major axis. Cor. 5. Pm: qm:: AC: BC. ae me ee ee ° . *. » - * ELLIPSE. | | 209 PROPOSITION XIII. THEOREM. The latus rectum 1s a third proportional to the axis major and minor. That is, Aa: Bb:: Bb: Ll . B. Since LS is a semi-ordinate to the axis, AC’: BC?:: AS .Sa@: LS’, Prop: XII, 23 -BC*-«ES*, Prop: lly 4 a Cor. 1. AGE BEsesBO “SDS And, Aa:Bb::Bb =: Li. ce hal PROPOSITION XIV. THEOREM. The. area of all the parallelograms, circumscribing an ellipse, formed by drawing tangents at the extremities of two conju- gate diameters, is “constant, each being equal to the rectangle under the axes. Let Pp, Dd, be any two conjugate diameters, SROX a parallelogram circumscrib- ing the ellipse formed by drawing tangents at P, D, p, d; then Pp, Dd, divide ie par- allelograms SROX. into four equal parallelograms. Draw PM, dm, ordinates to the axis; PF perpendicular to Ad. Produce CA to meet PX i in T and Sd in t. Then, CT -GAe: 3 CA «GM And, Ct 3.QAneus GAs Gm fs CT: C¢ :: Cm: CM But, CT : Ct :: TM : Cm, by similar triangles. ve ATS .Cm?t. Cree 2a * OM.MT=Om s - - + 2. Q) ApainnOM.; CA: : CA: CT “ CM: CA:: MA: AT, dividendo. Or, CM: Ma:: MA: MT, componendo. _ AM. Ma=CM . MT—Cm? -.- =. (2) But, AC?! BC*: : AM. Ma (Cm’) : PM’. Prop. XII. +. SSAC BES: ee DMs Similarly, AC : BC :: CM: dm Or, “BC :.dm :: CA: CM 210 CONIC SECTION. But, = COT 2. GAcsatwl Age CM “ag: Cote OU om Bat. PEs: sCT.. 4.dm.-«:Gd, for the triangle CdT=4 the parallelogram CPXd, Shy OO ot hace 4 OY Ue RY cL el Oe | *. rectangle PF . Cd=rectangle AC . BC or, parallelogram CX=rectangle AC . BC .. parallelogram SROX=4 AC. BC =Aa. Bd. Cor. By (2) Cm’?=AM . Ma =(CA+CM)’. (CA—CM) =CA’?—CM? “, CA?=CM?+Cm And similarly. CB’=PM?+dm’. PROPOSITION XV. THEOREM. The sum of the squares of any two conjugate diameters, is equal to the same constant quantity, namely, the sum of the squares of the two azis. That is, If Pp, Dd, be any two conjugate > B diameters, Pp? + Dd? = Aa + Bb’. Draw PM, Dm, ordinates the axis. Then, by Cor. to Prop. XIV, < Mf AC? + BC? = CM? + Cm? + 7 PM? + Dm . ”, = CP? + CD: : *, 4AC?-+4BC?=4CP?-++4CD? Or, Aa?-+Bb?=Pp? +Da?. PROPOSITION XVI. THEOREM. The rectangle under the focal distances of any point is equal to the square of the semi-conjugate. That is. if CD be conjugate to Sb. eae Ot Draw SY. HZ, perpendiculars to the tangent at P, PF perpendicu- lar on CD. Then by similar triangles SPY, Pie ELLIPSE. 211 Pr Sek seer base-SEa Or, SP : S¥ 22 AC : PF’. PE=AG) by Prap, VI ue aa epost : AM.Ma: AC’, -PM?:AM.Ma:: BC? : AC? Cos lL " Let P, M,, P, M,, ...... be ordinates to the axis from any point P, Py Sass Then by Prop. Preise a... Ma: : BC? : AC? P,M,* : AMj@M,a:: BC? ; AC’ Dis Py aan, tes A ee M,a: AM, .M,a, HYPERBOLA. 225 That is, the square of the ordinates to the axis are to each other as the rectangles of their abscisa. Cor. 2. By Prop. | | PM?: AM.Ma_ :: BC? : AC But AM = CM—CA, Ma = CM+CA, ee PVA? CMC A? ih BEF =: AC?. CR? : CA?.: CHR+-Cad? : dP’, since dP? = CD’. Cor. 3. Since by the proposition kat Zp PM: AM: Ma:: BC?: AC’, te | we have in the conjugate hyberbolas #\ | 6 i ow PROPOSITION XIII. THEOREM. The latus rectum is a third proportional to the axis major and minor. That is, Aa > Bb-: : Bo:; Li. Since LS is a semiordinate to the axis. \ AC? : BC?: : AS ..S@: Ls’; Prop. XII. — BC* ef C873 Prop. II. os RAC BCS BO MES or, Aad . Bb. 3: Bb? > Li. PROPOSITION XIV. THEOREM. The area of all parallelograms, formed by drawing tangents at the extremities of two conjugate diameters, is constant, each being equal to the rectangle under the axes. Let Pp, Dd, be any two conjugate diameters, WwXz, a parallelogram inscribed between the opposite and conjugate hyperbolas by drawing tangents at P,p, D,d; then Pp, Dd, divide the parallelogram WzXw into four equal parallelograms. Draw Pm. dm, ordinates to the axis ; PF perpendicular to Dd. Let CA meet PX in T and Wz int; Then Cis C2 32. GA- 3M 20 226 CONIC SECTIONS. Ct : CA :: CA: Cm, Prop XI: OT 2 Ops: Om: CM But CT : Ct :: MT: Cm, by similar triangles. MT: : Cm 221 0m:: CM . CM. MT = Cm? - = = = = = = (1) Again,” CM: CA 23:GA : CT « CM: CA :: MA: AT,dividendo : Or, “GMs MateeMac: MT, componendo : pa a CM. MT) Gale ae ee (2) But, AC?: BC?:: AM: Ma: PM? Or, ACG’ BU: Cn? : PM igh ACe:?BC 2: Cm > PM Similarly, AC : BC :: CM : dm Or, -BG.: dmiz: CA: CM But, CT xs CAs: CA]: (CM $e OT CA 2 BCs dm But, PE : CT?:% dm .Cd .. PRGA. BCS 3G .. Rectangle PF .CD = rectangle AC . BC or, Parallelogram CX = rectangle AC . BC *. Parallelogram WwXz = 4 AC. BC=Aca. Bb Cor. By (2), Cm = HM. Ma ces (CM—CA) (CM+CA). = CM’—CA CA? = CM’—Cm? yen similarly, CB’ = dm’—PM?. PROPOSITION XV. THEOREM. The difference of the squares of any two conjugate diameters, is equal to the same constant quantity, namely, the difference of the squares of the two azes. That is, if Pp, Dd, be any two con- jugate diameters, Pp?— Dd* = Aa’* — Bb’. Draw PM dm, ordinates to the axis. Then, by Cor. to last Prop. AC?—BC*=CM?+ PM?--(Cm*+d’m) =CP’—Cd@’ Ad—BbL’ =Pp*—Da’. HYPERBOLA. 227 PROPOSITION XVI. THEOREM. The rectangle under the focal distance of any point, is equal to the square of the semi-conjugate. That is, if CD be conjugate to CP, ey ees ke by =3, OD)": Draw SY, HZ, perpendiculars to the tangent at P, and PF perpendicu- lar to CD; Then by similar trian. SPY, PEF Peete Mada eb e PE OF be ae vas we DOs PE “EN = ACY by-Prop. VI. simslarly; HPs sy Hee CAC) ss PE Pape SY As ACs Ph CO DFs CO Eby Prope Lys But SY HAC Be by Praps vill Peerks vekb bes OL). PROPOSITION XVII. THEOREM. If two tangents be drawn, one at the principal vertex, the other at the vertex of any other diameter, each meeting the other’s diameter produced, the two tangential triangles thus formed will be equal That j is, triangle CPT = triangle CAK. Draw the ordinate PM ; then OVE AGA 22 Ch: CK, by similar triangles. a But, CM :. CA ::CA:CT a OF OE MEET 6) ma (Oa The two triangles CPT, CAK, have thus the angle C common and the sides about that angle reciprocally proportional ; these triangles are .*. equal. Cor. 1. Take each of the equal triangle CPT, CAK, from the common space CAOP ; there remains triangle OAT = OKP. Cor. 2. Also take the equal triangles CPT, CAK, from the common triangle CPM; there remains triangle MPT = trapez. AKP M. 228 CONIC SECTIONS. PROPOSITION XVIII. THEOREM. The same being supposed, asin last proposition, then any straight lines QG, QE, drawn parallel to the two tangents shall cut off equal spaces. That is, triangle GQE = trapez. AKXG triangle rgH = trapez. AKRr Draw the ordinate PM. The three similar triangles CAK, CMP, CGX, are to each other as c CAI CMA CGs ~. AKPM: trap. AKXG : : CM’—CA? : CG’—C A®, dividendo. But, PN. QG* :: CM’#CA?’ : CG’—CA’, .. trap. AKPM : trap. AK XG: : PM eet But, trian. MPT : trian. GQF:: PMia? Qe *.* the triangles are similar. ..trap. AKPM: trian. MPT: : trap. AK XG: trian. GQE, But, by Prop. X VII, Cor. 2. trap. AKPM = triangle MPT ; .. trap. AKXG = triangle GQE. And similarly,trap. AKRr = triangle rgE. Cor. 1. The three spaces AKXG, TPXG, GQK, are all equal. Cor. 2. From the equals, AKXG, EQG, take the equals AKRr, Egr ; there remains, RrXG, = rqgQG. Cor. 3. From the equals RrXG,rqQG, take the common space rqgvXG ; there remains, triangle vQX = triangle vgR. Cor. 4. From the equals EQG, TPXG, take the common space EvXG ; there remains, TPvE = triangle vQX. Cor. 5. If we take the particular case in which QG coincides with the minor axis, . The triangle EQG becomes the tri- angle IBC, The figure AKXG becomes the tri- angle AKC, .. triangle IBC = triangle AKC = triangle CPT. HYPERBOLA. 229 PROPOSITION XIX. THEOREM. Any diameter bisects all its own ordinates. That is, Q If Qgq be any ordinate to a dia- meter CP, Ui ug Draw QX, 2, at right angles to the major axis ; Then triangle vQX = trian- gle vgx ; Prop. XVIIL, Cor. 3. But these triangles are also equiangular ; Qu =! 19: Cor. Hence, any diameter divides the hyperbola into two equal parts. PROPOSITION XX. THEOREM. The square of the semi-ordinate to any diameter, is to the rec- tangle under the abscisse, as the square of the semi-conjugate to the square of the semi-diameter. That is, If Qq be an ordinate to any dia- meter CP, net: Ror inane a Gre Let Qq meet the major axis in E; Draw QX, DW, perpendicular to the major axis, and meeting PC in X and W Then, since the triangles CPT, Cv, are similar, trian. CPT: trian. CuK :: CP? Cv* or, triad, GPU scirap. PPK >: OR: Cv? —CP” Again, since the triangles CDW, vQX, are similar, ‘triangle CDW : triangle DEL ME A OD ts at): But, triangle CDW = triangle CPT ; Prop. XVIII, Cor. 5, And triangle Ree = tr apr TPE; Prop. XVIIL, Cor. 3. GRS asCNye Cv! CP? : vQ? Or, Qv? :Pu.vp: OP oy Phi MOM Cor. 1. The squares of the ordinates to any diameter, are to each other as the rectangles under their respective abscisse, Cor. 2. The above proposition is merely an extension of the property already proved in Prop. XII, with regard to the relation between ordinates to the axis and their abscisse. 230 CONIC SECTIONS. PROPOSITION XXI. THEOREM. If tangents be drawn at the vertices of the axes, the diagonals of the rectangle so formed are asymptotes to the four curves. Let MP meet CE in Q ; Then, MQ?: CM?:: AE? : AC’ 22 BOA >: MP? :CM?—CA”. . Now, as CM increases, the ratio of CM? to CM’—CA’* continually approaches to a ratio of equality ; but CM* — CA’ can never become actually equal to CM’, however much CM may be increased. Hence, MP is always less than MQ, but approaches con- tinually nearer to an equality with it. In the same manner it may be proved, that CQ is an asymp- tote to the conjugate hyperbola BP’. Cor. 1. The two asymptotes make equal angles with the axis major and with the axis minor. Cor. 2. The line AB joining the vertices of the conjugate axes is bisected by one asymptote and is parallel to the other. Cor. 3. All lines perpendicular to either axis and is termi- nated by the asymptotes are bisected by the axis. PROPOSITION XXII. THEOREM. If a line be drawn through any point of the curves, parallel to either of the axes, and terminated at the asymptotes, the rectangle of its segments, measured from that point, will be equal to the square of the semi-axis to which it is parallel. That is, SS the rect. HEK or HeK = CA’, and rect. AEK or hek = CA’. ZB For, draw AL parallel to Ca, and aL to CA. Then by the parallels. CA? :C or AL? :: CD’: DH’; and by Prop. XI, CA’: Ca? ::CD*—CA?: DE’ ; seby subpreo As Cate: CA? : DH" Et Ors Re But the antecedents CA’, CA’ are equal, } d Fr oo HYPERBOLA. 231 Therefore, the consequents Ca’, HEK must also be equal. In like manner it is again, by the parallels, CA’ : Ca’ or AL’: : CD? : DH’; CA’? :Ca:: CD?+CA?: De’; “. by subtr. CA? : Ca? : : CA* : De’-——DH? or HeK. But the antecedents CA’, CA’ are the same, .. the conseq. Ca’, HelX must be equal. In like manner, by changing the axes, is Ak or hek = CA’. Cor. 1. Because the rectangle HEK = the rectangle HeK, te Ee Feb: Beh. a And consequently HE : is always greater than He. Cor. 2. The rectangle KEK = the rectangle HEA. For, by similar triangles EA : EH :: Ek: EK. Scholium. It is evident that this proposition is general for any line oblique to the axis also, namely, that the rectangle of the segment of any line, cut by the curve, and terminated by the asymptotes, is equal to the square of the semi-diameter to which the line is parallel—since the demonstration is drawn from properties that are common to all diameters. Cor. 3. Hence it is evident that all the rectangles are equal which are made of the segments of any parallel lines, cut by the curves, and limited by the asymptotes ; and therefore, that the rectangle of any two lines drawn from any point in the curve, parallel to two given lines, and limited by the asymp- totes, is a constant quantity. GENERAL REMARK. Having thus discussed at length, the properties of the para- bola, ellipse and hyperbola, in the relations of their local and peculiar constructions, it may be observed, that there are many properties common to each—especially in the ellipse and hyperbola. These curves have many striking similarities in their determinations, although there is but little similarity in their construction ; for the axes, of the hyperbola are thrown without the curve, while in the ellipse they are within ; hence it should not be surprising, that the same or corresponding quantities, thus differently associated, should propagate by a somewhat similar condition of their mutations, curves so ap- 232 CONIC SECTIONS. parently dissimilar in their developments. The ellipse is a curve of limited extent returning into itself as the circle, but the hyperbola is unlimited in its construction, or its determin- ation, for its branches may be extended indefinitely ; the same may be observed in relation to the parabola, which may be indefinitely extended, and the branches of the curve become at length parallel to its axis ; this, however, is only in their in- finite extension. But the hyperbolic curve never approaches toward or evento a parallelism with the axis, but approaches infinitely toward its asymptotes; but without ever touching them, except in their infinite extension. The manner in which these curves are derived from the sections of a cone, as their name indicates, their origin will be shown in another volume, and their quadratures, and some other properties in relation to them, will be there discussed. HIGHER GEOMETRY MENG SU RAST ol Or N: FOURTH PART OF A SERIES ON ELEMENTARY AND HIGHER GEOMETRY, TRIGONOMETRY, AND MENSURATION, CONTAINING MANY VALUABLE DISCOVERIES AND IMPROVEMENTS IN MATHEMATICAL SCIENCE, ESPECIALLY IN RELATION TO THE QUADRATURE OF THE CIRCLE, AND SOME OTHER CURVES, AS WELL AS THE CUBATURE OF CERTAIN CURVILINEAR SOLIDS 5 DESIGNED AS A TEXT-BOOK FOR COLLEGIATE AND ACADEMIC INSTRUCTION, AND AS A PRACTICAL COMPENDIUM OF MENSURATION. BY NATHAN SCHOLFIELD. NEW YORK: PUBLISHED BY COLLINS, BROTHER & CO. No. 254 Pearl Street. 1845. Entered according to Act of Congress, in the year 1845, by _ NATHAN SCHOLFIELD, In the Clerk’s Office of the District Court of Connecticut. G. W. WOOD, PRINTER, 29 GOLD ST., NEW YORK. ——_— — ————————— PREFACE. Havine, in the former parts of this series, treated of the elements of geometry, trigonometry, conic sections, &c., it now remains for us, in accordance with our original design, to make such application of the former principles, as to elicit such other truths or principles as depend on their various combinations, and to investigate the relations of such subjects as pertain to the higher geometry. And, without attempting to give a full and perfect treatise on the subject, which would require volumes, we shall endeavor to present some portions of this ancient subject in a new dress; hoping, thereby, to render its beauties more plainly visible, and its oracles more intelligible. Some new solids are introduced into this volume, the most important of which is a class termed revoloids ; which, from their organization, seem to serve as a connecting link between rectilinear and curvelinear solids. The properties of those solids are discussed, and their surfaces and solidities are de- termined. Some new curves are also introduced and inves- tigated, among which is the revoloidal curve, whose quadrature is determined; and, from its relation to the circle, and also to rectilinear figures, we are enabled to approximate to the cir- cle’s quadrature to an indefinite extent. During the investiga- tion of this subject, other important properties of the circle will be developed, by which the area of the segment of a circle whose are and sine are known, may be computed with as little labor as that of the area of a triangle whose base and perpendicular are given. We have also introduced into this work a mode of con- lV PREFACE. struction for variable quantities, or such magnitudes as depend on variable factors ; and have adapted a notation, embracing some of the principles of the calculus, by which variable magnitudes may be algebraically discussed, and their condi- tions rendered intelligible. By this notation, some of the more difficult geometrical subjects are susceptible of the most ele- gant solution; and we are also enabled to get a definite algebraic expression for the circle’s quadrature, in terms of the diameter. The mensuration of such superficies and solids as depend on the higher geometry, follows at the close of the work. From the hasty manner with which a considerable portion of this work has been prepared, it can hardly be presumed to be entirely free from errors; but it is believed that if any errors exist, they are such as involve no important principle. The author, with these remarks, submits the work to the consideration of an intelligent public. CONTENTS. BOOK I. PAGE. On the Species and Quadrature of the Sections of Elementary Saga embracing the Oa bag rath and Gab etae Definitions, The Sections of a Cone, : The Sections of a Polyedroid, The Sections of a Prism, &c., . Quadrature of the Parabola, ; On the Ellipse, its Quadrature, &c., On the Hyperbola, its Quadrature, &c., Equations to the Conic Sections, BOOK II. Solid Sections, or Segments of Solids of Revolution, Ungulas, &c., Definitions, ; ; : ; ; i Cylindrical Ungulas and Segments, ; Conical Segments and -Ungulas, ‘ Segments and Ungulas of an Elliptical Cylinder, Spherical Ungulas and Segments, Parabolic Prisms and Ungulas, BOOK III. On Revoloids and Solids formed by the Revolution of the Conic Sra a Definitions, : : Quadrature of the Surface of a avalold determined, Revolvoids and Ungulas equivalent to a Sphere or Spheroid, ‘The Cubature of the Revolvoid determined, ; : Segments of Revolvoid, Parabolic Revoloid, its Cubature determined, ‘ Sections of Solids formed by the Revolution of the Conic Sections, Scholia and Formule, in relation to Cylindric and Conical Ungulas, . BOOK IV. On the Revolvoidal Curve, the Rectification of the a and other Curves, and on the Quadrature of the Circle, ‘ Definitions, : Projection of a Revolvoidal Curve, Equation to the Curve, The Ellipse, its Rectification, &e. ee Rectification of the Revoloidal Curve, On the Circle’s Circumference, and its Quadrature, Curve of the Circle’s Quadrature, . On the Quadrature of the Segment of a Circle, On Spirals—their eae so OLB. : 5 On the Cycloid, vi CONTENTS. BOOK V. On the Production and Resolution of Geometrical Magnitudes, . Cuap. I[.—Definitions and Principles, . ; : Production of Surfaces and Solids, Cuap. I].—On the Construction of Quantities Pevhose Hlerments are a ee Poe | aoe . 159 series either of constant or variable quantities, Explanation of Principles, and Notation, Construction of Variables, Construction of Curves from their Equations, | Quadrature of the Circle expressed algebraically, in terms of known functions of the diameter, -3 Equivalent Constructions for the Equations of Curvelinear : : at OL . 166 . 166 Solids, : : Cuar. Il.—The Differential and Integral Calculus, 4 Differential Calculus, . Integral Calculus, Cuar, 1TV,—On the Centre of Sureche aril Solids, the Virtual Centre AS aye br gs} a hey! 175 2179 . 180 . 181 Cuap. V.—On the Relations of Lines, Surfaces, and Solids, gene- . 182, 182 and Centre of cere OO cae Definitions, Virtual Centre of a Sy stem of Points, Virtual Centre of Surfaces, The Virtual Centre of Solids, The Virtual Centre of a Circular Arc, The Virtual Centre of the Surface of a Solid, rated by the motion of the Virtual Centre, . General Proposition, ; : MENSURATION OF SUPERFICIES. Circular Segments and Zones, . : To find the Circumference of a Circle, or any Are, Mensuration of the Ellipse, : ; Mensuration of the Parabola, Mensuration of the Hyperbola, . To find the Area of any Plane Surface by ‘Equi-distant Ordinates, MENSURATION OF SOLIDS. Mensuration of the Sphere and Revoloid, Segments of a Sphere or Revoloid, Segments of a Spheroid, . Mensuration of the Paraboloid, &c. opie : : 5 : Mensuration of the Hyperboloid, Sens : ; ‘ : ; Circular Elliptical and Parabolic Petey : - : Ungulas, . Rings, . Gauging of Casks, SPECIFIC GRAVITY OF BODIES. Table of Specific Gravities, Magnitude of Bodies, by their Weight and Specific Gravity, QUESTIONS FOR EXERCISE, Description of an Instrument for Measuring Distances and Heights by a single observation, Notes, y PAGE. . 145 . 145 . 146 159 160 169 . 186 Bib yy IoD . 202 . 207 . 210 . 212 . 214 . 216 . 218 ey bs) . 221 . 224 . 227 . 228 . 233 . 234 . 236 . 242, . 248 HIGHER GEOMETRY. PART II.-—-BOOK I. SPECIES AND QUADRATURE OF SUPERFICIAL SECTIONS OF ELEMENTARY SOLIDS. DEFINITIONS. 1. Superficial sections are surfaces formed when solids are cut by plane or curved surfaces. 2. If the cutting surface is a plane, the section is a plane seclion. 3. Superficial sections of solids take different names, ac- cording to the form of the solid in the plane of the section. 4. From the cylinder, we have the rectangle, the circle, and, as will be shown, (Prop. VIII. Cor.) the edlipse. 5. From the cone, we have five different figures, viz: a tri- angle, a circle, and, as will be shown in Propositions I, IL, and III., a parabola, an ellipse, and a hyperbola. 6. From the sphere, we have only the circle. Scholium. The parabola, ellipse, and hyperbola, will be more specially the subjects of this book. 8 SPECIES AND QUADRATURE OF PROPOSITION I. THEOREM. If aright cone BEG be cut by a plane Ap o, which is parallel to a plane touching the cone along the slant side BH, the sec- tion A po is a parabola. Let BEG be that position of the generating triangle which is perpen- dicular to the cutting plane Apo; An their common section, which is parallel to BE. Then, since the plane BEG passes through the axis, it is perpendicular to the base KoG, and to every circular section CPD parallel to the base; it is also per- pendicular to Apo. Hence, the common section PO of the planes A po, CPD, is perpendicular to BEG and therefore to An and CD. But, AN : ND: : BE: EG, which is_a constant ratio ; therefore, by the properties of the circle. AN: ND::CNx ND: NO’, since CN is equal and parallel to En, and constant. Hence the curve is a parabola whose axis is An. Cor. If L be the latus rectum of the parabola pAo, Lx AN NE a—GN XIND: PROPOSITION II. THEOREM. If a cone BEG be cut by a plane KAP through both slant sides the section is an ellipse. Let BEG be that position of the generating triangle which is perpen- dicular to the cutting plane: CPD any circular section. Draw AHK parallel to EG, and therefore bisect- ed by the axis BO. Then EN: CN:: EA: AK NA:ND:: EA: EG; | ENXxXNA :CNXND (NP’) -: FA’?: EGxX AK which is the property of an ellipse, one of whose axes is EA and the other a mean proportional between EG and AK. (Conic Sec- tions, Ellipse, Prop. XI) SECTIONS OF ELEMENTARY SOLIDS. 9 PROPOSITION II. THEOREM. If a right cone BED be cut through one side BE by a plane RAP which being produced backwards, cuts the other side DB produced, the section is an hyperbola. M Let DGEH be any circular section, BGH a triangular sec- tion through the vertex B of the cone parallel to the plane * Then, AN: ENS: BE: EF ; NM: ND:: BF: FD ANXNM: ENXND (NP"*):: BF’: EFXFD (FH?) which is the property of an hyperbola, whose axis major is AM, and whose conjugate axis is to AM as FH to BF. Cor. If GT, HT, be tangents to the circle at G, H; and planes passing through GT, HT, respectively, touch the cone along the lines BG, BH; also, if TB, the common section of the planes, meet AM in C, then the common section CO, CQ, of the plane RAP, extended to meet the tangent planes, are the asymptotes of the hyperbola. Draw BL parallel to DE, meeting AM in L: then the axes of the hyperbola being in the proportion of BF to FH, the an- gle GBH, or the equal angle OCQ is the angle between the asymptotes. Now, by similar triangles ALB, BFE, and CLB, BFT ; AL:CL:: TF: FE, and therefore AC: CL::TE:FE. In like manner, by similar triangles MLB, BFD, and CLB, BFT ; ML : CL: : TF: DF, and therefore CM: CL: : TD: DF. But, by the property of the circle. TE: FE: : TD: DF. 2 e210 SPECIES AND QUADRATURE OF Therefore, CA=CM. Hence C is the centre of the hyperbola, and CO, CQ, are the asymptotes. (Conic Sections, Hyperbo- _ la, Proposition XII.) Scholium 1. Let EHF represent an eee Say hyperbola, and AC, AD the asympto- tes ; let the two branches HE and HE be brought into the position He and Hf, so that the asymptotes become A’eand A’f, or till they become parallel to each other, and the curve becomes a parabola; the parabola then, may be regarded as an hyperbola, whose a- «5——7 7H; FD symptotes are parallel, and infinitely extended in each direction. If the extremities e, f of the curve are brought into the positions 7, 7, so as to incline toward the axis AB, so that the curve may again return into itself as its axis is extended, it then becomes an ellipse or portion of an ~ ellipse. These different figures are the result of the position of the plane forming the section through the cone. Scholium 2. A section of a polyedroid by a plane oblique to its axis may assume a combination of one, two, or three, of the varieties of surfaces. Thus, a section through a regular vertical quadredroid by a plane, parallel to a plane touching one of its sides, making an angle of 45° with the axis, toward the vertex, will consist of two parabolas on opposite sides of the same base, these will evidently be parabolas of equal type or similar parabolas, when the section passes through the centre of the solid, and as it approaches toward one of its sides, the parabolas become dissimilar. ' than 45° with the axis, the section will consist of the segments of two ellipses on opposite sides of the same ordinate, which ordinate is the line formed by the intersection of the cutting plane, with the plane through the centre of the solid perpendi- cular toits axis ; and if the cutting plane passes through the centre of the solid the two segments will be of similar type, but they will vary as the section recedes from the centre toward either side. If the plane should be passed through so as to make an angle with the axis less than 45°, then we should have hyper- bolas on the same base, these would be similar hyperbolas when the plane should pass through the centre, but would be- come dissimilar as it recedes from the centre. SECTIONS OF ELEMENTARY SOLIDS. 11 In like manner, ifa plane should be passed through the pentadroid AB parallel to one of its sides AD, the section would consist of two species, viz., that part which passes through the portion ABD would be a parabola, that passing through the part ABEC would be a segment of anellipse. But if the section should be passed so near to BE as to cut the base of the solid, the section would consist of a parabola, and a middle segment of an ellipse, and would have a rectilinear base. If the plane should be parallel to the side EB, then the section through the lower part ofthe solid would be a parabola, or a segment of a parabola, and the section through the upper part an hyper- bola, and in fine, if the plane should make an angle with the axis less than that of the side EB, then the section through both parts of the solid would consist of the dissimilar hyperbo- _ las or segments of hyperbolas. Leta plane be passed through a polyedroid of a greater number of sides oblique to the axis, and the section may be so made as to consist of segments of all the varieties of the conic sections, and may also, under certain conditions, have one, two, three, and at most four rectilinear sides, but it can have no more than two rectilinear sides, except where the section is parallel to the axis. PROPOSITION IV. LEMMA. From a rectangular prism there may be taken two pyramids of equal base and altitude with the prism; when there will re- main two other pyramids, each equal to half one of the lateral sides as a base, multiplied by one third of the distance of such base to the opposite side. For let the rectangular prism AH be fbi divided into twoparts, by passing the ,G77//77 planes EGDB through the opposite edges, [Sx and the portion ABDCGE will consist of the pyramid, whose base is ABDC, and vertex E, plus the triangular pyramid, whose base may be taken as GEC and ver- tex D, or DCG may be regarded as the | 7/7 7, base, and E the vertex. And since the a©“~ ———B other portion of the prism is similar to this, it can be divided in a similar manner. Hence the whole prism consists of two equal pyramids erected on the upper and lower bases of the prism + two other pyramids erected on the lateral sides as 12 SPECIES AND QUADRATURE OF bases, and whose vertices will be in an angle tormed by the intersection of its opposite side with one of the bases. Hence as in the proposition. Cor. 1. If a prism have a square base and a section be made through the prism parallel to the base, the sections through the pyramids erected on the upper and lower bases will be squares, and the sections through the pyramids, whose bases are the lateral faces of the prism, will be rectangles whose factors are the sides of the squares composing the sections through the former pyramids. That the sections through the pyramids, erected on the square bases will be squares, is sufficiently manifest; and since each of the other pyramids coincides with one of these along one of its slant sides, and with the other along another of its sides, it follows that the measure of its section through any parallel portion of the solid will be the rectangle of the edges of the section formed by the same plane through the two former pyramids. And since, if the section is taken in the middle, equidistant between the two bases, the sides of the sections through the pyramids with square bases are=half the sides of their bases, the sections at such place will each be=1 the section of the whole prism ; the sections of the two quadran- gular pyramids will be 2 squares=2 quarters of the whole section ; hence the sections through the two triangular pyra- mids are squares=to the former, and equal to each other, since the four pyramids fill the space, and constitute the whole prism. Cor. 2. Hence, also any section of a prism with a square base, made by a plane parallel to such base may be expressed by the square of a binomial, whose terms are composed of the sides of the sections through the quadrangular pyramids. Let hi or hs, the side of the square forming a section through the pyramid erected on the lower base, be represented by a; and let pd or dn, the side of the square forming a sec- tion through the pyramid erected on the upper base, be re- presented by 6; then will a+b=the line ho, the side of the whole section hodc, and a’ will represent the section through the pyramid ABDCE, 6’ will represent the section through the pyramid EF HGD, and ad will represent a section through each of the other pyramids, hence 2a will represent the sec- tions through both, and a’+2ab+0? will represent the whole section in whatever parallel the section is taken, always ob- serving that the values of a and 6 vary according to the sides of the respective sections, which they represent. SECTIONS OF ELEMENTARY SOLIDS. _—18 Cor. 3. If there be a series of numbers in arithmetical pro- gression, whose first term is 0 and last term z, and if the num- ber of terms is infinite between these extremes, then the sum of the squares of the series of numbers will be equal to one third of the square of the last term drawn into the series. For if an infinite number of planes be passed through the py- ramid, parallel to its base, and equidistant from each other through the sides of the sections made by those planes, they will be a series of numbers in arithmetical progression, whose first term is 0, and the last term may be called z. Now the sum of the sections drawn into their distance will represent the solidity of the pyramid, but the solidity of the pyramid is equal to one third of an equal ‘series of z, the last term or base of the pyramid drawn intothe distance or ratio ; or= to one third z drawn into the series. ¥Cor. 4. If there be a series.of numbers in arithmetical pro- gression increasing from 0 up to z, drawn into a similar series, decreasing from z down to 0, then will the sum of their pro- ducts be equal to half the sum of the squares of one of the series. For the sections through the pyramid HDGH, as we have seen, represent the rectangles of the sides of the corre- sponding sections through the two pyramids, formed on the two bases, and the sides of these sections are evidently, in each pyramid, a series in arithmetical progression ; one increasing from 0 to z, while the other decreases from zto0. Moreover, the sections.through the pyramid CDGE represent the solidity of that body, as the corresponding sections through the pyra- mid ABDCH, represents the solidity of that body. But the solidity of the pyramid CDGE, is equal to half the pyramid ABDCE, which as we have shown, may be represented by the sum of the squares of a series of arithmeticals, &c. Cor. 5. If in the expression a’+2ab+b’, a be made to pass successively through all the values from 0 up to z ; and bat the same time pass through all the changes from z to 0; then the sum of all the a’ will be equal to the sum of all the 2a), = the sum of all the 2’. Cor. 6. If a in the expression above, be made to increase from / successively to z, and at the same time } be made to pass through all the values from z to h, then the series repre- sented by 2ab, will be a mean proportional between those re- presented by a* and 0’, since in this condition the two pyra- mids represented by the series of a? and 8’, would be such as pertain to pyramids inscribed in a frustum of a pyramid, and erected on the two bases, and the portions represented by the 14 SPECIES AND QUADRATURE OF series of 2ab, would be such as pertain to the pyramids erected on the lateral sides as bases, and since the sum of these pyra- mids is equal to either of the others, when the two bases are equal, and because ab is a mean proportional between a’ and 5°, it follows that the series represented by 2ad is a mean proportional between those represented by a’ and 6°, which agrees with the property of the frustum of a pyramid found in the Elements of Geometry. Cor. 7. Let any plane KIHL be G N passed through a prism parallel to its base, cutting the pyramids ABDCEH, © EFNGD, DCGE, and EF BD in the sec- tions Hnot Krop, Ipot, and Lron. Then # since It is equal to Ln or ro, and Hx=on, it follows that as the section Hnot : Ipot: : Li »? Ipot : Krop : : Hnot : Lnor. Naa 4 And by addition: : (Knot+Ipot)=HIpn : KLnp. Henee Huot: HIpn: : HIpn : KIHL. Cor. 8. Hence, of two quantities, the square of the first is to the rectangle of the first and second, as the rectangle of the efirst and second to the square of the second, and as the sum of the first and second x by the first, is to the sum of the first and second X by the second. Also as the square of the first is to to the sum of the first and second, X by the first, so is the sum of the first and second X by the first, to the first and second X by the first and second. Thus let aand 6 be two given quan- tities, then will a? : ab:: ab: b’, 3: a’+ab: ab+b’, and a’: a’?+ab::a+ab: a’+2ab+4b’. Scholium. It has been shown (Prop. XXXIV. B II. Hi. Sol. Geom.) that the solidity. of a prismoid is equal to the product of the sum of the areas of the two ends+4 times AD and EH a mid- dlesection, equidistant between them Xj of the altitude; we may easily infer that this is also true of all prisms and pyramids and pyramidal frusta. Then since the prism AH is equal to the sum of the areas of the two bases+4.times the middle section hodc x } of the altitude AE, it follows that whatever parts make up this product are equal to the whole prism. First, then let us take the pyramid ABDCE = (the base ABDC-+4 higs, a middle section) x} AE, andalso the pyramid SECTIONS OF ELEMENTARY SOLIDS. 15 EFHGD=(the base EFHG+4gpdn) x 1 s AD; if from the expression for the whole E Wy g prism we take the expressions for the two [Rw pyramids erected on the bases we have, =e 4sencXiAE+4iopgX}Ak, that is the two , )e==")>.— remaining portions, are equal to four times === their middle sections, multiplied by } of their altitude. Hence we may inferthatif | / any portions of either of the pyramidsinto 4™ which the prism has been conceived to be divided, be cut off by a plane, or planes parallel to the base, the portions so cut off will be equal to the product of the sum of their two bases, + four times a middle section between them, X by 2 of the altitude of such portions. It may be ob- served that since a regular pyramid has but one base, its soli- dity is hence equal to the sum of this base + four times a sec- tion, midway between the base and vertice x } the altitude; and also in the two pyramids, whose bases are on the lateral sides of the prism, since they have no bases parallel to the middle section, their solidities for that reason are equal to four times their middle sections x 1 of their height. It may be further observed that each portion sgne, GE into which these pyra- mids are divided by the plane hodc is a wedge, whose base is the section forming the division. PROPOSITION V. THEOREM. If from the extremity of an ordinate to the axis of a parabola and perpendicular thereto, a line be drawn meeting another line, drawn from the vertex perpendicular to the axis, form- ing with the ordinate and abscissa a rectangle ABCD, and if a diagonal be drawn from the vertex A to the extremity of the ordinate, forming a right angled triangle ABC of the same base BC and altitude AB, then any line or ordinate drawn from the axis across the triangle, the parabolic area, and the rectangle, parallel to the ordinate, will be cut in con- tinued proportion by the sides of those figures. That is, EF: EG::EG: EH. Or EF, EG, EH are in continued proportioh. For by (Prop. VII. Cor. of Parabola) g¢ hence EF : BC: : EG’: BC’. : Or, EF : EH :: EG’: EH’, therefore (Prop. XXIV. B. I. El. Geom.) EF, EG, EH, are proportionals, or EF : EG: : EG : EH. 16 SPECIES AND QUADRATURE OF Cor. Let any number of ordinates GI be drawn across the exterior parabolic space parallel to the axis, and_ since EG? : BC?:: AE: AB, by equality we have Al’: AD*®: : IG: DC, and since this is true from whatever position on the line AD, the line IG may be drawn, it follows that any line IG drawn across the exterior parabolic space is proportional to the square of the distance of such line from the vertice A. PROPOSITION VI. THEOREM. If a parallelogram be circumscribed about a parabola, the area of the space exterior to the parabola will be equal to one third of the parallelogram, and the interior space will be two thirds of the parallelogram. Let ABCD be aparallelogram cir- cumscribed about the parabola AEB, and the space ECB, EDA ex- terior to the curve will be = } the parallelogram ABCD, and the space ek AEBA within the curve will be = 2 ABCD. A F B From the vertice E draw EF the axis of the parabola, which will divide the parabola and rectangle into two equal parts; it is to be proved that the exterior space ECB is = 1 EFBC. Draw the diagonal EB, and let an indefinite number of equidistant ordinates Ks’ be drawn across the triangle EBC, and exterior parabolic space EIBC, and those ordinates will represent their respective surfaces in the relation of their magnitudes respectively. Then since the distances of those ordinates on the line BC, estimated from the vertice B, are a series of numbers in arithmetical progression, the ordinates Kk drawn across the triangle, are also a series in arithmeti- cal progression, for Kk is proportional to BK in whatever position the ordinate Kk is drawn; and since it has been shown (Prop. V. Cor.,) that any ordinate IK drawn across the exterior parabolic space is proportional to the square of its distance BK. from the vertex B, it follows that the ordinate is proportional also to the square of Kk, the corresponding ordinate drawn across the triangle. But the ordinate EC or base of the triangle is equal and identical with the base of the parabolic exterior space ; hence each of the ordinates IK ter- minated by the parabolic curve is equal to the square of K&, the corresponding ordinate drawn across the triangle. SECTIONS OF ELEMENTARY SOLIDS. 17 Hence all the ordinates Kk may be represented by a series of numbers in arithmetical progression, whose first term beginning at B, is infinitely small or 0, and last term EC, and all the IK. will be a similar series of squares of those arithme- ticals ; but (Prop. IV. Cor. 3,) the sum of an infinite series of the squares of a series of numbers in arithmetical progression, whose first term is 0, and last term z, or EC, is equal to 4 EC XxCB, which is 1 of the rectangle EFBC. Cor. 1. Hence the parabolic segment EIBE, cut off by the diagonal, is equal to 3 of the rectangle, = } of the interior pa- rabolic space EIBF, = 1 the exterior parabolic space EIBC. Scholium. It has been shown in the argument above that the ordinates drawn across the exterior parabolic space are severally = the squares of the same ordinates drawn across the triangle EBC, this is true where EC=1, and for all other values of EC, they are in the same proportion. PROPOSITION VII. THEOREM. The area of the exterior parabolic space included in its cir- cumscribing rectangle is equal to the sum of tts base + four times the line or ordinate drawn parallel to the base, and equi- distant from the base to the vertex, multiplied by 1 of the al- titude. And the area of the interior space of a parabola is equal to the sum of its base, + four times the ordinate equidistant from this base to the vertex, multiplied by one sixth of the altitude. Let ABDC be a rectangle circumscribing the semi-parabola AcDB, and let Le be an ordinate drawn cross the exterior space ACDcA equidistant from A to C, and let M be an ordinate drawn across the interior space AcDB ina similar manner, then will the area ACDcA = (CD -P4tie) 1 AC. And the area AcDB = (AB+4cM) 3 BD. For draw the diagonal AD, then, since we have A shown in the argument to Prop. V1, that Le=Lh’, when cD= CD’, and since LA=1 CD: then if we call Lh, a, and he 2a, we shall have Le = a*, and CD = 4a’ that is Le = 3 CD or CD=4Le. Hence CD+4Lc=2CD, then by the proposition 20D x }AC = CDx 4 AC = the surface AcCDA, as was also found in proposition VI. 18 SPECIES AND QUADRATURE OF Again since LM =CD or AB, and Le =1 CD, cM= 3 AB. Then by the proposition (4cCM+AB) X } BD=4ABX}{BD= 2 (ABxBD)=the area, as found in the preceding proposition. Cor. 1. Hence the parabolic segment ADcA is equal to four times the middle ordinate ch X } the altitude AB. For since Lh = 1 LM, ch=Lc = } AB, and } ABX4=AB, and ABxX1 BD = the surface which agrees with Cor. 1 Prop. VI. Cor. 2. Also if another similar parabolic curve DeA be de- scribed on the opposite side of the diagonal, the space between the two curves would in like manner be found, = 4 times the ordinate ce X 1 of the altitude BD. Cor. 3. The same may be shown in reference to the area of the triangles ACD, and also of the rectangle ABDC, viz., their areas are equal to the sum of their upper and lower bases, + 4 times a middle ordinate, X } of their respective altitudes. Scholium. Let there be a rectangle CB, and a prism BK of the same altitude, and let the base of the prism be a square, one of whose sides is = AB the base of the rectangle. Let the prism be divided Hi into the pyramids BEGFD, DIKHG, DHFG, EGID, as in Prop. IV. Also let the parabolic curves. AcD AeD and the diagonal AD be drawn as in the proposition above. Then if a plane MzOP be passed through the prism parallel to its “ base, and if an ordinate LM be made to pass through the rectangle in the same plane, the plane will cut the pyramidal portions of the prism in the same relation, as the ordinate cuts the parabolic portions of the rect- angle ; viz., the section Mfsqg through the pyramid BEGCD will be to the section MnOP through the prism, as the ordinate eM through the exterior parabolic space ABDeA, to the ordi- nate LM through the rectangle ABDC ; and the section svO¢ through the pyramid DIKHG is to the whole section through the prism, as the ordinate Le through the exterior space ACDcA, to the whole ordinate LM through the rectangle, and so for the sections through the other pyramids, and ordinates through the interior parabolic segments, and this is true in whatever parallel position the plane forming the section and ordinate is drawn. For it has been shown (Prop. VI. Sch.) that the ordinate eM is equal to hM? ; the section Mfsq is evidently=the square of Mf; but Mf is= AM, hence Mfsqg =hM? ; therefore Mfsq SECTIONS OF ELEMENTARY SOLIDS. 19 will be expressed by the same numbers as eM, and this is true in whatever parallel position the plane is passed. | The same may also be shown in reference to the sections svOt through the pyramid DIHG, and the ordinates Le and LA, viz., the section svOt=O??=fn? =LA’ and Le=LA*. And since every ordinate LM through the rectangle, has the same relation to the rectangle, as every section through the prism, has to the prism, and each section Mfsg, svOt, has the same relation to the solid as each ordinate Le, cM, has to the surface ; it follows that the remaining sections, fnvs, stPg, have the same relations to their respective figures, and since the sec- tions Mfsq svOé are expressed by the same powers of the same factors as the ordinates Le and eM are expressed, ch and he must also be expressed by the same terms as the sections fnvs, stPq, viz., he=LAXhM, ch also=LAXhM, and ce= 2 (LAXhAM.) Hence if LA =a and hM=6@ then will Le=a’, ce=2ab and cM=0", and the whole ordinate may also be ex- pressed by a?+2ab+0’, which is the square of a binomial, the same as has been shown, (Prop. IV. Cor. 2) in reference to the pyramids composing a prism, and the expression is true in whatever parallel position the ordinate is drawn. It fol- lows therefore that the ordinates drawn across any of the pa- rabolic portions of the rectangle, have the same power of de- termining the area of the exterior or interior parabolic spaces, as the corresponding sections through the pyramidal portions of the prism have, in determining the solidities of those pyra- mids. But we have shown (Prop. IV. Sch.) that each portion of the prism divided as above, however selected or compounded, is equal to the sum of its two bases+4 times a middle section X ? of the altitude ; hence also the area of any parabolic portion or portions of the rectangle is equal to the sum of its bases + 4 times a middle ordinate X 3 the altitude. Cor. 4. The parabolic area included between two parallel ordinates is=the product of the sum of the two ordinates + 4 times an ordinate equidistant between them X 1 of the altitude of the parabolic segment. For let ABMec bea segment of the parabola included between the ordinates cM, AB, the axis MB and the curve Ac ; draw cS perpendicular to AB, and the parabolic segment ScA will be equal to (AS + 4 um) xi Se, and the rectangle SBMc is evidently=(S8+cM+4mp) 3 cS ; hence the whole space ABMc=(AS+SB+cM+4um-+4mp,) xiSce=(AB+cM+4up) 3 Se. Cor. 5. In like manner it may be shown that the space eMBA is=(AB+eM+4wp) 4MB, and hence also that any para- bolic portion AuchmA or uchm may be determined in the same manner. 20 SPECIES AND QUADRATURE OF Cor. 6. Hence generally, if a space be terminated by a parabolic curve on ene side, and either a parabolic curve or aright line on the other, the space included between two parallel lines, drawn across the figure cutting those sides, will be=to the product of the sum of those lines+4 times another line drawn across equidistant between the two multiplied by} the perpendicular distance of the two parallel lines. PROPOSITION VIII. THEOREM. If a circle be cut by a plane through its axis, and perpendicu- lars be drawn from every point in the circumference to the plane, the orthographic projection of the circle so drawn will be an ellipse. Let the circle ARML be inclined to the plane of this paper in such a manner, that the semicircle ARM may be above the paper, and the semicircle ALM below it, and let AM be the common intersection of the two planes. Let 9 z the semicircle ARM be projected downwards up- on the plane of the paper, by drawing perpendiculars QP, RB, from each point of the eircle, and let the » semicircle ALM be projected upwards, by drawing the perpendicu- lars gp, LO, &c. ; then the curve ABMO, marked out by this projection, will be as an ellipse. For draw QN, RC, at right angles to AM, and join PN, BC; then the angles QNP, RCB, will measure the inclination of the planes, and PN, BC will be perpendicular to their common intersection AM. Now QN:: PN: rad. : cos. QNP, and RC : BC: rad. : cos. (RCB = QNP) ; “. QN:PN:: RC or AC: BC. which agree with the pro- perties of the ellipse (Prop. XII. Cor. Conic Sec.) Ina similar manner it may be shown that the semicircle ALM is pro- jected into a semi-ellipse AOM ; and thus the whole circle ae is projected into an ellipse ABMO, whose major axis is Scholium. This proposition is manifestly true, when the plane of the projection does not cut the circle, or cuts it un- equally. SECTIONS OF ELEMENTARY SOLIDS. 21 Cor. Hence any section of a cylinder by a plane not per- pendicular or parallel to its axis, is an ellipse, or portion of an ellipse. | PROPOSITION IX. THEOREM. If on the major axis of an ellipse a circle be described, the area of the ellipse will be to that of the circle, as the minor axis of the ellipse to the major axis. ; Let ACBD be a circle described on the major axis AB of the ellipse AEBF, then will the area of the ellipse be to that of the circle as EF to EB. For it has been shown (Prop. XII. Cor. Conic Sec.) that any ordinate IL of the axis AB of the ellipse is to the corresponding ordinate GH of the circle, as EF to CD or AB, as the minor axis to the major axis. And since this is true in whatever position the ordinates to this axis are drawn; and because if we suppose an indefinite num- ber of equidistant ordinates to be drawn across the ellipse and circle, the sum of the ordi- nates in each will represent those figures in the relation of their areas, the sum of the ordinates in the ellipse will be to the sum of those in the circle, as EF to CD, and hence the area of the ellipse will be to that of the circle in the same ratio. Scholium. As the area of the ellipse bears this given ratio to that of its circumscribing circle, the quadrature of the ellipse must therefore depend on the quadrature of the circle. If *xX(AQ. AO, or CO,) or rAO*=the area of the circle, then «.AO.KO=the area of the ellipse. Hence the area of the ellipse is found by multiplying the rectangle under its semi- axis by the constant number 7, the ratio of the diameter to the circumference of a circle, which in the Elements of Geo- try we have found, developed to a certain order of decimals to be 3,1415926. But 3,1416 may be regarded as the value of « which is a convenient number to use, and is suffi- cient where extreme accuracy is not required. From this it also appears, that the area of an ellipse is equal to the area of a circle whose radius is a mean propor- 22 SPECIES AND QUADRATURE OF tional between its semi-axis; for the area of that circle is = x R’=7 Xthe square of /AOxEO = 7. AO.EO. Cor. 1. The area of an ellipse has the same ratio to the area of its circumscribed parallogram as the area of a circle has to its circumscribed square. For the area of the parallelo- gram circumscribing the ellipseis = 4 AOX EO; hence the area of the ellipse: area of the parallelogram: : +. AQ. EO: 4AO.EO:: 83,1416: 4::'7854: 1. Cor. 2. If a circle be described on the minor axis EF of the ellipse, then any ordinate PQ of the circle will be to the cor- responding ordinate Ie of the ellipse as EF to AB. Hence also the area of the ellipse is to that of the circle described on its minor axis, as the major, axis of the ellipse to its minor axis. Cor. 3. Hence any segment of the ellipse cut off by ordi- nates, either of the major or minor {axis is to a similar seg- ment of the circle, described on such axis, as its conjugate is to such axis. Thus the segment IAL of the ellipse AEBF is to the segment GAH of the circle described on the axis AB, as the ordinate IL is to the ordinate GH or as the minor axts EF of the ellipse is to the axis CD of the circle, or to AB the transverse axis of the ellipse. And the segment Ie is to the segment PQE as the axis AB as conjugate to EF of the ellipse, to the axis RS of the inscribed circle, or EF of the ellipse. Cor. 4. Since the area of a circle BESF is equal to its cir- cumference multiplied by half the radius EO, and since the area of the ellipse AE BF is = to that product increased in the ratio of EO to AO or CO, it follows that the area of the ellipse is equal to the circumference of its inscribed circle mul- tiplied by 3 the radius of its circumscribed circle, and that it is also equal to the circumference of the circumscribed circle multiplied by 4 the radius of the inscribed circle. Cor. 5. As a circle is to the square of its diameter so is any ellipse to the rectangle of its two axes, or the rectangle of any two conjugate diameters drawn into the sine of their included angle. Any two like segments or zones of the circle and ellipse, are alsoin the same proportion. SECTIONS OF ELEMENTARY SOLIDS. 23 Cor. 6. Hence we have the fol- lowing construction ; let ADE be an oblique segment of the ellipse AFBGA, cut off by an_ ordinate to the diameter AB, FG being _ the conjugate. Through the centre C draw aP perpendicular to FG meeting Aa and BP, in a and P, both parallel to FG ; then about the axes a P, FG, describe the ellipse aFPGa, meeting the ordinate pro- duced in eand d. Thenwill the right elliptical segment dae, be equal to the oblique segment DAE, and so will the whole ellipse ak PGa=AFBGA ; moreover their corresponding ordinates de DI: parallel to the common diameter FG, are everywhere equal, as are the like parts or zones contained between any two of such ordinates. And the same may be said of all ellipses of the same diameter FG, con- tained between the parallels aA, BP infinitely produced. . PROPOSITION X. THEOREM. All the parallelograms are equal, which are formed between the asymptotes and curve, by lines drawn parallel to the a symp- totes of an hyperbola. That is, the lines GE, EK, AP, AQ, being parallel to the asymptotes CH, Ch, then the parallelogram CGEK = paral- lelogram CPAQ. For, let A be the vertex of the curve, or extremity of the semi-transverse axis AC,: perpendicular to which draw AL or Al, which will be equal to the semi- conjugate, by definition XIX. Hyp. Also, draw HEDedA parallel to Iv. ahs bag. Then, CA?: AL? 3: CD CA? : DE’, and by parallels, At Abang Gers therefore, by substract. CA? : AL’ :: CA’* : DH?—DE? , or rectangle HE . HA ; consequently, the square AL’=the rectangle HE. EA. But, by similartrian. PA: AL:: GE: EH, and, by the same, QA: Al:: EK: Eh; therefore, by comp. PA. AQ: AL’?:: GE. EK : HE. Eh; and, because AL’=HE . EA, therefore PA. AQ=GE. EK. But the parallelograms CGEK, CPAQ, being equiangular, are as the rectangles GE.EK and PA . AQ. 24 SPECIES AND QUADRATURE OF And therefore the parallelogram GK = the parallelo- gram PQ. That is, all the inscribed parallelograms are equal to one another. Cor. 1. Because the rectangle GEK or CGE is constant, therefore GE is reciprocally as CG, or CG: CP: : PA: GE. And hence the asymptote continually approaches towards the curve, but never meets it ; for GE decreases continually as CG increases ; and it is always of some magnitude, except when CG is supposed to be infinitely great, for then GE is in- finitely small or nothing. So that the asymptote CG may be considered as a tangent to the curve at a point infinitely dis- tant from C. Cor. 2. If the abscissas CD, CE, CG, &c., taken on the one asymptote, be in geometrical progression increasing ; then shall the ordinates DH, EJ, GK, &c., pa- rallel to the other asymptote, be a de- creasing geometrical progression, having the same ratio. For, all the rectangles if CDH, CEI,CGK, &c., being equal, the ordinates DH, EI, GK, &c., are reciprocally as the abscissag CD, CE, CG, &c., which are geometricals. And the recipro- cals of geometricals are also geometricals, and in the same ratio, but decreasing, or in converse order. Be PROPOSITION XI. THEOREM. The three following spaces, between the asymptotes and the curve, are equal ; namely, the sector or trilinear space contained by an arc of the curve and two radit, or lines drawn from - its extremities to the centre ; and each of the two quadrila- terals, contained by the said arc, and two lines drawn from ats extremities parallel to one asymptote, and the intercepted part of the other asymptote. That is, The sector CAE =PAEG=BAEK, all standing on the same arc AE For, as has_ been already shown, CPAB=CGEK ; Subtract the common space CGIB, ee shallthe parallel PI=the parallel . Toeach add the trilineal IAE, SECTIONS OF ELEMENTARY SOLIDS. 25 Then is the quadril. PAEG=BAEK. Again, from the quadrilateral CAEK, take the equal triangle CAB, CEK, and there remains the sector CAE=BAEK. Therefore, CAE=BAEK=PAEG., PROPOSITION XII. THEOREM. Every inscribed triangle, formed by any tangent and the two intercepted parts of the asymptotes, is equal to a constant quantity ; namely double the inscribed parallelogram. That is, the triangle CTS=2 parallelogram GK. For, since the tangent TS is bisected by the point of contact E, and EK is pa- rallel to TC, and GE to CK ; therefore ye - CK, KS, GE are all equal, as are also CG,GT, KE. Consequently the triangle ¢ x = GTE = the triangle KES, and each equal to half the constant inscribed paralielogram CK. And therefore the whole triangle CTS, which is composed of the two smaller triangles and the parallelogram, is equal to double the constant inscribed parallogram GK. PROPOSITION XIII. THEOREM. If from the point of contact of any tangent, and the two inter- sections of the curve with a line parallel to the tangent, three parallel lines be drawn in any direction, and terminated by either asymptote ; those three lines shall be in continued proportion. That is, if HKM and the tangent IL be parallel, then are the parallels DH, EI, GK in x continued proportion. Pe Be’ C aks Be Got, M For, by the parallels, EI: IL ::DH: HM; and, the same EI: IL::GK:KM; therefore by compos. EI? : IL?:: DH. GK: HMK; but, the rect. HMK=IL? ; and therefore the rectangle DH . GK=EI’, or DH: El :: EI: GK. 3 &: » 26 SPECIES AND QUADRATURE OF PROPOSITION XIV. THEOREM, Draw the semi-diameters CH, CIN, CK ; (see last diagram.) Then shall the sector CHI = the sector CIK, For, because HK and all its parallels are bisected by CIN, therefore the triangle CNH=trian. CNK, and the segment INH=seg. INK ; consequently the sector, ClH=sec. CIK. Cor. If the geometricals DH, El, GK be parallel to the other asymptote, the spaces DHIE, EIKG will be equal ; for they are equal to the equal sectors CHI, CIK. So that by taking any geometricals CD, CE, CG, &c., and drawing DH, EI, GK, &c., parallel to the other asymptote, as also the radii CH, CI, CK ; then the sectors CHI, CIK, &c. or the spaces DHIE, EIKG, &c. will be all equal among themselves. Or the sectors CHI, CHK, &c. or the spaces DHIE, DHK, &c. will be in arithmetical progression. And therefore these sectors, or spaces, will be analogous to the logarithms of the lines or bases CD, CE, CG, &ec. ; namely CHI or DHIE the log. of the ratio of CD to CE, or of CE to CG, &ec. ; or of El to DH, or of CK to EI, &c. ; and CHK or DHKG the log. of the ratio of CD to CG, &c. or of CK to DH, &c. Scholium. If the common logarithms are multiplied by 2,302585093 their products will be the hyperbolic logarithms. PROPOSITION XV. THEOREM. If any number of ordinates AD, EF, &c. to GH, between the hy- perbola and the asymptote, are taken in geometrical progres- sion increasing, then will their distances AE, &c., on the asymptote be a series of geometricals decreasing, and AG will represent the sum of the decreasing series. “d » > SECTIONS OF ELEMENTARY SOLIDS. 27 That the geometrical a_? ordinates and their dis- tances on the asymptote are reciprocal geometri- cals is manifest from(Prop. XIV Cor.,) and since the distances AE, EK &c., on the asymptote, inter- cepted by those ordinates compose the whole series of decreasing geometri- cals, thesum of that series will be represented by AG, the part of the asymptote taken up by the series. Cor. 1. The last term Gi, is=AE X EF+GH. For the area ADFE = the area HGit, and if the distances on the asymp- tote are taken indefinitely small, ti may be regarded as =GH, and AD may be regarded as = EF, and if the hyperbola is equilateral, then the space ADFE will bb=AEX AD=GixGH converting this equation into a proportion, we have AK: Gi:: GH: AD, or GH: AD:: AE: Gi. Hence, Gi=AEX(AD or EF)+GH Scholium. It is evident that this proposition is true, however far the series may extend on the asymptote CA. Cor. 2. If the series commence at any point A on the asymp- totes, and decrease to infinity, CA will be the sum of the series ; hence the sum of an infinite series of decreasing geo- metricals is a finite number. PROPOSITION XVI. THEOREM. If any number of portions AK, EK, &c., to G, (sce diagram above) are the reciprocals of a series of ordinates AD, EF, KL, &c., to GH, in geometrical progression ; then will the area or space intercepted by AD and GH, between the curve and asymptote be = the space ADFE, multiplied by the number of terms of the series. For since the space intercepted by each two consecutive ordinates of the series is equal to ADFE, it follows that the whole space intercepted by all the ordinates, or the space ADHG is = ADFE x by the number of terms in the series. 28 SPECIES AND QUADRATURE OF Cor. Hence the area of the interior hyperbolic space HDB may be found, if the exterior space is known. For the interior space is equal the triangle HBD + triangle HnD + the rect- angle nDAG—the exterior hyperbolic space GHDA. Scholium. Let AD=), EF = e, GH = g, and AK the first term of a geometrical series=a ; the last term = z ; and AG the sum = s. Then we shall have a the first term, and z a the last term, and s the sum of all the terms of a geometri- cal series to find the ratio and number of terms. Then by geometrical progression will s=a-+- ar-++ar + ar® +. >. tap multiply by r srartar ape oe he Subtracting the first from the second, s (r—1)=ar"— a tt. bare a hare __a(r® — 1) tl And since, z=are— we r2z—da agecod multiplying by r—1, sr—s=rz—a : bigivy s—a By transposing and dividing;r= —— and in the equation z=ar*—* multiplying by 7 and dividing by a we have 2r Lo alg caer which equation is irreducible by the ordinary methods when zx is the un- known quantity, it being an exponential equation, and can be solved only by logarithms, which see. (Chap. V Trigonome- try) ; but if we proceed to raise 7 to some power, whose value is— the index of that power will be the value of 7 ; this can always be done to approximate exactness. Let the series of geometricals commence at the vertex H, making Gi the first term = a, and Jet n be a given number, then assuming any value tos, z, r, each of the others may be found by the formula above, and hence the area of the exterior and interior hyperbolic spaces may be found, as well as that of any section or segment. SECTIONS OF ELEMENTARY SOLIDS. 29 Scholium 2. If the portion of the hyperbolic curve is small, the ordinates AD, EF, &c., and consequently AE, EK, &c., will be very nearly in arithmetical progression as may readily be seen by the rapid manner in which the ordinates DN, DV, DH, DU, &c., approximate toward the curve as the hyperbo- lic arc is decreased, hence the hyperbolic area may be approx- imately obtained by first dividing it into portions having known ratios to each other, and finding the value of one when the whole will become known. PROPOSITION XVII. PROBLEM. Let it be required to find how far from G on the asymptote AC an ordinate must be drawn parallel to GH in order to inter- cept with GH an area =A 3 Take any distance Gi (see fig. to Prop. XV,) and since AG : it: : GH: Gi we have, the two parallel sides GH, zt, and their distance Gi of the quadrilateral GH to find its area, which call f, then == 2, let a represent the distance, Gi the first term of the geometric series, and GH ~+ tz = 7, the ratio. 7h — ] a Whence we have s =a (—— = the distance from G on the asymptote, that the ordinate must be drawn to intercept with GH, the area A. Scholium. Since the spaces intercepted by the ordinate GH or AD, with any base IN, are the hyperbolic logarithms of the ratios of AD to GH (P. XIII. Cor.) hence the area may be found by taking the logarithms of the ratio, and multiplying it into the base. PROPOSITION XVIII. THEOREM. If ail the ordinates to the diameter of an equilateral hyperbola be reduced in any ratio r, then the area of the hyperbola will be reduced in the same ratio. Let all the ordinates HI, NO, &c., of H the equilateral hyperbola HFI be re- [\ duced to hi, no, &c., forming a hyperbola hFi, then will the area of the equilate- ral hyperbola HII be to the area of the new hyperbola hFi as HI to hi, or as NO to zo. For, if an indefinite num- ber of equidistant parallel ordinates be drawn across the two hyperbolas, the 30 SPECIES AND QUADRATURE OF areas of each will be as the sum of the lengths of all the ordi- nates in each, but by hypothesis, each of the ordinates HI, NO, are to their corresponding ordinates ni, no in the given ratio r; hence, their sums in each, must also be in the same ratio. Cor. 1. Since the areas of similar figures, are as the squares of the lines similarly drawn in each; similar segments of simi- lar hyperbolas, are as the squares of their axes, or of their like diameters. Cor. 2. Hyperbolas of the same transverse axis and abscissa are to each other as their conjugate axes; but if their bases, or ordinates, and conjugate axes be the same, they will be as their transverse axes; and generally, hyperbolas having the same abscissa, are as the rectangle of their major and minor axes ; and, consequently, having the quadrature of any one hyperbola, we may from it find that of any other. Thus knowing the area answering to any abscissa in any one hyper- bola, we can find a similar abscissa in the other; then as the rectangle of the axes of the squared hyperbola, is to the rect- angle of the proposed one, so is the area of the former, to that of the latter. Scholium. 1. It may be shown thata parabola is always greater than a triangle of equal base and altitude, and that the hyperbola is always between the two; since from the nature of the sec- tion through the cone by which the hyperbola is produced, it may vary from a vertical section through the cone forming a triangle, to the section parallel to one of its sides forming a parabola, each of which extremes it can never reach, but may approach infinitely near. When the section passes through the vertice and base of a cone, the section is evidently a triangle, whatever angle it may make with the axis. Let the sec- tion be removed from the vertice by any quantity however small, and it becomes one of the curves described above; it may be an hyperbola or a parabola, either of which may agree infinitely near with the GE en B (yon FC triangle. Hence, the triangle and parabola are not opposite ex- tremes of the hyperbola as the limits of its dimensions, for the hyperbola and parabola may both terminate ina triangle at the same moment, or the two curves may become assimilated when they are infinitely removed from a triangular form ; nevertheless, when the hyperbola hh, parabola PP, and the triangle ACC are AS A A’ SECTIONS OF ELEMENTARY SOLIDS. 31 described on the same base C C, and of equal altitude B A, the hyperbola may always be described between the triangle and parabola; this arises from the nature of the sections from which they are formed. Let the axis be indefinitely extended as when the section becomes parallel to one of the sides of the cone, in which case the two asymptotes AC, AC of the hyper- bola CHC, become parallel to each other, as A’ e, A’ f, in which case the asymptotes may be conceived to vanish, and the curve, at this point, assumes other properties, as we have seen in our investigations ; the axis, after being increased to infinity in the direction, BA, vanishes or becomes changed, and is in- finitely extended in the direction B; but the same cause that brings the asympotes to a parallel position, may, by continuing, cause them to assume a greater distance toward A’ A’ than in the opposite direction, and the axis in such case becomes a definite magnitude, in which case the curve will be inclined inward, as at n 7, when, by extending the axis, it will return into itself and become an ellipse. Scholium 2. As it has been shown (Prop. IX Cor.6) in re- ference to ellipses, it may be also shown that all hyperbolas having the same centre and equal bases, and described be- tween the same parallels, although infinitely produced, are equal to each other, as are also their corresponding sections parallel to the bases, and likewise any frustum or segments intercepted by the parallels. The same may also be shown of parabolas of the same base and altitude, or those with the same base, and between the same parallels, that they are equivalent in area, &c. EQUATIONS TO THE CURVES FORMED BY SECTIONS OF THE CONE. 1. For the Ellipse. Let d denote AC, the semi-axis major or semi-diameter ; c = CM its conjugate ; « = AK, any abscissa, from the extremity of the diam. = DK the correspondent ordinate. Then, seers Xl Ellipse,) AC? : CM? :: AK . KB: DK, that is, d? : c? ::@ (d—x): y®, hence d? y? =c? (d—2?) or dy Uae the equation of the curve. And from these equations. any of the four letters or quan- tities. d, c, x, y, may easily be found, by the reduction of equa- tions, iyi the other three are given. Or, if p denote the parameter, = c? + -d by its definition ; then, (Prop. XII. Cor. 1)d:p: : 2 (d—2): y’, or dy’=p (dv—xz’) which is another form of the equation of ‘ curve. 32 SPECIES AND QUADRATURE OF Otherwise. Or, if d = AC the semi-axis;c= CH the semi- conjugate ;p = c? +d the semi-parameter ; x = CK the ab- scissa counted from the centre; and y = DK the ordinate as before. Then is AK = d—2x, and KB = d +2, and AK. KB = (d—2) X (d+z) = d?—x?. Then, d? :c? :: d?— «2: y?,and d?y? = c?(d’—2?), or dy =c./(d’—«"), the equation of the curve. Or, d: p::@—2: y’, and dy’=p (d?—2°) another form of the equation to the curve ; from which any one of the quanti- ties may be found, when the rest are given. 2. For the Hyperbola. Because the general property of the opposite hyperbolas, with respect to their abscissz and ordinates is the same as that of the ellipse, therefore the process here is the same as in the former case for the ellipse ; and the equation to the curve must come out the same also, with the exception of the signs which are sometimes changed from + to —, or from — to +, because the abscissz on the axis, lie beyond or without the curve, whereas they lie within it, in the ellipse. Thus mak- ing the same notation for the whole diameter, conjugate, ab- scissa, and ordinate, as at first in the ellipse ; then, the one abscissa AK being z, the other BK will be d + 2, which in the ellipse was d—— x; so the signofz must be changed in the general property and equation, by which it becomes da’: c’: : «© (d-2z): y*?; hence d? y’? = c? (dv+2’) and dy = c/(dz+2’), the equation of the curve. Or, using p the parameter as before, it is, d:p::x(d+z): y’, or dy’=p (dx+z’), another form of the equation of the curve. Otherwise, by using the same letters d, c, p, for the semi-axis, semi-parameter, and parameter, and z for the abscissa CK counted from the centre ; then AK=a—d, and BK=2x+d and the property d’: c’?: :(¢—d) X (a+d): y’, gives dy? = c” (x°—-d?) or dy = c./(x’—d’), where the signs of d? and z? are changed from what they were in the ellipse. SECTIONS OF ELEMENTARY SOLIDS. 33 Or, again using the semi-parameter, d : y:x*—d?: y’, and dy =p (x?—d’) the equation of the curve. But for the conjugate hyperbola, (Prop. XII, Cor. 3, Hyp.) as the signs of both 2? and d’ will be positive ; for the pro- perty in that case being CA’: CB’: : CD?+CA’: Dp’, itis d’: c::a+d?:y=Dp’, or d?y’=c" (x*+d’) and dy=c /(x’+a’), the equation to the conjugate hyperbola. Or, asd: p::a°+d’: y’, and dy’=p («*+d’), also the equa- tion to the same curve. On the Equation to the Hyperbola between the Asymptotes. Let CE and CB be the two asymptotes to the hyperbola dFD, its vertex being F; and EF, bd, | AF, BD, ordinates parallel to the asymptotes. Put L AF or EF=a, CB=z, and BD=y. Then, (Prop. XI Hyp.,) Ab. EF =CB.. BD, or @’=zy, the equation to the hyperbola, when the abscisse and 4 | \n ordinates are taken parallel] to the asymptotes. Sens 8. For the Parabola. If x denote any absciss beginning at the vertex, and y its ordinate, also p the parameter. Then AK: KD::KD:p:orz:y:: y: p;hence px=y’ is the equa- tion to the parabola. 4. For the Circle. Because the circle is only a species of the ellipse, in which the two axes are equal to each other ; therefore making the two diameters d and c equal in the foregoing equations to the ellipse, they become y’=dz—z’ y being the mean proportional between z andd— zx, when the abscissa x begins at the vertex of the diameter : and y* = d? — z’*, when the abscissa begins at the centre. Scholium. In each of these equations, we percieve that they rise to the 2d or quadratic degree, or to two dimensions ; which is also the number of points in which every one of these curves may be cut by aright line. Henceitis that these four curves are said to be lines of the 2d order. And these four _are all the lines that are of that order, every other curve being of some higher, or having some higher equation, or may be cut in more points by a right line: BOOK Ii. SOLID SECTIONS OR SEGMENTS OF SOLIDS OF REVOLUTION, CYLINDROIDS, AND PARABOLIC PRISMOIDS AND UNGULAS. DEFINITIONS. 1. Solid Sections or segments are the portions of a solid cut off, or out, from another solid by one or more plane or curve surfaces. 2. If any portion ofa cylinder oracone, is cut off by a plane which is not parallel to the base, the solid section, so,cut off, is called an ungula. 3. Ungulas cut from a cylinder are called cylindric ungulas. Thus the section ACDBEF cut off by the plane ACDB: as also the se- veral sections ACIHGM, GHIK and HIKLBD, are cylindric ungulas. | 4. Ungulas cut from a cone or conic frustum are called conical ungulas. Thus the sections ECDB and AFEGAB are conical ungulas. 5. Portions cut from a sphere by the intersection of two planes, are called spherical wedges or ungulas. Thus the spherical section ACBDAB is a spherical wedge or ungula, SEGMENTS AND UNGULAS. 35 6. A portion cut from a segment of a sphere by a plane perpendicular to the base of the segment is called a second segment. 7. Conical ungulas take particular names according ‘to the figure of the superficial section, viz., parabolic elliptical or hyperbolic. . 8. The portion of a cylinder or cone remaining after an un- gula is taken, is called the complement of the ungula, or ungu- lical complement, and its altitude is equal to that of the un- ula. Thus GNPKIHG (Fig. at Def. 3) is the complement of the ungula GKIH, and ADCEHA (Def. 4) is the complement of the ungula ECDB. 9. An ungulical supplement is what remains of the whole solid after an ungula is taken therefrom. 10. An elliptical cylinder is a cylindrical solid, every super- ficial section of which by planes perpendicular to the axis are equal ellipses. 11. An elliptical cone is one, every section of which per- pendicular to its axis are similar ellipses, and is the solid in- cluded between an elliptical base perpendicular to its axis, and a point as its vertex. 12. A cylindroid is a solid included between two bases of equal perimeter, one of which is an ellipse, and the other an ellipse of a different excentricity or a circle. 13. If a solid have two parallel bases, consisting of dissimi- lar ellipses of different perimeters, or one elliptical and one circular base, of different perimeters. The solid may be called a conoidal frustum. : 14. A parabolic prism is a prismatic solid, whose base is a parabola, and each of whose sections parallel to the base is equal and similar to the base. 15. A parabolic pyramid is a pyramidal solid, whose base is a parabola; and is the solid included between such base, and a point above as its vertex. 36 SEGMENTS AND UNGULAS. PROPOSITION I. ‘THEOREM. If a cylinder be cut by a plane parallel to its axis, the solid section so cut off will be equal to the area of its base multi- plied by its altitude, and its curve surface will be equal to the arc of its base multiplied by its altitude. ‘Let HFCGED, be a section of the cylinder, AD, cut off by a plane HFEG, parallel to the a axis of the cylinder ; then will the solidity of the section be equal to the area GD of the base mul- tiplied by the altitude DC, and its convexsurface q will be equal to the arc EDG of the base multi- plied by the altitude DC. For, if a plane be passed through a cylinder ,, parallel to the axis, it will divide the two bases proportionally, and every section dhec parallel to the base will be divided in the same ratio; hence the curve surface of the cylinder will be divided in the ratio that the cir- cumference of the base is divided. But the convex surface is equal to the circumference of its base multiplied by its alti- tude, (Prop. I. B. II. Hl. S. Geom.,) and its solidity is equal to the area of its base multiplied by its altitude, (Prop. I. B. IIL El. S. Geom.) Hence the curve surface of the portion so cut off, proportional to the section of the base, is also equal to the arc of its base multiplied by its altitude; and its solidity, for the same reason, is equal to the area of its base multiplied by its altitude. PROPOSITION Il. THEOREM. If a plane cut a cylinder diagonally, passing through the oppo- site edges of the two bases, then the cylinder will be divided into two equal ungulas, and the curve surface of each ungula will be equal to the perimeter of its base multiplied by half its altitude ; and the solidity of each will be equal to the area of the base multiplied by half the altitude. Let ALDMA be a plane passing diagon- ally through the cylinder BC, cuttingthe op- » < posite edges of the two bases at A and D, and the two ungulas ADB, ADC will be equal in surface and solidity, and the curve surface of each is equal to the circumference of its base multiplied by half its altitude, andy their solidities are equal each, to the area of its base multiplied by half its altitude. For, the two ungulas are symmetrical (Def. 19, B. II. HZ S. Geom.,) being so- =< lids, similarly formed, in opposite sides of BN. the plane ALDMA as a base, and hence SEGMENTS AND UNGULAS. 37 are equal each to each. Moreover, let a plane be passed through the cylinder, parallel to,.and at equal distances be- tween the two bases, forming the superficial section ILKMI, and the two ungulas will be cut by that plane in the same ratio, since they are similar solids, and since they are cut by a plane parallel to and at equal distances from their bases: hence the solid section, or partial ungula ALMI, is equal to the partial ungula DMLK, both in surface and solidity; and for the same reason, their complemental ungulas ALMKC, ILMDB are equal, each to each, both in surface and solidity. Therefore, the cylindric section KB between the two parallels ILKM and BEDF, is equal to the ungula ABEFDA, both in surface and solidity; but the cylindric surface of KB is equal to the circumference BEDFB of the base multiplied by IB, equal to half the altitude of the ungula ABD; and the solidity of KB is equal to the area BEDF multiplied by 1B, equal to half the altitude of the ungula. Hence, &c. PROPOSITION III. THEOREM. [fa cylinder be cut diagonally by a plane which bissects the base, the ungula cut off by such plane will be equal to its cylindric surface multiplied by one-third of the radius of the circle of the base. Let AFEC be an ungula, cut off from the , cylinder ABDC by the plane AFH, bisecting the base CFDE in FE, then will the ungula AFEC be equal in solidity to its cylindric surface multiplied by one-third of the radius IC of the base. For, conceive the ungula to be divided in- to cylindrical elementary pyramidals (Prop. XIIL. Schol. 1 and 4, B. Ill. L7. 8. Geom.) by planes parallel to the axis of the cylinder, and p all passing through the centre I of the base, F - then these will all be perfect pyramidals, since lines drawn from every point in the cylindrical base to the centre I, lie wholly in the solid, or in the plane surfaces of the solid; and since all parts of the solid are included between the cylindrical base of the ungula and centre I, in right lines. And these pyramidals (Prop. XIV. B. III. Ei. S. Geom.) are each equal to its cylindrical base multiplied by one-third of its altitude, viz., one-third of CI. Therefore, the whole ungula, being made up of all the pyramidals, is equal to the sum of all their bases multiplied by one-third of the common altitude CI. 38 SEGMENTS AND UNGULAS. Cor. 1. If a line ec be passed along the axis of the cylin- der, and the edge of the ungula, through its whole extent, be- ing always perpendicular to the axis, in every position ec or gh, or an the solid included within the surface described by the motion of this line, together with the ungula, is equal to the cylindric surface of the ungula multiplied by half the radius of the base. For let the cylindric surface be divid- ? ed at pleasure, by planes passing through @ the axis, and the several divisions will , a all be elementary portions of the cylinder, || and (Prop. II], Cor. B. Ill, Hl. S. Geom.) each will be equal to its cylindrical base x multiplied by half its altitude, or half the C B radius of curvature of the cylinder. dl Cor. 2. Hence the section included between the ungula and the axis of the cylinder is equal to half the ungula ; since the ungula, is equal by the proposition to its cylindrical surface, multiplied by one-third of the radius of the base, and the two sections together, are equal to the same sur- face multiplied by 1+-4=4 the same radius. PROPOSITION IV. THEOREM. From the complement of two similar ungulas whose bases are together equal to the base of the cylinder, a cone may be taken of equal base and altitude to that of the cylinder, when there will be left a residual portion equal to its cylindric surface multiplied by one-third of the radius of the base. Let DEFC be the complement of the two similar ungulas DEF A, and CEFB whose bases AEF and BEF are together equal to the base of the cylinder, and there may be taken a cone DGCHL of equal base and alti- tude with the cylinder, and the portion of the cylinder remaining will be equal to its cylin- dric surface multiplied by one-third of the radius IF of the base For since each of the ungulas are cut off by planes pass- ing from the centre I, of the lower base of the cylinder, and through opposite edges of the upper base, those planes just SEGMENTS AND UNGULAS. 39 pass along the sides, of the cone touching it on the two opposite sides, through its- whole length, and be cause the cone terminates in a point I in the centre of the lower base, there exists a portion FIDHC on one side of the cone, and a similar portion EIDGC on the other side, in- cluded between the surface of the cone, and that of the cylinder; and these two portions consist of regular pyramidals with cylindric bases and their vertices all centre in I, at the vertex of the cone ; for if lines be drawn from every point in their cylindric surface to the centre I, those lines will pass through every point in the solid portions to which those sur- faces belong. And because the cone DGCHL is supposed to be taken from the complement, these pyrimidals are elementary portions of a conesected cylinder, (Prop. XHI. Sch. B. UL. Hl. S. Geom. and Def. 2,). Hence (Prop. XII. Sch. 5, B. OL, £7. S. Geom.) they are equal to their cylindric bases multiplied by one-third of the radius IF. Cor. The opposite ungulas DEFA, CEFB and the com- plements IFDHC, IEDGC of aconesected cylinder are seg- ments which are in the same proportion to each other in their cylindric surfaces, as in their solidities. PROPOSITION V. THEOREM. The solidity of an ungula, whose base is less than half the base of the cylinder, is equal to its cylindric surface, multiplied by one-third of the radius of the base of the cylinder, minus the pyramidal segment of a cone, whose base is the base of the ungula, and whose vertex, is the point where the diagonal plune produced forming the ungula, would cut the axis of the cylin- der or its axis produced. Let BSHK be an ungula whose base SHK va is less than Jhalf the base of the cylinder, and its solidity will be equal to its cylindric sur- y face, multiplied by one-third of the radius of Le the base of the cylinder, minus the pyramidal ie segment KSHOof acone whose baseisKSH, JN [pp yy the base of the ungula ; and whose vertex | is O, the point where the plane BSH pro- duced to IE cuts the axis of the cylinder. My For the solidity of the ungula BFED, is equal to its auntie dric surface multiplied by one-third of the radius of the base of the cylinder (Prop. III,) and the section KSHFED minus 40 SEGMENTS AND UNGULAS. the conical section KSHO is a portion of a coneseeted cylinder, which (Prop. XIU, B. Ill, Hi. S. Geom.) is equal to the product of its cylindric surface multiplied by one-third of the radius of the base, and because this section of the conesected cylinder is formed by a plane passing through the centre O, cutting it in such manner that every point in its cylindric surface may be connected by right lines with the vertice O, these lines being included in the same solid ; this section is also equal to its cylindric surface multiplied by one-third of the distance OD of this vertice from the cylindric surface; hence, the other two portions of the ungula, viz., the conical KSHO, and the ungula BSHK are equal to the remainder of the cylindric sur- face of the ungula BFED, viz., the cylindric surface of the small ungula BSHK multiplied by one-third of the radius OD of the base of the cylinder. Now, if from this product we take away the conical segment SHKO we shall have the ungula BSHK. Hence, &c. Scholium. If on the diagonal plane BSH of the ungula as a base, the pyramidal BSHP be described, P being the vertice of such pyramidal situated in the centre of the circular sec- tion of the cylinder, and in the plane KSH of the base of the ungula produced, the sum of this pyramidal and ungula is equal to their cylindric surface multiplied by one-third of the radius of the base of the cylinder ; and the pyramidal BSHP is equal to the conical portion, KSHO. For the pyramidal BSHP together with the ungula BSHK, constitutes the regular elementary pyramidal, BSHKP, with a cylindrical base, which (Prop. XIV. B. IIL, #2. S. Geom.) is equal to its cylindrical surface multiplied by one-third of the radius, PK. of the base, and because the sum of the ungula, BSHK, and pyramidal section of a cone KSHO, is equal to the same product (Prop. V.) it follows that the pyra- midal section BSHP is equal to the conical portion KSHO. PROPOSITION VI. THEOREM. The solidity of an ungula KLMD whose base is greater than half the base of the cylinder, is equal to its cylindrical sur- face multiplied by one-third of the radius of the base of the cylinder, plus a conical segment, whose base is the base of the ungula, and whose vertice is the point S, where the plane ELM cuts the azis of the cylinder. SEGMENTS AND UNGULAS. 4} Or the solidity of the ungulu is equal to the same product plus the pyramidal ELMO, whose base is the elliptical sections ELM and whose vertice is QO, the centre of the base of the cylinder. For through the point S pass the plane IF KC, and the ungula will be divided into two portions, one of which, EFCK, is equal (Prop. II.) to its cylindrical surface multi- plied by one third of the radius of the base of the cylinder; and the portion FCK LMD, from which, if we take a conical sec- tion LMDS, we shall have a portion of a conesected cylinder remaining which is equal to its cylindric surface multiplied by one third of its radius. Hence, the whole ungula ELMD is equal to its cylindric surface multiplied by one third of the radius of the base of the cylinder plus a segment of acone, whose base is the base of the ungula and vertice, the point where the plane ELM cuts the axis of the cylinder. Again, if we pass the two surfaces MOK and LOE meet- ing each other in the right line OH, those surfaces will cut out the pyramidal ELMO, whose base is the elliptical section ELM, and whose vertice is the centre of the base of the cylinder, which if we take from the ungula, will leave a portion which is equal to its cylindrical surface multiplied by one-third of the radius of the base of the cylinder, since it con- sists of a portion which may be divided indefinitely by planes passing through the vertice O, and each of those portions so divided will be perfect pyramidals, and will remain elemen- tary to the whole section. (Prop. XIII, Schls. B. Ill, £2 S. Geom.) Hence, &c. Cor. 1. Hence the pyramidal ELMO, is equal to the coni- cal segment LMDS. Cor. 2. As the plane section ELM is an ellipse, the surface of the pyramidal ELMO contiguous to the other portions of the ungula so divided is a conical surface ; for if the several points in the elliptical curve be connected with the vertice O, those lines must include a conical surface, since the section of a cone by a plane passing through both sides is an ellipse. Cor. 3. Ifthe ungula is so cut as to include the whole of the base BLDM, the circumstances will still be the same, and the ellipse of the section ELM becomes perfect, and the pyra- midal ELMO becomes a perfect oblique cone, and is equal to the conical body LMDS, which, in such case, becomes a per- fect right cone, whose base is the whole base BLDM of the cylinder, and whose altitude is half the altitude of the cylinder. 4 42 SEGMENTS AND UNGULAS. PROPOSITION YH. THEOREM. If in acylindric ungula ADB, whose base is equal to that of the cylinder, a cone be described on the same base and of an equal altitude with that of the ungula, the cone will be equal to two-thirds of the ungula; and each solid section of the cone made by planes parallel to the axis of the cylinder, and perpendicular to a plane passing through the axis of the cone and cylinder, will be equal to two-thirds of the corresponding section of the ungula cut off by the same plane. Let the oblique cone ABD eee of equal base and altitude to that of the ungula ADDNB, be supposed to be inscribed in the ungula, and the soli- dity of the cone will be equal to two-thirds of the ungula, and each solid sec- re! I P tion IRcD, PveD of the cone made by planes parallel to the axis of the cylinder, and perpendicular to a plane passing through the axis of the cone and cylinder, are equal to 2 their corresponding sections [RMbD, PuNiD of the ungula cut by the same planes. For, conceive an indefinite number of parallel planes IRM), NiPv, to be passed through the cone and ungula inde- finitely near to each other, and the two bodies will be divided into an indefinite number of strata, the sum of which, in each solid, will be in the relation of the magnitudes of the two bodies ; and the corresponding strata in each solid will be in- the relation of the superficial sections contiguous to such strata, in each, since by hypothesis the strata are indefinitely thin, so that no appreciable space intervenes between them ; hence, if the solidities of any stratum, or number of contiguous strata in each solid be represented by the superficial sections passing through such stratum or associated strata, they will exist in each in the relation of the superficial sections in each, made by the same planes. Now the superficial sections IRe, Pve &c., made by planes parallel to the side AB of the cone, are all parabolas, (by def.) and all the corresponding sections of the ungula are rec- tangles circumscribing the several parabolas, since each side of those sections Mb, Nr of the ungula are parallel to the opposite sides RI, vP, passing through the base. But the area of the parabola (Proposition VI, Book I,) is equal to two-thirds of its circumscribing rectangle or parallelogram, and as each of the several rectangles circumscribe their re- Ata te Ne | SEGMENTS AND UNGULAS, 43 spective parabolas, throughout the whole extent of the two solids, it follows that not only is the cone equal to two-thirds of the ungula, but also that each solid section, or segment of the cone, cut by the planes RMOI, &c., is equal to two-thirds of the corresponding segment of the ungula cut by the same plane. Cor. Hence, as in Prop. VIII, Cor. B. Ill, El. S. Geom. a cone is equal to one-third of its circumscribing cylinder of the same base and altitude. Scholium. If the cone and ungula are cut by planes pa- rallel to the plane ADDN, the superficial sections of the ungula will be ellipses, or portions of ellipses, and the corresponding sections of the cone will be parabolas. PROPOSITION VIII. THEOREM. Two cylinders erected on the same base, and beiween the same parallel planes are equivalent or equal in solidity. Let ACDB and AEBD be two cylinders erected on the same base BD, and included between the same parallel planes BD and EAC, and the two cylinders will be equivalent. For let the cylinder ABDC be divided _,, ‘ into sections BDF f, FfgG by planes pa- rallel to the base, and let each of the sec- tions so divided, be removed from their positions in the cylinder ABDG, so that their several bases shall agree with, and become sections of the oblique cylinder AEBD ; viz., let the solid sections fF'Gg, be removed to the position Hhot, and let the other sections be similarly posited in reference to the two cylinders, so that the bases Hh, Mn, &c. of the several solid sections may be in- cluded in the oblique cylinder AEBD, and these several sec- tions will still be equal in their altitude to AB, the altitude of the cylinder from whence they are severally derived. Let, now, the number of these solid sections be indefinitelyfincreas- ed, and the altitude of each will be indefinitely small ; and hence the right cylinder will become identical with the oblique cylinder AEBD, and will still have the same altitude AB. Cor. 1. Since the base of the oblique cylinder AEBD is a circle, every section parallel to the base is likewise a circle, and equal to the base. Cor. 2. Hence every section of the oblique cylinder made by the planes perpendicular to its axis, is an ellipse. Therefore, if referred to its axis, it becomes an elliptical cylinder, but when referred to its base, it is circular. 44 _ SEGMENTS AND UNGULAS. ‘PROPOSITION IX. THEOREM. Cylindrical ungulas of the same base and equal altitude are equivalent. Let ABD and EBD, be two ungulas described on the same base BD ; and let their common altitude be AB, and the two ungulas will be equivalent, or equal in solidity. For each section FC or IL of the x two ungulas parallel to the base BD, have the same relation to the re- spective sections of their cylinders ; viz., the ungulical section FC has the same relation to the cylindric section Fh, as the ungulical section IL has to the cylindrical section IN ; and the equality of this relation remains through the whole alti- tude AB of the ungulas, and by Prop. VIII, it appears that two cylinders erected on the same base, and between the same parallels are equal in solidity. Therefore, the ungulas erected on the same base, and of the same altitude, are equi- valent. Scholium. The above proposition is manifestly true, whether the base of the ungulas is equal to that of the cylinder in which it is erected, or less than that base. Cor. 1. As the same arguments would apply to conical un- gulas, it may be inferred that conical ungulas of the same base and equal altitude are equivalent, if the cones from which they are taken are similar. Cor. 2. Since cylinders with equal bases are proportional to their altitudes, cylindrie ungulas on the same base are pro- portional to their altitudes. PROPOSITION X. THEOREM. Ifa cone be cut by a plane passing through the centre of the base, the solidity of the ungula formed by such plane will be equx! to its curve surface multiplied by one-third of its dis- tance from the centre of the base. Let the cone ABC be cut by the plane DEF, cutting off the ungula EFDB ; also by the plane EDG cutting off the ungula GEDB, or the ungula GEDF, and the solidities of the several ungulas will be equal to their curve surfaces multiplied by + of the distance LL of ce surface of the cone to the centre of the ase. SEGMENTS AND UNGULAS. eae For the sections may each be divided into an_ indifinite number of pyramidals or elements (Prop. XIII and XIV, B. Ill. Hi. S. Geom.) by dividing their curve surfaces as bases, and by passing the planes of division through the centre I of the cone’s base, at which point the several vertices of the py- ramidals all centre ; and because those pyramidals are all perfect; viz..as they include all the space intercepted between their bases and vertices in right lines, each one is equal to its base, multiplied by one-third of its altitude, and since the alti- tude is equal in each, and equal to the distance of the curve surface of the cone from the centre of its base, IL is that altitude, and hence IL is the common altitude of each of those pyramidals; hence the sum of the pyramidals con- stituting any section, is equal to the sum of their bases, con- stituting the curve surface of such section multiplied by one- third of IL. Cor. Hence, if a cone be cut by a plane passing threugh the centre of the base, the solidity of the cone, and its convex surface is divided in the same ratio, since the cone (Prop. [X. B. Ill., Hl. S. Geom.) is equal to its convex surface multiplied by 1 of its distance from the centre of the base. PROPOSITION XI. THEOREM. If a cone be described in a cylinder, and two parabolic ungulas be cut by planes passing through the centre of the cone’s base, the ungulas so cut, will be equal to two thirds of the unguli- cal complement of the cylinder, of equal base and altitude to the parabolical ungulas. Let BDS be a cone described in the cy- linder ABDC, and let EF Bd, and EF DM be two equal parabolic ungulas, described on and including the whole base of the cone, and the two ungulas will be equal to two- thirds of the ungulical complement BEDF-NP of the cylinder, of equal base and altitude. F For let the cone BDS be brought in the position BDA. and the parabolic ungula IEFD of the cone BDA will be equal to two-thirds of the semi-ungulical complement NPEFD of the cylinder, (Prop. VII ;) and because conical ungulas on the same base and equal altitudes are equivalent (Prop. IX Cor.,) the ungula IEF'D of the oblique cone BDA is equal to the un- 46 SEGMENTS AND UNGULAS. gula EFDM of the right cone BDS; hence the parabolic un- gula EFMD is equal to two-thirds of the semi-ungulical com- plement NPEFD of the cylinder ; and the two similar ungu- las dEFB, MEFD are, together, equal to 3 of the ungulical complement BEDFPN of the cylinder. Cor. 1. Hence the conical complement EFdMS of the parabolic ungulas=the com- plement PMNd EF +the cone PMNds, is equal to two-thirds of the cylindrical ungula APNB. For if the whole cone BDA or BDS is=2 of the ungula BDA, (Prop. VIII,) and if the two conical ungulas dEFB, MEFD, as shown above, is = 2 of the com- plement P NBD, then must the complement EFdMS= 2 the cylindrical ungula APNB | Cor. 2. The small cone dMS, whose base passes through the vertices of the ungulas, is = to 4, of the cone BDS, since the diameter of its base is.necessarily = } that of the larger, and since they are similar solids ; for similar solids are to each other as the cubes of their like sides, (Prop. XXXV,B. Il, El. S. Geom.) and cube of 1 is 1, cube of 2 is 8 ; hence 178: : cone dMS: cone BDS. PROPOSITION XI, THEOREM. If a right cone be cut by a plane perpendicular to the plane of its base, the convex surface of the cone, and the plane of the base, will be divided in the same ratio by the cutting plane: Let DEF be a plane cutting the cone \ ABS, perpendicular through the base AEBF, = and the convex surface of the eone will be Ue divided by the cutting plane, in the same | ratio that the surface of the base is divided. [fe WA acy H F For since it is shown (Prop. XXII, B. Il, El S. Geom.,) that the convex surface of a pyramid and its base is divided in the same ratio, by a plane perpendicular to its base, and because a cone may be considered as a pyramid, whose cone vex surface consists of an indefinite number of planes, which are indefinitely narrow, it follows that if the convex surface and base of a right cone are cut by a plane perpendicular to the base, those surfaces will be each divided in the same ratio. Cor. 1. Hence if any portion of the base of a right cone is taken, the portion of the curve surface perpendicular above it, will be to the whole curve surface of the cone, as such portion of the base is to the whole base. SEGMENTS AND UNGULAS. AT Cor. 2. Hence, also, if a regular polygon, for instance a. square EF HG, be described in the base of the cone, and if on each side of this square, a plane be raised perpendicular to the base, the portion of the conical surface, cut off toward the axis, is to that of the rectilineal polygon EFHG, which cor- responds to it perpendicularly below, as the surface of the cone is to the area of its base; or as the slant side AS of the cone is tothe radius of the base; and, in fact, whatever figure be inscribed in the base, if we conceive a right prisma- tic surface raised perpendicular from the perimeter of the figure, it will cut off from the conical surface a portion which will be to it in the same ratio. Scholium. The solid sections DEFB, LGE, &c., are hyper- bolic ungulas, (Def. 7.) And if ungulas DEFB, LGE, &c., are taken from the cone, the remaining portion or compliment will be equal to its curve surface multiplied by.one-third of the distance of the curve surface of the cone from the centre of the base, + the surface of the plane hyperbolic sections ~ multiplied by one-third of their respective distance from the same centre of the base. Cor. Hence the portion of the cone included between the centre of the base, and that portion of the convex surface left by the ungulas, since it is equal to its convex surface, multi- plied by one-third of its perpendicular distance from the centre, the quadrature of which we have shown to be attainable, be- comes known in absolute terms, or its cubature is attained without regard. to the circle’s quadrature. PROPOSITION XIII. THEOREM. If an elliptical cylinder, and a circular cylinder, have equiva- lent bases and equal altitudes, they are equal in solidity ; and any ungulas similarly cut from each, with equivalent bases and ahitudes. are equivalent. For it has been shown (Prop. II. B. Ill. Zl. Geom.) that the solidity of a cylinder, with circular base, is equal to its base multiplied by its altitude ; and because this is true of any prism, whatever be the form of its base, (Prop. XVI. B. Il. El. 8. Geom.) it must be true of a cylinder with an elliptical base. Therefore, an elliptical and a circular cylinder of equivalent bases and equal altitudes, are equivalent. 48 SEGMENTS AND UNGULAS. Again, let GEFB be an ungula cut from a circular cylinder ABDC; and if the base of this cylinder, including that of the ungula, is drawn out or elongated in the direction from BA toward DC, so as to become elliptical ; and if every section of the cylinder parallel to its base should become equal and similar ellipses, then the cylinder becomes an ellipti- cal cylinder. E Now, if planes should be passed through the cylinder par- allel to its axis, and in the direction AC or BD of its elonga- tion, in passing from a circular to an elliptical cylinder, this transformation will have been effected by an elongation of these planes proportional to the elliptical elongation of the base, or of the solid ; and, as this would be true in every par- allel plane, it follows that the elongation of those planes may be regarded as a measure of the ratio of enlargement of the solidity, by the same means. And as every such parallel sec- tion becomes enlarged in the same ratio, any specific portion of such section must suffer the same specific enlargement. And as the increase of any solid sections through which any portions of the plane sections pass, may be measured by the increment of those planes, it follows, that both the ungula and its complement are each increased in the ratio of the in- crements of the parallel sections passing through each in their enlargement. And the ungula which was cut from a circular cylinder, becomes the ungula of an elliptical cylinder, which ungula has become enlarged in the ratio of the enlargement of its base; and the solidity of the ungula from the elliptical cy- linder, is to the solidity of the ungula from the circular cylin- der, as the base of the former to the base of the latter. And the same would be true, if instead of an enlargement of the circu- lar cylinder to form the elliptical cylinder, it should be con- tracted in the direction AC, so as to give it eccentricity in the other direction ; but, by hypothesis, the cylinder and an ellipti- cal cylinder have equivalent bases; hence, ungulas similarly cut from each, are equivalent. Cor. 1. Hence, also, if a cone with a circular base, and one with an equivalent elliptical base, have equal altitudes, their solidities will be equivalent; and ungulas with equivalent bases and equal altitudes cut from each, are equivalent. Cor. 2. Ungulas, whose bases are the like parts of circular or elliptical cylinders, are as their altitudes; and it having been shown that they are also as their bases when their alti- - $< SEGMENTS AND UNGULAS. 49 tudes are equal, it follows that they are generally as the rect- angle of their bases into their altitudes. Let ABEC and ABED be two ungulas cut from any cylinders, cir- cular or elliptical, such that AB shall be the same in each, and such that ID, the altitude of the base in | the first, shall be equal CE, the alti- D®auy titude of the second, and IC the al- titude of the base of the second shall be equal to the altitude ED of the first,-and the two ungulas so described will be equivalent. PROPOSITION XIV. THEOREM. The solidity of a cylindroid is equal to the product of the sum of the areas of the two ends, and four times the area of a pa- rallel section, equally distant between the two ends multiplied by 1 of the height. Demonstration same as for the prismoid, Prop. XXXV, Cor. B. Il, El. S. Geom., which see. PROPOSITION XV. THEOREM. If an ungula is cut from a sphere by two planes which inter- sect each other, not in the centre of the sphere, then the solidity of the ungula will be equal to its spherical base multiplied by one- third of the radius of the sphere, plus the products of the super- ficial sections of the ungula multiplied by one-third of the per- pendicular distances of the planes of those sections from the centre of the sphere, estimated from the sides of those planes op- posite the ungula. Let ABDE be an ungula cut by the planes HEI and FDG intersecting each other in the line AB, not passing through the centre of the sphere, then // 1 will the solidity of the ungula ABDE be 4 ;* equal to the aiherteal aurtice ADBEA, [psy by — - 3 (3) = the elliptic ungula EFCB; putting A for the height of the ungula, D and d for the diameters of the base and end, or top, of the frustum, respectively. 94 ON REVOLOIDS AND UNGULAS, | pee D—d conic frustum, the remainder will express the complemental If this value be taken Fro ht ha = the whole — 2 —d’) x n — D circular 3) RB fond BD ; segment, whose height isp - + py X (fre oeel BD—D+d th segment whose height is——— rr a TH a x —nd? + Dx 3 segment whose height 1 iy ak bed xX seg- BD—D ise ment whose height is “ped . ~ 1 cane) If the points D and A coincide, the section EFC becomes, a whole ellipse, and the formula above, become } Dhn X D*—d./Dd Dita. wat eee elliptical ungula ACB - - - - - (5) And the complemental ungula ACG = dhn X D vy Dd—d’ TADEST Fe ITAY, ae oc woth tbat Io MetabpecadtD If the angle CDB be equal to the angle VAB, the section will be a parabola, whose axis is CD, and base EF =2y (AD x DB) =2 v(D—d) x d, and its area, by prop. VI, B. I, =2DCXEF =4DC vy (D—d) 1) d; and therefore the ex- an Dd—ad? presssion becomes De Fie (Dxsegment EBF ——, De” + DCO VDd—d@ = the ee ungula EFBC - - - (7) If this be taken from 1 hn ee, the remainder will ex- press the megs tk ungula x bee i Viz. 3 yh X($dVDd—a@ 55 stp 7 X segment whose height is D? 1h or HD 1 x(( (D—d) 4./d (D—d) —nd*+ D Xsegment whose aD height is—7) - - - - -- +--+ +--+ * 5 ON REVOLOIDS AND UNGULAS. 95 If the angle CDB exceed the angle VAB, the N section will be a an hyperbola, whose transverse (a CGx CD axis is CN, but the transverse axis = FG pees ; dxXCD CP being drawn parallel to VA, or ppp DR DB and the conjugate also =GC v PD =d/H-7_DB and the area of the hyperbolic section substituted in the gen- eral formula (1) will give the solidity of the hyperbolic ungula. Scholium 3. If from the perimeter of the section EFC there be projected, the surface EIF by y perpendiculars AP,CI, &c., to the base of the cone, the surface EBFIE on the base, will be to that, perpendicular above, viz., ECFB on the surface of the cone as OB to AV; as the radius of the base to the slant height of the cone, (Prop. XII, B. II.,) moreover the area of the section EFC is to the area EIF as DC: DI; hence the 4 convex surface FCEB = S = = ms TS Ae ee ee a cn : And the area FIE Sia FCEF DC Let ECEF=B and the expression for the surface of the base DI FIE of the ungula FIEC becomes Da * DI Ean a) Letthe base EFB of the ungula EFBC be called A, and the area EBFIE will be Mae x B- - - - - - (8) CB DI And consequently S = TB * A— Da * B - - - (4) If from ne x base AFBE of the cone) = the convex sur- face of the whole cone, there be taken that of the ungula found above ; the remainder ay Xx (PAEF +p, x FCEF) - (5) will express the convex surface of the remaining part EFCVA of the cone. And if from the value Jast found, there be taken OB V x circle CG, = aS xX circle GC the convex surface of the + 5 T4 is 96 ON REVOLOIDS AND UNGULAS. % cone GVC, the remainder on x (TAEF x DC x FCEF — circle GC) - - - = gir (6) will express that ue the SOUIICHiCeE EFAGC. If the section FCE be an ellipse, the surface of the ungula 3 : I will be Nea X (circular segment FBE — = X elliptical seg- VB GC’ x DI ment ee ; and hence OB AB? x (GC—AD) wie “Ap * (segment circle AB, whose height is AB X x EBF— = / = x (eer x D2 DB (D—d) ~ Winn oh 5 Mote ; ; DB-(Dowa) * / DBO oii) x seg. cir. AB, whose B—D--d heightis D x hi. huge ee yi eg eee OL And the value of the surface EFCVA, (Formula 5,) will VB become, OB * (circ. seg. FAE X _ x ellip. seg. FCE) (8) .. _ v(4h?+(D—d)’) ad. 2D+d—AD Or itis = —~ yg, (PAE + 7 X dA oe 7 d—A S=B x seg. circle AB, whose height is D x——7— ce (DY And the value of the surface of the complemental ungula, | VB ; (Formula 6,) will become 6p B (cir seg. FAK + pa * ellip. seg. FCE — circle CG) = v(t +(D~) 2 1 — D— AD & x =m Jun AD * 88: circle AB, whose x (-n@+ FAE + —AD height is D x “72 oo) 4.) Ce a i a ETH) When D coincides with A the expression will become, VB j _ n/(4h?+(D — d)’) D+d x (D:——* ya) for the convex surface of the ungula RE ie eye = 8m ee ye mole ee era a) VB GC And Gg X Aly(AB x GO) x n=2n x VBx AIA/ 7p 2 , e ; ON REVOLOIDS AND UNGULAS. 97 4h? D—d)’) D+d a BS 0! a /Dd for the oblique cone AV. os i ws = Also a x (AIV(AB x bani _ / (4h? + (D—d)?) eo. Dr ar Xx ee /(Dd—d?)) for the complemental elliptic ungula ACG. - - - - - (18) If DC be parallel to AV, or the section a parabola; since its area Bis 4 DC x DF = 4 DC y(AD x DB,) the general formula for the ungula will become Ze x seg. (FBE— 4 DI y(AD x DB)) = Ae x [seg. FBE to height DB 2 (D — d) y(d(D— @))) for the convex surface of the parabolic ungula FEBC. - (14) And the expression in Formula 5 will become, 214.([D—.qd)2 op X (eg. FAE + $DI v(AD x DB)) = en x [(seg. FAE to height AE + $(D—d) v(d(D—d))] for that of the part AEFCV. Also, that in Formula 6, will be ve x seg. FAE + 4DI 2 eS 2 v(AD x DB) — AD: x 2) eee ment FAE to height AD + 2 (D—d) yv(d(D—d))’— nd, for that of the complimental parabolic ungula FAEDG. (15) If the angle CDB be greater than the angle VAB, or the section be an hyperbola, its area being found, and substituted for B in the general formulz, will give the surfaces ofthe hy- perbolic ungulas. If the hyperbolic section be perpendicular to the base, DI will vanish, and the expressions will become, vee ae ae) x seg. of cir. AB, whose height is 3 a for the curve surface of the perpendicular ungula CIB. (16) 4h? +(D—d ee x seg. of the circle AB, height that of the remaining part FAYE Ge Lic il Ree RS 8 Ty 8 v (4h? +(D — d) ) 18? a d pets — nd*,) for that of the complemental perpendicular un- ee ALGO St) ea a ONS! oe (18) he xX seg- , for X (seg. of the circle AB, whose height is 98 ON THE RECTIFICATION AND BOOK IV. ON THE REVOLOIDAL CURVE, THE RECTIFICATION OF THE ELLIPSE, AND OTHER CURVES, AND ON THE QUADRITURE OF THE CIRCLE, &C. DEFINITIONS. 1. Tue revoloidal curve is the curve forming the contour of one of the facial sides of a revoloid; since this designation may apply to any revoloid, therefore, if the revoloidal curve is mentioned without reference to the species, the curve of a right revoloid is understood. 2. If the revoloidal surface is extended on a plane, its con- tour is called a plane revoloidal curve ; and the surface is called a plane revoloidal surface. 7 3. The vertices of a plane revoloidal surface are the two angular extremities, as D and E. 4, The vertical or transverse axis of the plane revoloidal surface, is the right line drawn through the vertices, as DE. 5. Its conjugate axis is a line drawn at right angles to its transverse, which it bi- sects, terminating in the curve, as AB. 6. A quadrant of a revoloidal surface is a portion cut off by the two axes, as ACD or BCD. E 7. Any area bounded partly by curves and partly by right lines, is sometimes called a mixtilineal area or space. QUADRATURE OF CURVES. 99 PROPOSITION I. THEOREM. The vertical length of one of the plane surfaces of a quadran- gular revoloid,is equal to the semi-cireumference of its in- scribed cirele, and the length of any double ordinate to its con- jugate diameter, is equal to the arc of that circle cut off by such ordinate toward the extremity of the conjugate diameter. Let ADBE be a plane surface from a quadrangular revoloid, and AFBG its in- scribed circle, and the vertical length DE of the revoloidal surface will be equal to the semi-circumference FBG, and the length of the double ordinate HI will be equal to the arc LBM cut off by such ordinate. For since (Def. 8, B. III,) the vertical section of a right revoloid through the centre of its opposite sides, is a circle, and since in a quadrangular revoloid this cir- cle is such as may be described on a dia- meter equal to the conjugate axis of the revoloidal surface, and because each of the facial surfaces of a revoloid extends from one vertice to the other, passing through half the circum- ference, it follows that its length, DE, is equal to half the length of that circumference which is also equal to the semi- circumference FBG. Again, since any ordinate HI drawn parallel to the trans- verse axis DE, is the representative of a parallel to a vertical section through the line DE, it follows that the section formed . by a plane passing through the ungula of which this face is the surface, is a segment of a circle whose chord terminates in the curve forming the edge of the ungula, and the section of this ungula is similar to the section of its contiguous ungula by a plane perpendicular to this section; which section, through the contiguous ungula, may be represented by a segment LBM ; for a quadrangular revoloid has its sides at right an- gles to each other in a plane perpendicular to the transverse axis. Hence, if HI is equal to the are containing a segment equal to the segment LBM. it is therefore equal to the length of the arc LBM. . A 100 ON THE RECTIFICATION AND PROPOSITION Il. PROBLEM. To make a plane projection of the facial surface of a right quadrangular revolovd. With a radius EA, equal the radius of the circle forming a vertical section of the revoloid, describe a circle ACBD, and draw the diameter AB, and from E perpendicular to AB draw the lines EF, EG, each equal to one-fourth of the circumfer- ence of the circle, ACBD, or equal to one-fourth of the cir- cumference of the revoloid. Divide these lines into any num- Th ber of equal parts, as 1, 2, 3, 4, &c., on the line EF. In like manner divide each quadrant of the circumference into the same number of equal parts, 1. 2, 3, 4, &c. ; through the divi- sions on the circumference draw lines from 1, 2, 3, &ce., parallel to DC, and through the divisions on the line EF draw lines both ‘ways parallel to AB, as 7g, 6f, 5e, &c., and where these lines meet the former corresponding lines through the divisions cor- responding to the same numbers, will be points in the curve forming the boundary of the surface, through which, if a curved line a, b, ¢, d, e, f, g, &c. is drawn, this line will represent the revoloidal curve, and the space enclosed will represent the plane surface of a right quadrangular revoloid. Scholium 1. The nature of this curve is such, that, as it pas- ses off at the vertices, it reproduces itself again, passing into % 4 QUADRATURE OF CURVES. 101 another curve of the same character but of opposite curva- ture, for where it passes the vertices F and G, the signs be- come changed from positive to negative and from negative to positive, so that the curve is reproduced indefinitely as the axes EF and EG are continued. Scholium 2. The revoloidal curve passes into and becomes identical with the circle while passing the extremities of the diameter, but its fluxion carries it out of the circle as it leaves these points, and it becomes incorporated with and identical with a right line as it passes off at the extremities of the axis, but its fluction carries it out of the right line as it becomes extended. For let the diameter AB be produced each way to H and I, so as to be equal to FG, and from the extremities of these lines draw HF, IF, IG, GH, forming a square circumscribing the revoloidal surface; let HF be extended to M, and IF be extended to L, then these lines so produced will cross each other at right angles in I’, and form an angle with the axis FG of 45°. Let the revoloidal curves extend to P and P; now if the vertice F be brought to its natural position, on the re- voloid, these curve lines evidently cross each other at right angles also, and at the point of contact form an angle of 45° with the axis, which is the same as that formed by the right lines ; hence these curve lines agree with the right lines, HM and IL, at that point both in position and inclination, and there- fore are identical. And also, as the revoloidal curve passes into and occupies the space of the circle at the extremities of the diameter, A and B having, in its original position, formed a part of the circle at that point, and the position of that point not having been changed in referenve to the axis or diameter AB, it fol- lows, that it is still equal to and identical with the circle at those points, but its fluction, or the law of its propagation, _ causes it to leave the circle after passing those points, Cor. Each of the ordinates through the quadrant, AEF, pa- rallel to the axis, EF, is equal to the portions Al, A2, A3, &c., of the arc of the quadrant, intercepted by those lines respec- tively toward the point A. Scholium 3. Hence, this curve is generated by the locus of the intersection of two right lines, AC and EF equal the ra- dius and semi-circumference of a circle moving uniformly from any point IX or A, perpendicular to each other, through their respective lengths. 102 ON THE RECTIFICATION AND Let A be the origin, and since CF equal F the quadrant AE, hence CF=17, AC=r. Let arc 4’ equal any arc Ah, measuring the angle ACA, then since Ah=CP it is also E é’=y; hence, the equation to the curve is Pp y = arc 6/ G Let F be the origin, and we have y=QP =hG=sin. é’=sin. ACE. Then the equation, considering F as the it t origin, is y=sin. 4 PROPOSITION III. THEOREM. The contour of a plane revoloidal surface from a right revoloid is equivalent to the perimeter of an ellipse, formed by a verti- cal section through the angles of the revoloid. For the section of a revoloid through the angles is an ellipse by definition, and this ellipse terminates the facial surface of the revoloid when in its proper position. Now if the cylin- dric surface of the revoloid is extended on a plane, its parts are not altered in relation to each other ; its vertical length on the plane is equal to its length on the cylindric surface of the revoloid ; and its conjugate suffers no change, being a right line parallel to the axis of the cylindric surface while on the revoloid and a right line still when extended ; for the surface may be extended in like manner as we would unroll a piece of cloth, or a piece of paper, made. to agree with its surface, which suffers no contortion of any of its parts in the change, but the whole surface is the same in reference to its edges after the change as before, and as each facial surface of the revoloid extends through half the circumference of the circle of the revoloid, viz: from one vertice to the other, each side of the surface is terminated by one-half of the ellipse formed by a section through the angles, and as the angles of the revo- loid cause generally the ellipses to cross each other at the ver- tices,. forming a vertical angle also on the facial surface, the other side is bounded by one-half of a similar ellipse, so that the whole perimeter of the facial surface of a revoloid is equal to the perimeter of an ellipse by a plane passing through the angles of the revoloid. Cor. Hence, the perimeter of a right quadrangular revo- loidal surface is equal to that of an ellipse whose conjugate or minor axis is equal to that of the transverse axis of the re- voloid, and whose major axis is in the ratio to its minor axis as the /2: 1. QUADRATURE OF CURVES. 103 For the plane passing through the angles of the revoloid forming the ellipse, cuts the plane forming a circular section through the centre of the sides at an angle of 45°, therefore, the transverse axis of the ellipse is as the diagonal of a square of which the vertical axis of the revoloid or its diameter forms a side, - > PROPOSITION IV. THEOREM. If there be described two ellipses concentric with each other, on axes of which those of the outer one exceed those of the inner one by N, then the two ellipses will not be equidistant through- out, but will be nearer to each other at points in the curve be- tween the vertices of the axis, than at the vertices. . Let AB, ED, and ab, ed, be the two axes of two concentric ellip- ses, and let the axis ab=Q, and ed=R, and let the axis AB=Q +N and ED=R+N, then the distance between the two ellip- ses, between the vertices E and A will be less than at those ver- tices. For, draw the two equal con- jugate diameters, HG, FL, and also the two Al. fl, Then, HG?+FL’, or 2HG’?=AB?+ED? (Prop. XV, of the Ellipse,) and 2hI’=ab’*+-ed? Now, because the sum of the squares of the axes AB’-++ ED” are not greater than HG’?+F'L’, those squares cannot be pro- portional (Prop. XVIII, B. I, El. Geom.) hence also. (Prop. XXIII. B. L., #1. Geom.) the axes themselves cannot be pro- portional. Now it is evident, that when the axes AB and ED are nearly equal, then also they will very nearly form the extremes of a proportion of which the two diameters HG, FL are the means ; which is the more nearly true the nearer the two axes are to an equality, or the nearer the ellipse approaches to a circle, and hence they are more disproportional, the greater the ec- centricity of the ellipse. Now it is evident, that the inner ellipse, a, e, 6, d, is more eccentric than the outer one, AE.BD, since the two axes of the inner one are less than those of the outer one by the same constant quantity N, (by hypothesis,) hence the conjugate di- ameters Al, fl, are more nearly equal to HG and FL than ab to AB, or thaned to ED. Hence the elliptical curves are 104 ON THE RECTIFICATION AND nearer each other at any point F or k between the vertices of the major and minor axes, than at those vertices. Cor. 1. Hence, if about an ellipse there be described a curve concentric thereto and equidistant throughout, such curve will not itself be an ellipse, but if it is described very near to the elliptical curve and equidistant, it may be regarded as an ellipse without much error in so considering it, and this will be more nearly correct the more nearly the ellipse approaches to the circle. Cor. 2. If about any curve except the circle, another curve be described, in such manner as to be equi-distant in all its parts from the former, these two curves cannot be similar curves, neither in properties nor in figure, all which may be shown by the same reasoning as in the proposition, viz: a curve described equidistant from a parabola, either within or without, cannot be a parabola, and a curve described equi- distant from a hyperbola, cannot itself be a hyperbola ; all of which is evident from the properties of those curves, but when they are drawn exceedingly near to those curves they may be regarded as curves of similar character as those near which they are so drawn. PROPOSITION V. THEOREM. If there be described two curves, one within and the other with- out an ellipse, of such kind that they shall both be equi-distant from the ellipse, or the ellipse shall be midway between the two, then will the space included between the two curves so described be equal to the circumference of the ellipse mulii- plied by the distance between the two curves. | Let FGHK, fghk, be two con- centric curves described so as to be equi-distant from the elliptical circumference AEBD, one with- out and the other within, so that AEBD shall be midway between the two, then will the curvilinear space Ff, Gg, Hh, Kk, Ff, be equal to the circumference AEBD multiplied by their common distance Ff, or Gg. For, describe about each of those curves polygons FbmG, &c., Anok, &c., fpgg, &c., such that their corresponding sides will be parallel, and draw bp, mq, &c., and the surface included QUADRATURE OF CURVES. 105 between the outer and inner polygon will be divided into the trapeziums Ffpb, bpqm, &c., each of which is equal to the half sum of its parallel sides multiplied by their distance Ff; but ~ any side, An of the polygon described about the middle curve is equal half the sum of its corresponding parallel sides of the trapeziums; hence the trapezium Ffpb is equal to An x Ff, and the trapezium bpgm is equal to no X-Ff; and since this is true for each of the trapeziums, it follows, that the sum of all is equal to the sum of all the sides of the polygon described about the ellipse AEBD, multiplied by the common distance Ff. And this would be manifestly true, whatever be the number of the sides of the polygon described about the ellipse, but when the number of the sides of the polygon is indefi- nitely increased, the polygon becomes a curve similar to that about which it is described. (Prop. XII, Cor. 4, B. V, Ei. Geom.) Hence as in the proposition. Cor. It is evident, also, that if there be described curves within and without a parabolic or any other curve, so as to be equi-distant from it, then the space included between the outer and inner one will be equal to the curve situated midway between them, multiplied by the distance between the outer and inner curves, and the same may be aflirmed of the revo- loidal curves so drawn. Scholium 1. The polygon described about two eccentric curves drawn so as to be equidistant throughout, are not simi- lar polygons, since the figures about which they are described are not similar. (Prop. [V., Cor. 1.) Scholium 2. The last two propositions suggest a method of rectifying the elliptical circumference, and also of finding the lengths of other curves whose quadratures are correctly or approximately known. For, if about an ellipse AEBD, whose major axis AB=P, and whose minor axis ED=Q, another ellipse FGHK be des- cribed, whose major axis FH=P-+N, and whose minor axis GH=Q+N, and if another concentric ellipse be also des- cribed within the former, whose major axis fh=P— N and whose minor axis gk=Q — N, we shall have an elliptical rng Ff, Gg, Hh, Kk, whose area may be found by subtracting the inner ellipse from the outer one, then if this area is divided by the distance of the inner and outer circumference the quotient will be the elliptical circumference AEBD. 106 ON THE RECTIFICATION AND PROPOSITION VI. PROBLEM. Let it be required to find the distance Mm, between two concen- tric ellipses, the difference of whose major and minor axes are each equal to a quantity N. Let HGPK, bghk be two concentric ellipses, the dif- ference of whose axes, HP —bh, is N, also GK—gk =N; draw the equal con- jugate diameters ML, NI, ml, ni; draw the right co- ordinates MN, mn, draw mt parallel to HP, and join Mm, which will be the distance between the ellipses at the points M, m; we shall have sC?=13PC’ and sM’=}CG’, also © EC?=1AC? and st??=3Cg"* (Prop. XXI. Cor. 1 of Ellipse.) and sC—EC=sE or mt, and sM—st or mE=Mt. Hence VM??4+-mé* =Mm=the distance of the two curves at the points M, m. Let the axis HP=25, and GK=19, and if N=, then bh—=23', and gk=—1F and sC=8.83175 EC=8.13172, hence im=.70003 sM=6.71751 and mE=6.00999, hence Mi=.70752 and Mm=.995302. Scholium. 1. But because the line Mm is not perpendicular to the curve AFRD at the point of contact, but very nearly per- pendicular to MW,, it is therefore greater than the true distance of the curves, which we will sappose is wv, to find which, we have ; Se YC EG? and WC= C YE=YC—CE. and Ws=WC —Cs mY =Vmb?-+EY*, MW= Vv Ms?+ Ws? Therefore, we have the sides of the right-angled triangles mEY, MsW given to find the angles W and Y, the difference of which, when found, is equal to the angle made by the lines MW and mY with each other produced. Then we have a right- angled triangle Mm, MW produced. and mY produced right- angled at M, to find the sides MW produced, and mY pro- duced, and also the base wv of an isusceles triangle having the same vertical angle, which line uv will be the shortest dis- tance required. QUADRATURE OF CURVES. 107 Scholium 2. Let the equal conjugate diameters IN, LM, in, im, be drawn, and the difference of the semi-diameters C1—Ci will be very nearly equal to the distance It of the curves at the points 1,7. But IN= ViPH?+!GK%, and in= Vi1bA+2¢k', hence, ViPH?+:GK?— V1 6h*+1¢h?=the distance Ii very nearly. Let the axis AR=24, and FD=18; then, if we make PH =25, and GRh=19, IN will==22.2036. Also, we may have bh=23, gk=17; hence, in or ml=20.2287. 22.2036 — 20.2237 ; Therefore, ———-———-———— =.9899 =the difference CI and Ci, which is nearly equal the true distance of the curves through the point A, which will be more accurate as the axes appruach equality, and will be approximately true till the ec- centricity of the curves becomes very great. PROPOSITION VII. PROBLEM. To find the length of the ell ptical circumference, approximately. It has been observed, (Prop. V. Scholium,) that the cir- cumference of an ellipse inscribed between two other concen- tric ellipses, is equal to the area or space included between the two divided by their distance from each other ; but since the distance of the curves is not constant in every part, we must take their average distance If Gg. PA, (see diagram to Prop. VI.) be the distance of the two extreme ellipses through the lines of their axes, and uv the distance through the point e, then the average distance will be very nearly equal }Gg+ uv; hence, if the area of the ring within the exterior and interior circumferences is divided by }Gg-+1upv, the quotient will be the length of the whole circumference AFRD ; or if a quadrant of the ring is divided in like man- ner, the quotent will be the length of a quadrant FeR of the circumference. If we assume the axes AR=24, FD=18, as in the last pro- position, and the axes of the other ellipses as there assumed, we shall have for the area of the greater ellipses, PH X3GH X7r=12.5X 9.5 x+=373.06381; and the area of the smaller ellipse $2kX 1bhX i=11.5X8 5X* =507.09042, there- fore the area of the ring is equal to 65.97339. ( : Let this be divided by et at the average distance as found in the last proposition, and we have 66.128 for the cir- cumference of the ellipse AFRD. Let the area be divided by the distance as found in (Sch, 108 ON THE RECTIFICATION AND 9899+ 1 2. Prop. V,) and we have65.97339~+ aa the elliptical circumferance, which is true to four places of figures, and very nearly to six; hence, this mode of compu- ting the elliptical circumference is sufficiently accurate for any ordinary calculation. = 66.3115 = Scholium. The length of the parabolic or hyperbolic arc may be approximately determined in the same manner as is here suggested for the ellipse. PROPOSITION VIII, PROBLEM. Let it be required to find the length of a revoloidal curve. Let ADBE be a revoloidal curve from a right quadrangular revoloid whose length is required; let two other concentric curves KLMR, FGHI, be described on each side of the first and equidistant there- from, and if these two last des- cribed curves are at a small dis- tance only from the former, they will be very nearly revoloidal curves likewise. Let aa, the axis of the revoloid from which the surface ADBE is supposed to be taken, equal P; and since, by hypothesis, ADBE is the surface of a right quadrangular revoloid, the conjugate is also equal to P, and the vertical length DE of the revoloidal surface will be equal }+XP ; since DE is equal to half the circumference AaBa of the revoloid. Now, let N be the distance AK, or AF, that the concentric curves are proposed to be drawn; and since the angles DLn and LDn are each=45° in a right quadrangular revoloid, make the semi-circumference of the revoloid from which the surface KLMR is taken, equal daP+2V Dn? + Ln? =30P+2V2N?=!n', and make the circum- ference of the revoloid from which the surface FGHI is sup- posed to be taken, =}+P —2VDn?+ Ln?=!0¢P—2V QN = 14, then may P’=the diameter of the greater revoloid, and P” equal the diameter of the lesser, and P+2N equal the conjugate axis KM of the larger surface, and P—2N equal the conju- gate axis of the smaller. ms QUADRATURE OF THE CIRCLE. 109 And we have (Prop. III. B. IIL) the area of the surface KLMR=(P+2N)XP’, and the area FGHI=(P —2N)xP”, and (PP’+2P’N) —PP”"+2P"N equal the space KF, LG, MH, RI between the inner and outer curve. Let this area be divided by the distance KF or N, and the quotient will be the length of the revoloidal curve ADBE very nearly. Scholium 1. Since the major and minor axes of the outer and inner curves vary in very nearly the same ratio, it follows that the outer and inner curves must be very nearly similar to the central one where the distance KF is small, even if the surface possesses a considerable degree of eccentricity, pro- tracted in the direction KM ; but the greater the eccentricity when protracted in the direction LR, the greater is the simila- rity of the concentric figures, and the greater is the accuracy with which we can rectify the curve. 2. Since the perimeter of a revoloidal curve is also the pe- rimeter of an ellipse, such as is formed by a vertical section through the axis and the angles of a revoloid, this mode of rec- tifying the revoloidal curve furnishes also a mode of rectifying the elliptical circumference ; and reciprocally. PROPOSITION IX. PROBLEM. To find the vertical length of the revoloidal surface, and consequently the circumference of the circle. Let ABC be the quadrant of 4 one of the facial surfaces of a quadrangular revoloid, and BCD a quadrant of the inscribed cir- y cle; AC the semi-transverse, M2 and CB the semi-conjugate axis ; extend CB to FI’, making it equal to AC, draw AF, and the trian- angle ACF will be equal to one- fourth of the square circumscrib- ing the whole revoloidal sur- face. From the extremities of © the radii CD and CB, draw the gB F lines DE and BE, perpendicular to those radii respectively, then will BCD be a square circumscribing the quadrant equal to one-fourth of the square circumscribing the circle. On the line BE take any distance Ba, and from the point a, draw the line ab parallel to the semi-conjugate, BC ; then set off from C on the transverse CA, the distance Cc equal the are 110 ON THE RECTIFICATION AND Bi cut off by the line ab; and draw cd also parallel to BC, cutting the curve AB, and from d draw dg parallel to AC, then will the rectangle CBab equal the portion of the revoloidal surface CedB. (Prop. III-Cor.1, B.III,) and cd=ib=cosine of the arc Bi, and Ch=ig=sine of the arc. The portion of the revo- Joidal surface CBdc may consist of the rectangle Ccdg, and the segment Bdg ; the former of which is equal to the product of the line Cc or dg, into the cosine ib or cd; and the latter may be divided into the two portions Big, a segment of the circle, and the trilinear space Bid, the space included between the circle and the revoloidal curve; the former of which is equal to the difference between half the product of the arc Bz, or its equivalent Cc, into the radius CB; and half the rectangle of the sine ig, into the cosine 7b; and the latter may be approxi- mately estimated by considering, that the distance between the circle and the curve at any point, and in direction parallel to the base line id, is equal to the difference between the arc of the circle included between such point, and the radius CB, and its sine. Thus, the distance id=arc Bi or dg, its equivalent, — gi, its sine. Hence, it will be perceived that its value con- verges rapidly as we approach the semi-transverse AC. Though the ratio of this space is constantly changing with regard to its linear dimensions, as the vertical length of a conjugate section is varied; yet, when the distance Ba is taken very small, its area may be regarded as equal to one-half the product of the base 7d, into its vertical height perpendicular to that base, viz., into gb. Let x=the arc Bi=Cc, and s=sine of the arc Bi, and c=cosine. Then will ce=the rectangle Ccdg, and jrz—%3cs=the segment Big, fire) re en fos Hi we ae eee and... .7s=the rectangle BCba. Hence, cx+tra— hes+3ra—'rs—icxr+ics=rs. By transposing and condensing icx+ra=2rs a. are 3s Therefore, he PART OEE Ta (1) Or if we make the curvilinear space idB=aiB, making the mixtilineal space Ccdb equal to the arithmetical mean between the mixtilineal space diBCc and diaBCc, we shall have xc+ kra—tsc=the space diBCc and xc + rs —cs =the space diaBCe half the sum of which make=the rectangle | CbaB=rs. the space Bid, QUADRATURE OF THE CIRCLE. 11 re a Qex+-3 eee eat =rs and Qex+tre—{cst+rs. 3CS+Ts Hence, So a a 2.) Again, draw fG parallel to 4 CBor DE, cutting offany arc DL, and set off from A the dental AH, equal the length of the arc y DL. and from the point H, draw b the line Hé, perpendicular to the axis AC, to cut the curve in v, and the trilinear space AHv will be equal to the rectilineal DEG, (Prop. II, and Cors. B. HI.) Now, if we compute the area = of this trilinear space AHv in c 2B F terms of its sides, those sides may be rectified by its known quadrature. The length of the side AH equal the are DL, and the area AHv is less than half the rectangle of AH into Hv, and greater than }vH*; but will be very nearly un arithme- tical mean between the two. Let AH Sr=are DL, Hv=s=sine of z, DG=v=versed sine of x. te oes Then 9 area AHbp, and vr=area DGfE; but it has been shown aks the area AHv=area DGfE. Lites SL Hence, ur=* 5 Or, 4vr=2z'+2s. 2 Ce Ss Completing the square vr sta = 4dr ‘a Nhs Hence, . Ce X/ oo ae - - (3.) The smaller the arc is taken, the greater the accuracy ; for when the arc is taken very small, the sine and arc are very nearly identical; when no appreciable error would occur in so considering them, and the area AHv would still be found between half the square of AH, and half the rectangle of AH into Hi; and if the difference of x and s becomes =o, then 2 e the expression for the surface becomes barely a ’ - ¢? ’ - 112 ON THE RECTIFICATION AND and pa V re) =! eo ee eS (4,) Let the two arcs, DL, iB, be taken such that the diagonal At is nearly equal the arc 2B, and if we make AH¢ equal the rectangle DGfE, and nBCe equal the rectangle aBCe, the re- sult will be a close determination of the length of the arc of the circumference ; for the area AH, in such case, is very nearly as much in excess, above the area AH», as the area nBCc is in defect of the area dBCc, and if these arcs are taken very small, any error may be rendered evanescent. For this purpose, let z=the arc iB and z'=the arc DBL. Then let cr—ics+-j}rz=rs : J les-ttrs =- — -« wa MS OST OR: which reduced gives z cLir (5.) And obit a! Then Se OUT te eee we ow MO Whence, if we take an arithmetical mean between the re- sults of these two equations, we shall arrive at a very approx- mate determination of the circumference, if the arc is taken very small. For an example, let us take an arc= z,;1,; part of the cir- cumference. We have, by trigonometry, the sine of that arc =.00025566346=s, and its cosine=.99999996732=c. Hence, by formula 1, we have ars — r-+ic Therefore. 00025566346 X 3= 3rs=.00038349819 and 1+9999996732+2=r+ 7c=1.4999998366 : | And nae .0003834981 9 1.4999998366=.00025566346278 2 hence x=.00025566346.278, which multiplied by 24576, gives the whole circumference=628318526133 when the diameter is=2, or 3.1415926306=7, when the diameter is 1, which result is true to eight places of figures, but the 9th should be 5 instead of 3. If instead of this arc, we take that of ;515; of the circumfer- ence, the sine of which is=.000157079632033, and the cosine =.999999987462994 3rs=,.000235619448050 r-+ic=1.499999993831497 3 3rg 2°* _-—000157079632676 r+ ze 000 40 whence 7=.000157079632676 x ae 3.1415926535.2, | * . yo , QUADRATURE OF THE CIRCLE. 113 which is true to eleven places, but the last figure should be 8 instead of 2. Hence, it will be perceived, that, by taking a very small arc, we can rectify the circumference to any degree of exactness. Cor. 1. Since in formula (6) }2’=vr; if r=1, then v=}2? (1.) Hence, the versed sine may be taken as half*the*square of the arc, which is approximately true wheng/the arc, is. small, and if very small, may be taken as an accurate determination, but as the arc is increased, if we proceed as in formula (3,) where rv=12z*+1sz, we have the value of'v or the versed sine, equal to | of the square of the arc+1 the rectangle of the arc and sine; which is approximately true for any arc of the quad- rant, and may be taken for an accurate determination when the arc is small. Hence this general formula v=12°+1s8e - -- - (2) 2 gis r+ic deduce expressions for the sines, of small arcs, in terms of the cosine and arc; and also for the cosine in terms of the sine and arc. For from this equation we obtain Cor. 2. Hence, also, from formula (1,) z= we may s=2z-+ler - - + +. + + - + (1) 3 and c=——2 = weaerat) Payee ne Wad eT eey) Scholium. 1. If s is taken equal to the sine of 30°, we shall avoid in some measure the inconvenience of using imperfect decimal terms; for the sine of 30° is equal to half the radius, and we then have only one decimal term entering into the ex- pression, viz., the cosine which is, 866025, &c.; but we have a more difficult determination of the area of that portion of the revoloidal curve existing without the circle, viz., the space Bid; which, if determined, would lead to the true determination of the arc of the circumference. 2. If the ratio between the quadrant of the circle and the portion of the revoloid without the quadrant, viz., if the portion ADB is determined, then the ratio of the circumference may also be determined. Thus, if }r*=the area of the quadrant CDB, and z=the area of the space included between the quadrant and the curve, viz., ABD, and if the ratio between these terms are known, that is, if pee Wet ee then trx+z=r (Prop. Ill, B. ML.) by uniting extremes and means, 114 ON THE RECTIFICATION AND irnx=2m z= — tra By substituting this value of z in the former equation irnz =mr* — rmx; by dividing and transposing }nv-+}mz=mr ; pee Oe AT et / Se hence of the circumference. 3. Or if the ratio between the quadrant of the revoloidal surface and its circumscribing parallelogram is determined, the circumference may be computed also. For the revoloidal quadrant being equal to the square of the radius, (Prop. III. B. III,) it is in the same ratio to its circum- scribing parallelogram, as the radius to one-fourth of the cir- cumference, seeing the parallelogram is equal to the product of the radius by one-fourth of the circumference. PROPOSITION X. THEOREM. If a quadrant ABC of a plane revoloidal surface from a right quadrangular revoloid be described, and a rectangle ACBD circumscribe the quadrant, then if the axis AC be divided into any number of equal parts, and if ordinates be drawn from the points of division across the rectangle, then the parts of those ordinates included between the axis AC, and the revoloidal curve BA, will be equal to the sum of a series of sines of arcs of the quadrant EB, of a circle in arithmeti- cal progression, equal in number to the number of the ordi- nates, and the parts of those ordinates intercepted by the curve AB, and line CD ; will be equal to the sum of a similar series of versed sines, and the sum of the whole ordinates will be equal to an equal series of radii of the quadrant EB. Draw any ordinate as KG, and from m, where it cuts the curve, draw mM paral- lel to AC, and let it cut the circumference of the circle described on the ordinate CB, in x draw nh parallel to mK, and the line mK=the line nh, is the sine of the arc nl, and because (Prop. II,) the line AC=the quadrant EB of the circumfer- ence, and the line mM=KC is also=the arc nB, the line KA equal the arc nE; take Kd=KA, and from d, draw the ordi- nate dH parallel to the former ordinate, and from the point 27, where it cuts the curve, draw iL, and QUADRATURE OF THE CIRCLE. 115 from the point a where the latter line cuts the quadrant, draw ab, which will be the sine of the arc aE=to the part id of the ordinate dH ; and because iL is equal to aB (Prop. 1.) and mM =nb, and AC=EB, and because AC —mM=mM —iL, En =na; and since the same may be shown in reference to any other ordinate, drawn from a point on the axis AC, whose dis- tance from d toward C is equal to Kd or AK, it follows that the parts of equidistant ordinates drawn across the rectangle intercepted between the curves AB, and the axis AC, are equal to the sum of a similar series of sines of arcs of the quadrant EB taken in arithmetical progression; which is the first branch of the proposition. And since av=1H, is the versed sine of the arc aB corres- ponding to the sine aL with its complement ab, and nw=mG versed sine of the arc 7B, its complement being mK; and since they have been shown to be at distances from each other pro- portional to the arcs En, na, &c. ; and since the same may be shown in reference to the portion of any ordinate intercepted by the curve AB, and line BD, wherever drawn, it follows that the sum of the portions of the equidistant ordinates intercepted by the curve AB and line CD, is equal to the sum of a similar series of versed sines of arcs taken in arithmetical progression ; which is the second branch of the proposition. And since the lines CA and BD are parallel by hypothesis, the ordinates are all equal in length, and equal to the radius CB, hence the whole series of ordinates is a similar series of radii. Cor. 1. From the preceding demonstration, it appears that any line or ordinate drawn from any point on the axis AC, parallel to the conjugate CB, and terminating in the curve AB, is equal to the sine of the arc on the quadrant BE, cut off by a line drawn from the point of termination of ‘such ordinate in the curve parallel to the axis AC. Cor. 2. Hence, also, the sum of the series of sines is to the sum of a similar series of radii as the square ECBF, described on the radius, to the rectangle CB, AC, of the radius and are of the quadrant. Since the square ECBF is equal (Prop.{lil. B. III.) to the surface ABC. “ee 116 ON THE RECTIFICATION AND PROPOSITION XI. THEOREM. The area of a plane revoloidal surface is to that of its circum- scribed parallelogram, as the sum of an indefinite sertes of sines in the circle, to the sum of an equal series of radit. ‘ Let ACBD be a parallelogram circum- scribing the quadrant of the revoloidal surface ABC, and let EBC be a quadrant of the circle, then will the area of the quadrant of the revoloidal surface ABC be to that of a parallelogram ACBD, as the sum of an indefinite series of sines of the quadrant BCH, to the sum of an equal series of radii. For, since all ordinates vb, dH, &c., cut the surfaces of those figures in rela- tion of their magnitudes in the sections C M iit vcs through which such ordinates pass; and since (Prop. XI,) if ordinates be drawn through those figures equidistant from each other, the portions of the ordinates intercepted by the curve and axis, are equal to the sum of a series of sines of ares in arithmetical progression for the whole quadrant equal in number to the number of the ordinates, and if these ordinates are equidistant from each other, the sum of the portions passing through either surface, drawn into their common distance, may be taken for the surface ; and since the distance of the ordinates is equal by hypothesis, both for the parallelogram and revo- loidal surface, the portions of the ordinates intercepted by each, will be in relation to their surfaces respectively, when their num- ber is indefinitely increased, and their distance becomes indefi- nitely small. Hence, as the sum of a series of sines of arcs of the whole quadrant taken in arithmetical progression, is to the area of a quadrant ACB of the revoloidal surface, so is the sum of an equal series of radii to the area of the parallelogram ACBD; and what has been shown for one quadrant of the revoloidal surface, is also true for the whole. Cor. Hence, the space BDA without the revoloidal surface, is to the revoloidal surface, as a sum of an indefinite series of arcs in arithmetical progression to the sum of a similar series of sines. QUADRATURE OF THE CIRCLE. 117 PROPOSITION XII. THEOREM. If the quadrant AB of a revoloidal curve. be made to revolve about its axis AC, and if a plune hemisphere of a quadran- gular revolovd be described about the solid so generated, hav- ing the same axis AC, then the revoloid will be to its circum- scribing prism, as the sum of the squares of a series of sines _ of the quadrant, to the sum of the squares of an equal series of radii. Let an indefinite number of planes be passed through the revoloid perpendicular to the axis, and at equal dis- tances from each other, and the sections made by these planes will all be squares, (Prop. 1, Cor. 4 B. III.) and their sides will all be equal to the ordinates Hm, hn, &c., drawn through the intersec- F C B tion of such planes with the vertical sections; and hence the side of each parallel section, is equal to twice the sine nh of the quadrant corresponding to such section, being = nn. Now let each of those parallel planes be extended to HG pass- ing through the prism, and it is evident that each of the sec- tions of the prism will be squares, whose sides are severally equal to twice the radius CB, of an inscribed circle. Now the magnitudes of these solids through each section, are evi- dently in the relation of the magnitudes of such sections ; and if the number of these equidistant planes are indefinitely in- creased, and they are indefinitely near together, their sum will represent the whole of each of the solids in the relation of their whole magnitudes, and since each conjugate section of the revoloid is the square described on double the sine, answer- ing to the ordinate in reference to the quadrant CEB, and each section of the prism is the square described on the line HG, equal twice the radius, it follows that the solidity of the revo- loid, is to that of the circumscribing prism, as the sum of the squares of a series of the sines of the quadrant, to the sum of the squares of an equal co of radii. Cor. Hence, the solidity of the space between the surfaces of the revoloid described as above, and that of its circum- 118 ON THE RECTIFICATION AND scribing prism, is to that of the revoloid, as the sum of the squares of a series of versed sines of the quadrant, to the sum of the squares of an equal series of sines, and this space is to that of the prism, as the sum of the squares of the versed sines, to the sum of the squares of an equal series of radii. Cor. 2. Let the prism and also the revoloids, be divided into four quadrants by planes through the vertical axis, and passing through the centres of the opposite sides; and the so- lid so described. will be truly represented by the value of the conjugate parallel sections, passing through them, viz: the seginent of the revoloid, will be represented by the sum of the squares of the sines; the space between the revoloid and sur- face of the prism, by the sum of the squares of an equal series of versed sines, and the prism by the sum of the squares of an equal series of radii. Cor. 3. Since the revoloid has the same ratio to its circum- ‘scribing prism, as the solid of revolution about which it is des- cribed, has to its circumscribing cylinder; the solid formed by the revolution of the revoloidal quadrant AB, will be to its circumscribing cylinder, as the sum of the squares of a series of sines of the quadrant, to the sum of the squares of an equal series of radii. ; PROPOSITION XIII. PROBLEM. Let it be required to find the circumference of the circle from the ratio of the sum of the series of sines for every minute of the quadrant to the sum of an equal series of radit. The number of the series of sines to every minute is 5400 = the number of minutes in the quadrant, which is the num- ber of radii to be compared with the series of sines, and if the radius = 1, then 5400 is the sum of the series of radii; and the sum of the series of sines to every minute is by Trigono- metry = 3438.2467465. And (Prop. XI,) the area of the revoloidal surface is to that of its circumscribing parallelogram, as the sum of an in- definite series of sines to the sum of an equal series of radii; but the series of sines to every minute being a definite number, and such that the surfaces between the lines may be render- ed appreciable, they do not represent those spaces or tra- i QUADRATURE OF THE CIRCLE. | 119 peziums in their exact ratios, but represent the longest sides of those trapeziums, making up the revoloidal surface, but in or- der that they may be true indices of those trapeziums, they should be such as pass through the centres, when each would be reduced, by a quantity equal to half the difference between itself and the next greater one, and as the sum of all their differences, is evidently equal to the radius, halt of the sum of their several differences is equal to half the radius ; therefore, the sum of the natural sines must be reduced by that quantity, viz: 3488.2467165 — ,5=3437.7467465, when if we make r = radius and x= 1 of the circumference, we shall have, (Prop. IX., Sch. 3.) r:2:: 3437.7467465 : 5400 Hence, x=1.570.796337= 1 the circumference when the diameter is 2, which is true to 8 places of figures, viz: to 1,5707963, but the 9th figure should be 2 instead of 3. Cor. 1. Because the cosine of 60° is equal to half the ra- dius, and because the surface of any conjugate section of the revoloid is equal to the radius multiplied by the cosine corres- ponding to each section, (Prop. XI, \ or. 1.) the sum of the sines for 60° is equal to half the sum for 90° or the whole quadrant; and consequently, is equal to the sum of the series, for the arc from 60° to 90°. Cor. 2. Hence, by proceeding as in the proposition, using the sum of the sines for an arc of 60° in comparison with a corresponding portion of the circumscribing parallelogram, the ratio of the circumference of the circle to its radius, may be determined by this arc, in the same manner as by the whole quadrant. Scholium. It will be perceived that the sum of all the natu- ral sines of the quadrant, to any number denoting the series, may be calculated by reversing the operation, viz: w:r:: mr: sum of series of sines minus tr, when nm = the number denoting the series: and this may evidently be effected for the whole quadrant or any portion of it. ie ” 120 ON THE RECTIFICATION AND PROPOSITION XIV. THEOREM. If there be any number of equidistant ordinates of different lengths drawn from aright line AB, and terminated by the vertices of a polygon, then the area comprehended between the greatest and least ordinate, and the right line AB and polygonal line CDEF is equal to the sum of all the middle ordinates + half the sum of the extreme ordinates drawn into the common distance AG. For the quadrilateral ACDG is = 1 (AC+GD) AG, Bs the quadrilateral EDGH = 1 (GD+HE) AG or GH, and the quadri- lateral HEFB =! (HE+BF) HB; hencec by addition we have (2 AC+ GD +EH+; BF) AG. A H B PROPOSITION XV. THEOREM. If a right line AK be divided into any even number of equal parts AC, CE, EG, gc. ; and at the points of division their be erected perpendicular ordinates AB,CD, EP, &c., termi- nated by any curve BDPS ; and if a be put for the sum of the first and last ordinate AB, SK, b for the sum of the even ordinates CD, GH, LM, FQ ; and c for the sum of all the rest, EP, IR and NO ; then (a+4b+2c) x 1 of the common distance AC will be the area, ABSK very nearly Through the first three points Q@ Ss BDP, let a parabola be con- Fee | ceived to be drawn, having its ae axis parallel to the ordinates ; 3 the parabolic area ABPE, (Prop. iy VI. Schol .B. 1) will be (AB+ x 4CD+EP) x 4+ AE = (AB+ 4CD+EP) x 4+ AC; and when z P ¢ aes ie yal ee Ge the points of BDP are at no CO GL Tan ue great distance from each other the parabolic curve will very nearly cioncide with any other regular curve, drawn through the same points. Let us now take the ordinates EP, GH, IR ; then will (EP +4GH-+IR) xX } EG=the area EPRI; and (IR+ 4LM+NO) oD i‘ | QUADRATURE OF THE CIRCLE, 121 x 1 [L=area IRON. Also (NO+4FQ+KS) x1 NF=area NOSK, whence by addition, we have [(AB+KS) + (4DC+4 Ree tot AE cae Ag BNO Doe ig (21 225-20) xt : Cor. This theorem may be applied for computing the con- tents of solids, by using parallel sections instead of the ordi- nate as will appear in Prop IV., Corellaries and Scho- ium, B. I. i , Schol. It is evident that the greater the number of ordinates and the nearer the points DBP, &c., are to each other, the more nearly will any curve, drawn through them, agree with the parabola, and hence the greater accuracy will be ob- tained ; the same remark will also apply to solids. Cor. 2. Hence if the area of any space AB, KS is known, and the ordinates AB, IR,KS, the value of the line AK may be determined. For if AB + KS = aand IK = 3, we have, by considering the curve as a parabola, whose axis is parallel to KS, the area A =(a+ 4b) x + AK, let AK = p, hence we have A 1M, ln — -—_# ee 2 Bs u th tec ae) +, dict ahodetad 6: (1) Or, if the number of ordinates is increased, we have by the proposition, A= (a + 4b + 2c) i p+ n, w being the number of divisions in the line AK, or the number of ordinates 3nA less 1, and we have p = eee - - ” (2) PROPOSITION XVI. PROBLEM. To find eleven equidistant ordinates to hyperbola between the asymptotes, and by means of those ordinates to find the area. me Taking the equation a?=zy, and assuming a = 10, and the first value of z, or the distance from the centre to the first ordinates = 10, and if the lower distance is 1 = d, we shall have for the ordinates: 10 10 10 10 10 10 10 10 10 10 10 100 TE PaT ae 18) Tae pero MPs ALS) 79" 20 10 =10 the sum of the first and last or a= 55 + an 1,5 9 I22 ON THE RECTIFICATION AND the sum of the even erates o =7 +a e+e sta 3 io 3.459539 The sum of all ie ber ee ordinates or € =< 10 w+ iti at ~+= = 2. 72817445 Therefore, by ‘the Bade (a + 4b + 2c)id=6,9315021, is the area required. PROPOSITION XVII. PROBLEM. To find the value of x by equi-distant ordinates to the revoloidat curve. If A = the area of a quadrant, 4_8 AIR of the revoloidal eurve, anda | |S = the minor semi-axis, Al = the ra- dius of a circle inscribed in the curve, and 1 = the semi-major axis, IR ; and if @ = an ordinate EN to the major axis, equi-distant from I to the vertex R, then will the area of the curve, considering it as a parabola, be = a+4 bx453 and if A equal the area of the revoloidal uadrant, the semi-transverse = sal =< x = Les 1} 4 Fain 1(q nae) — a+4ab (1) Let | a=1,and b= /1=70712, &c. and Alawiiltas foe 1 P31 GB7D969 oe ELT ry ae ae this is true to two places of figures only, but the third should be 7 instead of 6. Let now two other ordinates, CL, GP, be taken in addition to this, equidistant therefrom, and from the extremities of the axis, 3An t and we have by Prop. XVI, Cor. 2, rey Pa a i _ Let the two ordinates b be the sines of 22° 30’ = .382683, and 67° 30’ = .923880, and c = sine of 45° = .707107, and we have a= 1, 4a = 5.226252, 2c = 1.41424,A =r? = 1. u LN ne Taga Fa 1.5705848, é&c. = — ene 7 4b4- 20) 71640466 NP Toa which is true to four places, viz: 1,570, but the fifth should be 7 instead of 5. hence we have QUADRATURE OF THE CIRCLE. 123 Let, now, seven ordinates be taken between the two extremes, and the distance of the ordinates will be reduced . Thus the ordinates will be the sines of 11° 15’, 22° 30’, 33° 45, 45°, 56° 15’, 67° 30’, 78° 45’, 90° = .195090. .382683. 555570, .707107, .831470, .823880, .980785, 1. sin. 11° 15! = .195090 __ jain, Of 0). 33° 45! = .555570 G7 sin. GO° eatenh } ty 56°15’ = .831470 78° 45’ = .980785 2.562915 sin. 22° 30’ = .382683 i C. = 45° 0’ = .707107 67° 30’ = .923880 20.13670 Hence, a=1, 4)=10.251660, 2c=4027340, and n=8. 3n T ~ ° = ee) STP ete L = Therefore, in the formula FRC On eth nee have } « 1.5707833, which is true to five places, but the sixth should be 9 instead of 8. By comparing each of the results obtained above with the true numbers, we shall have the ratio of its approximation. Thus, the difference between the first result and the true number is, .0035724 that of the second, .0002115 that of the third .0000180. Hence, it will be seen that the result approximates in nearly a geometrical progression, to the true value of 7 as we in- crease the number of ordinates, or as the distance between the ordinates is decreased. We pay, therefore, determine the number of ordinates that must bé taken, in order to give an accurate result to any number Of decimal places ; for it will be perceived that the ratio of the above variations are nearly 16 to 1. Hence, we may sfely infer, that it will approximate at the rate of three decimal figures in every two subdivisions. Instead of computing the value of } « from the ordinates drawn in the whole quadraxt, we may take any small arc of the quadrant, and having found its quadrature let it be called A’; and if we proceed’as before, by drawing one ordinate equi- distant from its extremes, we shall have, according to the for- 1 / mula, tee = x’ = the assumed arc, which, multiplied by the number of times this arc is contained in the quadrant, will Se) 124 ON THE RECTIFICATION AND give the same result as though the ordinates are taken for the whole quadrant. 1A | Let the formula row; = «' be applied to a segment of the revoloidal surface whose arc is 30°, and whose greater ordi- nate is the radius. Here, A’=sin. 30° X r=}; hence, A’=14A, a=r + sin. 60° = 1.866025 = sin. 75° = .965926 6A! 3A ‘ " | BEES Ae OTTy =o siguagabampatiess ass PEE radius 1; hence the arc of 90°, or £7 = 1.57076010; which is true to five places of figures. Let there be two other ordinates drawn across the segment, 3nA! ——$—$___— = / ’ eee 18 Neieenae then by the formula, q44h42¢ 77? Since = gr+3, we shall 9nA! have Arne ae Ir, and r= 4 sin. 90° — .100000 = 1,866025 fits 67° 30’ — .923880 = ) sin. 82° 30’ — .991445 ( = 1.915825 c== sin. 75° = .965926 Hence, a+4b+2c = 11.459157 and, 9nA’ = 18, 9nA! AMD lak 1.57079616. which is true to seven places, which is as far as the sines are calculated, on which the value is predicated. Let the first and second results be compared, and we shall have the difference of the frst result and the true value, 1.57079632 —1.57076010 | = .00003622 that of the second, 1.5'7079632 —1.57079616 | = -00090016 Divide the first error by the last, aad we have the quotient = 226 = the ratio of the approximation, or the proportional accuracy of determination, by varying tne number of the or- dinates. It will be perceived, therefore, that this portion of the revoloidal curve approximates much faster than the whole quadrant, and is, therefore, more nearly simila: to a parabola than the whole curve ; it may hence be inferred, that if any small segment of the revoloidal surface is taken adjacent to QUADRATURE OF THE CIRCLE. 125 the conjugate diameter, such segment will be very nearly a portion of a parabola, and that, by so considering it, the value of the length of the arc may be determined with any required degree of accuracy. For, in taking the whole quadrant, we found the ratio of convergency, by doubling the number of the ordinates, to be 1 to 16; and in the segment embracing the arc of 30°, adjacent to the conjugate diameter, we find the ratio of convergency to be 1 to 226; and if an arc is taken still smaller, the ratio of con- vergency will become proportionally greater. Let any segment of the revoloidal surface be taken, and if the value of its arc, or the value of +, be computed by any number of ordinates, and if the number of ordinates is then increased so that the common distance ‘is reduced one-half, and the value of 7 is again estimated by the increased number of ordinates, and if the variations of the two results from the true value be compared with each other, they will show the ratio of convergency of the process for that arc by increasing the number of its ordinates, or the rate of approximation by any specific increase of the ordinates for such arc or segment. Hence, we may at all times determine the value of 7 to any required degree of exactness ; for if we wish to determine its value to any given number of decimal places, we have only to assume some given arc and find its rate of convergency, then take such an arc as, according to this rate, will give the re- quired result. The arc of 90° gave the true result only to 2 places, that of 30° to 5 places, with three ordinates; and we may expect a still greater ratio of convergence for a smaller arc; let us take an arc of 10°; we may, according to this ratio, only have the value to 8 places, and by proceeding to decrease the arc, we should, by taking 1°, have the value to 16 places; but since the curve approaches more and more to a similarity with that of a parabola as we approach the vertices of the conju- gate axis, the ratio of convergency increases also rapidly as we approach that point, or as the arc assumed is decreased ; so that, by taking an arc of one minute of a degree, the accu- racy of determination would extend to many places of deci- mals; andif the are should be reduced still further, to seconds | and fractions of a second, the result would come out true to several hundred decimal places; all of which is manifest by pursuing the investigation. Let a distance be taken on the axis equal the arc of 2 mi- nutes of a degree from the conjugate diameter. Then having the sine 89° 58’ = .9999998308 that of 89° 59 = .9999999577 = 126 ON THE RECTIFICATION AND the cosine of 2 is .00058177637, which, multiplied by radius, is the value of A’ Wye have, in the formulae e BYE, in the formula 775 6A’ = .00349065837 a+ 4b = 5.9999996616 = 0005817764172 = «’ = the arc of 20° 2’, which A’ a+4b multiplied by 2700, the number of such arcs contained in a quadrant, we have the value of 1 * = 1.57079632644, which is true to 10 places, or as far as the sines on which its value is predicated. For more extended investigations on this subject, see notes. and PROPOSITION XVIII. THEOREM. If a circle be described, and from its centre a line equal to one- fourth of the circumference be drawn perpendicular to a ra- dius, the triangle formed by connecting the extremity of this line with the extremity of the radius, will be equal to the quad- rant of the circle. Let ABE be a semicircle, described on the radius AC, from the centre C draw the line CD equal to one-fourth of the circumference perpendicular to the radius AC; join DA, then will the triangle ACD be equal to the quadrant AEC. For, according to Prop. XV, Cor. 1, B. V. El. Geom. the area of a sector of a cir- 8 C A cle is equal to half the product of the are of the sector mullti- plied by the radius. Now, the quadrant AEC is a sector of the circle, and the triangle ACD is equal to half the product of the arc AB, or the line CD multiplied by the radius AC; hence the triangle ACD is equal to the quadrant AEC. Cor. Hence we may infer that AF’, the segment of a circle cut off by the line DA, is equal to the portion of the triangle DEF cut off by the arc EF; for the triangle ADC is equal to the quadrant AEC, and if the line AD cuts off a segment AF from the quadrant, then it necessarily includes an equal space DEF within the triangle and without the quadrant ; otherwise the triangle ADC, could not be equal to the quadrant AEC. QUADRATURE OF THE CIRCLE. 127 PROPOSITION XIX. THEQGREM. If a circle be described, and | from the centre two radial lines be drawn perpendicular to each other, equal w length te the are of the quadrant ef the circle; and if the extremities of these radial lines be connected with anether line as hypothenuse, forming with those dines a triangle, and if this third line is bisected by enother radial line from the centre, bisecting also the arc of the quadrant ; and if a line be drawn from the point of the hypothenuse cut by the last mentioned radical line, to the extremity of the radius, or the point where one of the _firstmentioned radial lines cut the circle ; then will the triangle formed by these three lastmentioned lines, be equal io the sector included between the two radial lines forming sides of this triangle ; and the lastmentioned line will cut off a segment from the ‘sector without the triangle, equal to the portion included in the trian- gle without the sector, and a perpendicular let fall from the point of bisection of the hypothenuse on the radius, is equal in length to the arc of the circle included between the bisecting line and radius. From the centre C of the circle AB draw the radial lines CD and CG, each equal in length to the arc of the quadrant AE, forming a right angle at C, draw DG; draw also CH bisecting DG in H; and draw HA; then will the triangle ACH be equal to the sector ACT, and the segment Ab cut off from the sector by the line HA, will be equal to the portion HTd included in the triangle ACH, but without the sector ACT; from the point H of the intersection of the line CH with DG, draw HI perpendicular to AC, and the line HI will be equal to the arc AT included between the radial lines AC and CH. Draw the line AD forming with the radius CA, and the ra- dial line CD the triangle ACD ; and (Prop. XVUL) the triangle thus formed is equal to the quadrant AEC; draw also HI per- pendicular to CD; and because HL is parallel to AC, the two sides DG and DC will be cut proportionally by the line HL, (Prop. XIV, B. IV, El. Geom.) ; so that if the line DG is bi- sected in the centre at H, the line CD is also biseeted in the centre at L, so that LC or HI=!CD. Now, the two triangles ADC and AHC, having a common base, viz., AC, are as their altitudes, (Prop. VIII, Cor. B. IV 128 ON THE RECTIFICATION AND iil. Geom.) ; but the altitude of the triangle AHC is HI, equal to LC, equal to half the altitude CD of the triangle ACD ; therefore, the triangle ACH is equal to half the triangle ACD; but the triangle ACD is equal to the quadrant ACE; therefore, the triangle ACH is equal to half that quadrant, or is equal to the sector ACT, which is the first branch of the proposition. Now, because the triangle ACH is equal the sector ACT, the segment Ab cut off from the sector by the line AH, is equal to the portion HT included in the triangle and without the sec- tor, (Prop. XVIII. Cor.) which is the second branch of the proposition. And because the line CD is equal to the are AE of the quad- rant, and because the arc AT is equal to half the are AX, it is also equal to half the line CD=CL=HI, which is the last branch of the proposition. Cor. If BA be extended to a, so that Ca shall be equal CH, and if a line Habe drawn and bisected by Ce, and a line be drawn from e to A, forming the triangle ACe, this triangle so formed will be equa] to the sector of the circle intercepted by the lines CA and Ce; and the line Ae will cut off a segment of the circle without the triangle equa! to the space included in’ the triangle without the sector, and a perpendicular ei let fall from e on the radius AC, is equal to the are intercepted by the lines CA and Ce. And the same may be inferred from any further divisions or subdivisions of the circle. PROPOSITION XX. THEOREM. With the radius CA let there be described a circle AEBF, and from the centre C draw the line CD perpendicular to the diameter AB, or radius CA, and equal in length to the arc of the quadrant AX, let the line CD be divided into any number of equal parts, as 1, 2, 3,4, &c., and let the arc of the quadrant be divided in like manner into a similar number of equal parts 1, 2,3, 4, &c. Krom the divisions on the line CD draw the lines la, 2b, 3c, &c., parallel to the radius CA ; and through the divisions 1, 2, 3, &c., on the are draw the radial lines Ca, Cb, Ce, Gc. Then, if from the points of in- tersection a, b, c, &c., of the radial lines, with their respec- tive parallel lines according to their respective numbers, lines be drawn as fA, to the extremity of the radius CA, then will this line, together with CA and Cf form atriangle which is equal to the sector CA6, included within the radial line Cfand the radius CA. QUADRATURE OF THE CIRCLE. 129 And the segment Am cut off from the sector by the line fA is equal to the portion {m6 included in the triangle without the sector; and if lines be drawn from the several points of in- tersection of the parallel and radial lines perpendicular to the radius as {K, the lines so drawn will be respectively equal to the ares intercepted by the radial lines from the extremities of which they are drawn, and their radius CA. For each of the radial » lines Ca, Cb, &c., cut the arc AE in the same ratio that the corresponding pa- rallel lines la 2b, &c., cut ar the perpendicular CD ; thus ww} ° the radial line Cg, passing 2% through the point or divi- lf] sion 7 on the arc, cuts off _«/,7/ one division from the arc, and intercepts with the ra- dius.CA all the rest, and the corresponding line 7g pa- rallel to the radius CA cuts off D7 on the perpendicu- lar CD, so that if the whole line CD is equal to the are ¥ AE, the portion C7 of the line CD = Hgis equal to the are A7 of the circle. And the radial line C6f cuts the arc AE in the same ratio as the line {6 parallel to CA, cuts the line CD, viz., the radial line C6f cuts the circle through the division marked 6, and the line f6 parallel to CA, cuts the line CD in the division marked 6, so that if the whole line CD equal the arc AE, then will the portion C6 of that line= fIK be equal to the portion A6 of the are. Now, the area of every sector of the arc AC6, is equal to the arc of the sector multiplied by half the radius CA ; but the triangle AfC is equal to the line fK _ multiplied by half the base or radius CA ; therefore, ‘the triangle AfC is equal to the sector AC6, and hence the seg- ment Am, cut off from the sector by the line Af, is equal to the portion fm6 included within the triangle, but without the circle, and, as has been shown, fK=C6 is equalto the arc A6. And as the same holds true in each of the points of intersection a, b, c, d, &c., it follows that the result corroborates the affirma- tion expressed in the proposition. Cor. If a curve line be drawn through the several points a,b, c,d, &c., anda triangle be formed by, two lines from any point in the curve drawn to the two extremities of the radius, the 130 ON THE RECTIFICATION AND triangle so formed, will be equal to the sector included between the radius and the other line terminating in C, and the same relation of areas and lines will exist with regard to the trian- gles and lines drawn from any point of the curve, as though they were drawn through the points a, b, c, d, &. Scholium 1. The curve BD may be described about the quadrant BE in a similar manner; and since from any point in this curve, if a triangle is constructed on the radius as a base, this triangle is equal to the sector of the circle included between the radius and one of its sides, the curve may be called the curve of the circle’s quadrature. This curve varies from the revoloidal curve, inasmuch as the revoloidal curve is formed by drawing a line through the points of intersection of a series of lines parallel to the radius drawn through their re- spective divisions on the perpendicular, with an equal series of lines perpendicular to the former, drawn through their re- spective divisions on the arc (see Prop. II.) Scholium 2. This curve is generated a‘ by the locus of the intersection Q. of the £ two right lines CF,HG; HG being made to pass uniformly along from A to C, being always perpendicular to AC, while «6 CF revolves about the centre through the arc ED. Let the origin be at A. Let AH |= 2, HQ = y AC = DE = ,' angle ACQ =4, tS = sine 6 =s, tM = cosine 6= cc Then c:s::9/—a:y E m1 c Sr'-—S$x i Hence y = tah which is the equation to the curve. ot PROPOSITION XXI. THEOREM. Uf from the extremity of the radius of a circle, a chord be drawn cutting off any segment less than a semi-circle, and if from the centre of the circle a secant be drawn through the opposite extremity of the segment, and if the secant be produced so that a line drawn from iis extremity, perpendicular, to meet the diameter produced, shall be equal to the arc of the seg- ment, then the area of the segment will be equal to that of a triangle formed by the chord of the segment, and the part of this secant line without the circle, and aline joining the opposite extremities of this line with that of the chord. QUADRATURE OF THE CIRCLE. 131 For it is evident from the converse of Prop. X VIII Cor., that if CE be so drawn that the perpendicular Ez shall be equal to the arc Aa, then the triangle CEA will be equivalent to the quadrant AaC ; take away the triangle ACa, then there will re- main the triangle Aak = the segment Afa. First, let Aa be a chord cutting off the segment Afa ; from the centre C through the extremity of the segment at a, drawa secant line Ca produced to E, so that the perpendicular Et on the diameter, shall be equal to the arc Afa cut off by such secant ; then will the area of the seg- ment Afa be equal to that of the trian- gle AaE formed by the chord Aa, with the part aE of the secant without the circle, and the line AE joining the opposite extremities. Secondly, let the chord Ab extend E into the second quadrant, cutting off the segment AD#, draw Cbd, and ex- tend it to,E, so that the perpendicu- Jar EF drawn from the point E to the diameter AB, produced, shall be equal to the arc ADod cut off by the chord Ad, or the secant CE, and the area of the segment ADb will be equal to that of the triangle ADE formed by the chord Ad, the part of the secant 0H, and the line AE join- ing their opposite extremities. For, in the triangle ACE, the area is equal to the base AC, the radius of the circle multiplied by half the altitude EF, or the arc ADb of the segment, but the sector ADOC is equal to the radius AC multiplied by half the are ADb ; hence the triangle ACK: = the sector ADOC, therefore, if we take the triangle ACd from each, we shall have the segment ADb = the triangle ADE. PROPOSITION XXII. THEOREM. If from the extremity of the radius of a circle, a chord be drawn cutting off a segment greater thana semi-circle, and if through the opposite extr emit y of the segment, a secant be drawn from the centre of the circle, and if the secant be produced, so that a line be drawn -from its extremity on the diameter, produced, if necessary, shall be equal to the arc of the segment, then the area of the segment will be equivalent to the triangle formed by the radius, secant line, and a line joining the opposite extremities of these lines, plus a tri- 132 ON THE RECTIFICATION AND angle formed by the radius and the chord, with the hne- joining the opposite extremities of those lines. For, let the chord extend into the third quadrant, cutting off the segment ADBdA, greater than a semi circle, draw Cd, and produce it to E, so that the perpendicu- Jar EF, on the diameter AB pro- duced, shall be equal to the arc ADBd of the segment, and join AK, and the area of the triangle ACE, plus ACd will be equal to the segment ADBdA. For the area of the triangle ACE is=the base AC, or radius, multiplied by half the perpendicu- Jar FA, which is equal to the arc ADBd, by hypothesis ; but the area of the sector ADBdCA is equal to the radius CA multiplied by half the arc ADBd, hence the q triangle ACE = the sector ADBdCA. — Add to both the tri- angle ACd, and we have the segment ADBCA = the triangle ACE + the triangle ACd. Secondly, let the chord Ae extend into the fourth quadrant. cutting off the seg- ment ADBFE. Through the point « draw the line CE produced, so that the line EF perpendicular to BA produc- ed, shall be equal to the arc ADBFe. Then the segment ADBFe will be equal to the trian. ACE + the trian. AeC. For the area of the trian- gle ACE is equal to the base AC, or radius multiplied by half the perpendicular FE, which is, by hypothesis,* equal the arc ADBF ; but the. area of the sector CADBFeC is equal to the radius AC multiplied by half the are |// ADBFe ; hence the triangle ACE is equal the sector © E A QUADRATURE OF THE CIRCLE. 133 CADBFeC. Add to each the triangle ACe, and we have the segment ADBFe = the triangle ACE + the triangle ACe. Scholium. It may be observed, that, as the termination of the segment approaches the point B, or as the segment becomes equal to the semi-circle, its equivalent triangle becomes infi- nitely extended in the line Ali, and at the same time the sine becomes infinitely small, and while it passes the point B, the sine is equal to 0, and AE is infinite. The same may be said as it approaches the point A on the fourth quadrant, and be- comes equal to a complete circle. PROPOSITION XXIII. THEOREM. The area of a segment of a circle is equal to half the product of the difference between the arc of the segment and its sine multiplied by the radius. First, let Afa (see first diagram to Prop. X XI) be a segment of a circle cut off by the radial line Ca, and produce it to E, so that the perpendicular Ez on the radius will be equal to the arc of the segment ; join EA and (Prop. X XI.) the area of the triangle AEKa will be equal to the segment Afa ; from the point a draw the perpendicular as, which is the sine of the arc Aa of the segment, and the area of the triangle AKa will be equal to (Et—as) !AC (Prop. XX XI, B.1V, Hl. Geom.) Hence, the segment being equal to the triangle, A Ea is equal to half the pro- duct of the difference of the arc, and its sine multiplied by the radius. Secondly, let the chord Ab (see second diagram to Prop. X XI) extend into the second quadrant, draw CE, making EF = the arc ADb; draw AH, and the triangle AEKb will be equal to the segment ADd; draw bs, the sine of the arc ADb of the seg- ment ; then will the area of their triangle AKd be equal to ' (EF -- bs) AC; (Proposition XXXI, Book IV, EHlements Geom.) ; hence the segment, being equal to the triangle, is equal to half the difference of the arc and its sine multiplied by the radius. Thirdly, let the chord Ad (see first diagram to last prop.) extend into the third quadrant, cutting off the segment ADBd, greater than a semi-circle. Draw Cd and extend it to KE, making EF equal to the arc ADBd ; join IA, and the area of the segment ADBd will be equal to the triangle ACK+ACd, (Prop. XXII) ; draw ds per- pendicular to the radius CB, and this line will be the sine of the arc ADBd ; and since it is a sine of an arc greater than a semi-circle, its value is to be considered negative, by Trigono- metry. ov 134 ON THE RECTIFICATION AND From EF substract — dS, and we have EF + dS, which. if we multiply by AC, will give the area of the triangle AEC +ACd, which (Prop. XII.) is also equal to the segment ADBd. Hence the area of the segment ADBd is equal to the differ- ence of the arc, and its sine multiplied by the radius. Fourthly, let the chord Ae (see 2d fig. last prop.) extend into the fourth quadrant, cutting off the segment ADBFe; draw EC so that EF shall be equal tothearc ADBF¢e of the segment joinEA, and the area of the segment ADBFe will be equal to that of the triangle ACE + the triangle ACe (Prop. XII); draw es, the sine of the arc ADBFe, which, by Trigonometry, is nega- tive, being below the diameter, and in the fourth quadrant, which sine substract from the line EF, and we have EF + es, which, multiplied by half AC, gives the value of the segment ADBFe. Hence we have, as before, the segment equal to half the product of the difference of the are of the segment, and its sine multiplied by the radius. Hence we may infer, that a circular zone or a portion of the circle included between two segments is equal to half the product of the difference of the arc of the zone, and one of the segments included, and the sine of such arc, — the dif- ference between the arc of the included segment, and its sine multiplied by the radius. For if ABDE bea zone, and ABL L be a segment of the circle C, the seg- ment EALBD equal to the zone ABDE, and segment ABL; but these segments is are respectively equal to the excess of rs their several arcs above the sines mul- . tiplied by half the radius. Therefore, E D the zone ABDFE, is equal to the differ- ence of the excess of the arcs EALBD, and ALB, above their respective sines, multiplied by half the radius. The same may. be shown in relation to the area ABDFE in reference to its external segment EDF, or AEFDB, regard being had to the positive and negative va- lue of the sines. : r Scholium. Let A represent the area of the segment of a cir- cle ; let « = the arc of the segment, and s = sine of that arc. and let 7 = radius, and the segment may be expressed as in the following formula =1(re— rs) - - “ (1) Ifr = 1, then the expression becomes A = 1 —s) it ms ‘ (2) whence the segment is expressed in terms of the arc and sine QUADRATURE OF THE CIRCLE. 135 Let A’ = the area ofacircular zone ABDE and x’ = the are EKALBD s' = the sine of that arc. Then will the area of the zone be expressed by ' A'=1(re'—r1s')—7(727—1s)_- - : - (3) If r=1 A’=}(a'---s')—--}(a---s) - 4 - : . (4) The arc may be expressed in terms of the numbers represent- ing the segment, and the sine of the arc, by the formula, « = pd ag tt : - - - - - : > (5) And the sine, in terms of the arc and segment by BEV s Se) DIA: - - - (6 Sch. That the area of the segment of 5) a circle is equal to half the difference of the arc of the segment, and its sine X by the radius may also beshown as follows. 4 B Let AE be the segment of a circle, ats. whose radius is AC, draw ES perpendi- cular to the radius, and ES will repre- sent the sine of the arc AE; let AC be represented by r, and ES by s, and the arc AE by 2, then will 3 re = the sector ACE, and irs = the triangle AEC, and + rx —trs = the difference of the sector and triangle = the segment AK, viz., the seg. is = half the difference of the arc, and its sine multiplied by the radius. Also, in the segment EFB let CB, be represented by r, and the arc EFB by 2, and ES will be the sine of the arc EFB, 1 ra = the sector ECBF, and irs = the triangle ECB. irz —zrs = their difference = the segment EBF. PROPOSITION XXIV. THEOREM. The area of the space intercepted by a tangent and secant with- out a circle, is equal to half the product of the difference of the tangent and arc, intercepted by the secant, multiplied by the radius. Let ATF be the space intercepted by the tangent AT, and the secant CT, with- out the circle; and the area of that space will be equal to half the product of the difference of AT and the arc AF, multi- plied by the radius AC. ? For the arc of the triangle ATC is equa to ;AT x AC, and the area of the sector ACF is equal to half the arc AFXx AC. Let t=the tangent AT, and «=the arc AF, r=AC. # r 136 ON THE RECTIFICATION AND Then we have for the triangle ACT, rt, and for the sec tor AFC, 1rz; if from the triangle we take the sector, we have Lrt—1rz=the space AFT, hence as in proposition. Scholium. Draw Fs, FA, and Fv; then Fs will represent the sine of the angle ACF’, and sA or Tv, the versed sine; hence the traingle ATF equals one-half the product of the differenee of the tangent and sine multiplied by radius equals one-half the pro- duct of the tangent and versed sine ;-draw As’ perpendicular to CF, and this will also represent the sine of the angle ACF on the radius CF; hence, also, the triangle ATF is equal to one-half the difference of the secant CT and radius multiplied by the sine. PROPOSITION XXV. THEOREM. If about a plane revoloidal surface from a quadrangular revo- loid, a square be described, the space enclosed by the square, but without the revoloidal surface, will be to that contained by the square, as the sum of an infinite series of segments of the quadrant of the circle, whose arcs are in arithmetical pro- gression, to the sum of a similar series of sectors subtended by the same arcs. ae Let ACD be a quadrant of a plane revoloidal surface, and let HCD be a triangle forming one-fourth of a square circumscribing the whole re- voloidal plane surface. Divide the axis CD into any number of equal parts as 1, 2, 3, 4, &c., and through the points of division draw ordinates 72 la, 2), 3c, &c., parallel to HC, and g ZAM ee these ordinates will cut the surface YZ fi of the quadrant of the revoloid and H A Cc. its circumscribing triangle, in the relations of their magnitudes, through the portions where such ordinates pass; and if the number of those ordinates be indefinitely increased, the sum of those drawn across the triangle HCD, will be to the sum of those drawn through the revoloidal surface ADC, as the area of the triangle to the area of the revoloidal surface. Now, the ordinates al, 62, c3, &c., are equal to the arcs represent- ed by numbers on the arc AF of the quadrant, corresponding to those on the axis, (Prop. II, and Cors.) ; and the portions 17, 16, m5, &c., of those ordinates, are severally equal to the sines of those arcs, (Prop. XIII, Cor. 1,); hence the portions of those ordinates, intercepted by the curve AD and line DH. are 5 QUADRATURE OF THE CIRCLE. 137 severally equal to the difference of the arcs and sines repre- sented by these ordinates. And since we have shown (Prop. XXIII.) that the area of the segment of a circle is equal to half the difference of the arc of the segment and its sine multiplied by radius, and since the area of the sector of a circle is equal to half the arc of the sector multiplied by radius, it follows that the sector and segment containing the same arc, are to each other, as the arc is to the difference of the arc and sine. Let a equal the arc, s equal the sine, r equal the radius. Then tra=the sector, and tra— $rs=the segment containing the same arc, which are to each other in the ratio of , tra: 4ra— Grs, or of a: a—s. And since this is true for a segment and sector, contained by any arc in that quadrant, that is, the sector of a circle is to the segment containing the same arc as the arc is to the arc mi- nus the sine, or as the ordinates g7 to gi, or as f6 : f 1, &c.; and as this relation evidently exists in reference to each of the par- allel ordinates, and as the ordinates represent arcs of the cir- cle in arithmetical progression, it follows that those ordinates drawn across the triangle, may also represent a series of sec- tors of a circle, while the portion of those ordinates intercept- ed by the curve AD and line HD, may represent a similar se- ries of segments of the circle. And as we have shown above that the surfaces HCD and HAD are to each other in the re- lation of the ordinates passing through each, it follows that the trilinear space ADH will be to the triangle HCD as the sum of an infinite series of segments of the circle whose arcs are in arithmetical progression, whose first term is equal to the common difference, and whose last term is the quadrant of the circumference to the sum of a similar series of sectors with the same arcs. And since the whole plane revoloidal surface consists of four quadrants, ADC, and its circumscrib- ing square consists of four triangles, HCD, it follows ‘that the whole revoloidal surface will be to the whole circumscribing square in the same ratio. Cor. The revoloidal quadrant ACD, may be represented by a similar series of isosceles triangles formed by the chords of the series of segments with two radii drawn from the extremi- ties of those chords. For we have shown that the series of sectors may be represented by the area HCD, while the series of segments are represented by HAD; if we take the seg- ments from the sectors, we shall have the triangles, evidently equal HCD — HAD=ACD. 10 ", ¢ 138 ON THE RECTIFICATION AND PROPOSITION XXVI. THEOREM. The area of a quadrangular revoloidal surface, from a right quadrangular revoloid, is to that of its circumsribing square, as the sum of an infinite series of sines of the quadrant whose arcs are taken in arithmetical progression, and whose com- mon difference is equal to.the first term, to a similar series of ares of such sines. v Let DHEI be a square circumscribing the plane revoloidal surface DAEB, and let ADC be a quadrant of that surface, the triangle HDC being its corresponding portion of the square ; divide the semi-axis CD into any number of equal parts, as 1, 2, 3, 4, &c., and through the points of division draw the ordi- nates la, 2b, 3c, &c., parallel to CH, those lines will each be equal to such arc of the quadrant AF as corresponds to the E point of division on the line CD, in reference to similar divi- sions of the quadrant AF’; thus the line 1a equal the arc F1 on the quadrant, since it is evidently equal to 1D on the axis CD; and (Prop. II, Cor.) 1D=1F; also, if we take the line 2b=2D, this is also, for similar reasons, equal to the arc 2F; and hence each line parallel to CH, terminated by the lines CD, HD, are equal to the arc, corresponding to the divisions on the line CD, through which it passes. Again, the portion of these lines or ordinates intercepted by the axis CD, and the curve AD, (Prop. X, Cor.) are severally equal to the sines of the arcs, corresponding to the divisions from which they are drawn on the axis. And since, by hypo- thesis, the arcs are taken in arithmetical progression, they QUADRATURE OF THE CIRCLE. 139 must be equidistant; hence, if the number of such ordinates are indefinitely increased, the sum of the portions of them in- tercepted by the axis CD, will represent the area or the sur- face ADC, as the whole ordinates drawn across the triangle HDC represent the area HDC. Therefore the surface ADC is to the surface HDC as the sum of the series of sines, to the sum of a similar series of arcs taken as above. Now, since this is the case with one quadrant, it must also be true in rela- tion to the whole revoloidal surface and its circumscribing square. Cor. 1. Hence, if a square KLMN be described on the di- ameter of the inscribed circle, the square so described will be to the square DHEI as the sum of the series of sines of the quadrant, to the sum of a similar series of arcs of the sines ; since the square KLMN (Prop. III, B. II,) is equal to the area of the revoloidal surface DAEB. Cor. 2. Each of the ordinates al, 62, c3, &c., drawn across the triangle HCD parallel to HC, is equal to the are of the cir cle represented by the corresponding numbers in the divisions of the quadrant AF. Thus, al equal the are 1F, 62 equal the arc 2F, and g7 equa) the arc 7F or the quadrant. ON SPIRALS. ; There is one class of curves which are called spirals, from their peculiar twisting form. They were inven by the an- cient geometricians, and were much used in architectural or- naments. Of these curves, the most important as well as the most simple, is the spiral invented by the celebrated Archi- medes. This spiral is thus generat- ed: Let a straight line SP of an indefinite length move uni- formly round a fixed point 8, and from a fixed line SX, and Jet a point P move uniformly also along the line SP, start- ing from 8, at the same time that the line SP commences its motion from SX, then the point will evidently trace out a curve line SPQRA, com- mencing at 8, and gradually extending further from S. 146 ON THE QUADRATURE OF CURVES. When the line SP has made one revolution, P will have got to a certain point A, and SP still continuing to turn as before, we shall have the curve proceeding on regularly through a series of turnings, and extending further from S. To examine the form and properties of this curve, we must express this method of generation by means of an equation between polar ordinates. Let SP=r7r,SA= 6, ASP = @; then since the increase of r and éis uniform, we have SP:SA:: angle ASP: four right angles: : 6 : 20 bd b ate — 6 j — Bic a, if a “e From this equation it appears that when SP has made two revolutions or 6== 47, we have r = 2b, or the curve cuts the axis SX again at a distance, 2SA ; and similarly after 3, 4,7 revolutions it meets the axis SX at distances 3 ; 4,7 times SA. Let any number of concentric circles be described, whose radii IA, IC are in arithmetical progression, and if the cir- cumference of the outer circle is divided into any number of equal parts, and radii are drawn from each of the points of division in the circumference of the circle to the centre I, lines drawn from the centre in such manner as to pass through the points of intersections, of the several radii with the curves in consecutive order, will be spirals similar to those of Archimedes. If the curve line pass from ete I, through rpB, or through A EST rgA, the curves IrgA, IrpB KE so 4S will be spirals; so also the =f JSC Aga TAS, lines IsdmD, IsdnD, and IwhfC A Z PE TeX ce and spirals; all generated by / / (FLEES ‘ the same laws, but with differ- AGA ales Nal LN B ent ratios of their angular, { \ \\' xt LSPA Phi compared withtheir rectilenear (CY) LHe motion ; or their circular, with \, Vat Se) “anh ff their radial motion. The two Gaps pee SPY spirals IsdmD, IsdnD, com- ne ee a ie mencing on the line IC, and De terminating at D, form the heart-like figure IsdmDndsl. These spirals, and especially the one with a heart-like form, are extensively used in mechanical operations ; to communi- cate a uniform rectilenear, reciprocating, from a rotary motion; it is therefore important that they receive some consideration. - ON THE QUADRATURE OF CURVES. 141 The area of any portion included between the spiral and its circumscribing arc of the circumference terminated by the radius, its origin, is equal to two-thirds the sector, having the same arc of the circumference as its base. For let fa, Ib, Ic, &c., be the radii of circles in arithmetical progression, then will the arcs of these circles intercepted by two radii, be also in arithmetical progression, and since the value of 4 also increases uniformly in arithmetical progression along with r or , hence the value of the intercepted parts of the several arcs will be a series of arithmeticals multiplied into another corresponding series of arithmeticals, therefore their products will be a series, of the squares or a series of numbers proportional to a series of squares of a series of arith- meticals. It has been shown (Prop. IV, Cor. 3, B. I) that the sum ofan infinite series of the square of a series of numbers in arith- metical progression increasing from 0, is equal to 1 of the last term multiplied by the number of terms. But the arc intercepted by the spiral with the radius, as its origin is‘ the last term, and the radius represents the num- ber of terms ; hence the area IrpBCI is equal to 1 of the pro- duct of the are BC x IC; and the area IsdmDBCI is = 4 arc CBD xX IC; also the area IvhfCADBCI is =} circumfer- ence ACBD x IC. But the area of the whole sector IBC, ICBD, or ICBDAC in either case is equal to half the product of their respective arcs, multiplied by the radius; hence the space intercepted by the spiral in each case is } that of their respective sections. Cor. 1. Hence the space IrpBI, is =} the sector ICB; the area IsdmDI is = + the sector ICBD or } the semi-cir- cle ; and the area IvAfCI is = 4 of the whole circle, Also the heart IsdmDndsI = 1 of the whole circle. Scholium 1. The spiral of Archimedes is sometimes used for the volutes of the capitals of columns, and in that case the following description by points is useful. (See first diagram. Let a circle ABCD be described on the diameter CSA, an draw the diameter BD at right angles to AC ; divide the ra- dius SC into four equal parts, and inSB take SP=1SC, in SA take SQ='SC, and in SD take SR=2SC; then from the equa- tion to the curve these points belong to the spiral ; by subdi- viding the radius SC and the angles in each quadrant we may obtain other points as in the figure. In order to com- plete the raised part in the volute, another spiral commen- ces from SB. 142 ON THE QUADRATURE OF CURVES. § Scholium. These spirals are of the same kind, as those formed by winding achord around a conical spire, from the vertex to the base, in such manner, as to encircle the spire at qual distances ; the exact length of such curve is of difficult determination. . The same spiral would be represented by the convolutions of a conical screw ; also, by a screw represented on a disc. Ifthe origin of the spiral is C f any point M, not in the cen- re of the concentric circles, then the area AFCM123A between the spiral and the outer circumference is=1 of the , product of the are ACB through which the curve would have passed from the centre I, mul- tiplied by the radius — } of the arc LMxIM — } (are LM + arc BC) x MC. D If PQM be a triangle, whose base PQ = the semi-circum- ference ACB = the angular space passed through by the two spirals MA, MB, then either portion PCM, QCM of the tri- angle may be expressed by } PC or 1 ACXCM = ! arc CBx IC—; arc LMxIM—t(arc LM + are BC) x MC, from this P Cc Q - substract the expression for the area included within the spiral, and the arc AB, and we have? ACB x IC—_1LM x IM= the difference of the areas. ON THE CYCLOID. — 148 THE CYCLOID. if a circle EPF be made to roll in a given plane upon a straight line BCD, the point in the circumference which was in contact with B at the commencement of the motion, will, in a revolution of the circle, describe a curve BPAD, which is called the cycloid. This is the curve which a nail in the rim of a carriage-wheel describes during the motion of the carriage on a level road. The curve derives its name from two Greek words signifying * circle formed.” The line BD which the circle passes over in one revolution is called the base of the cycloid ;if AQC be the position of the generating circle in the middle of its course, Ais called the vertex and AC the axis of the curve. The description of the curve shows that the line BD is equal to the circumference of the circle, and that BC is equal to half that circumference. Hence also if EPF be the position of the generating circle, and P the generating point, then every point in the circular arc PF, having coincided with BF, we have the line BF = the arc PF, and FC = thearc EP or CQ; hs QM parallel to the base BD. Let A be the origin of the rectangular axes, AC the axis of x, and O the centre of the circle AQC. Fer AMi== 2, AU= a, MP =y, angle AOQ = 4: ae by the similarity of the position of the two circles, we ave PN = QM, ond PQ = NM ; ~. MP= PQ + QM = NM+ QM= FC + QM = arc CQ + QM that is, y = ad + asin. 6 = a (6 + sin. 4) (1) x= a@—acos. 6 = avers. 4 - (2) The equation between y and zis found by eliminating 4 be- tween (1) and (2) | al : 4 144 ON THE CYCLOID. a—x /2 axr—x? cos. 6 = 2 Sine = a a andy = aé-+ asin, 4 —liqe-- x = acos. ( z + V¥ ax—x? But we can obtain an equation between x and y from (1) alone ; that is from the equation, AP = are CQ + QM. For arc CQ = a circular arc whose radius is a and versed sine x | ° z i = Wes 4 a circular arc whose radius is unity and vers. sin. tix —1! a —l1 x mst rie ey . Y =a vers. ie; +/9 a aa? If the origin is at B, BR = a and RP = y, the equations are «= aé---a@ sin. 4 = a--- a cos. 4, We shall not discuss these equations at length, as the me- chanical description of the curve sufficiently indicates its form. The cycloid, if not first imagined by Galileo, was first ex- amined by him ; and it is remarkable for having occupied the attention of the most eminent mathematicians of the seven- teenth century. Of the many properties of thiscurve the most curious are, that the whole area is three times that of the generating circle, that the arc CP is double of the chord of CQ, and that the tangent at P is parallel to the same chord. . Also that if the figure be inverted, a body will fall from any point P on the curve to the lowest point Cin the same time; and ifa body falls from one point to another point, not in the same vertical line, its path of quickest descent is not the straight line joining the two points, but the arc of a cycloid, the concavity or hollow side being placed upwards. BOOK. V. ON THE PRODUCTION AND RESOLUTION OF GEOMETRICAL MAGNITUDES, CONSIDERED AS LINES, SURFACES, AND SOLIDS, EXISTING IN THEIR SPECIFIC RELA- TIONS OF FORM AND PROPORTIONS. CHAPTER 1. DEFINITIONS AND PRINCIPLES. Art. 1. We have hitherto referred lines, surfaces and solids, in all their varieties of figures and species, to some specific quantities and relations which were cognizable in such mag- nitudes, and whose properties were rendered evident to our consideration. Magnitude we have compared with magni- tude; figure with figure; and we have thereby established their relations, under arbitrary considerations. We will now consider magnitudes in the relation of their organization, or in the relation of their laws of production ; and instead of referring magnitudes to specific magnitudes ar- bitrarily chosen, we will refer them to others, only in the rela- tion of their laws of generation. 2. Since a point by definition is locality without extension, any number of associated points cannot possess magnitude, hence a magnitude is not a multiple of one, or any number of points. 3. Neither can any number of lines, however associated, constitute a surface, since lines are supposed to possess no breadth or thickness, one of which is essential to a surface; for if one line does not possess breadth, neither can any num- ber of associated lines; and if a line be multiplied by any ab- stract number, since it is expressed only in relation to its length, it can only be multiplied or increased in that relation. 4. So, also, if a surface be multiplied by any number, in itself considered, the product cannot be a solid; for since the surface possesses no thickness, it does not possess the charac- teristic of the solid, and hence any number of such surfaces, or multiple of such surfaces in themselves considered, cannot be a solid. 5. The distance between any two points is a line. For a point being locality without extension, if there be two locali- 146 PRODUCTION AND RESOLUTION OF ties, they must be seperate from each other, and their distance from each other is necessarily extension in space, which agrees with our definition of a line, viz., “extension in one dimension.” 6. Space is a medium in which all positive objects, and all local relations exist ; its existence is only indicated by its uni- versal property of extension; it is infinitely divisible in each or all its three dimensions of extensions, and infinitely exten- sible. 7. Any definite portion of space, or any extension in space, is magnitude. Magnitude may possess extension in one, two, or three di- mensions, but space can properly exist only in its three dimen- sions of extension ; if it can be divested of extension in one dimension, it can in another, and so on till its extension is ex- tinguished. Magnitude may be properly applied to extension in whatever degree it exists; but space cannot properly exist independent of its three dimensions, which are its essential properties. ! 8. If there be two points A, B, the first point A, drawn through the distance AB, produces or, describes the line AB; that is, the distance from AtoB bo 11 154 PRODUCTION AND RESOLUTION OF ordinates drawn across the triangle EGD, be a series decreas- ing in the same time from IG or s to 0; call this series b: then the series represented by a. drawn into that represented by b, will generate the solid EGCD. 25. Let the ordinates drawn across the triangle ABC be a series decreasing in arithmetical progression from z, or AB to 0, and let those drawn across the triangle BCD bea sim- ilar series of ordinates increasing from 0 to z, then may the series of proportional means between the corresponding terms, represent a series of equidistant ordinates, drawn acrossa semi-circle, whose diameteris z= DB, and the rectangle of the corresponding terms. may represent a similar series of equidistant ordinates, drawn across such portion of a parabola whose axis is AB, as is intercepted by the curve, and a diagonal from the vertex B to the extremity of the base, the axis of the parabola being equal to <. For the ordinates dn drawn across a ¢ semi-circle are severally mean propor- —=====4>=> tionals, between the abscissze of the diameter; that is, any ordinate dn, isa mean proportional between dB and dD; now if their be a series of ab- scisse dB taken in arithmetical pro- gression increasing, then their cor- responding abscisse dD, dD, &c.,, B will be a series of decreasing arithmeticals ; and if BD is equal toCD or AB, then will the abscissee dD, dB be severally = to their corresponding ordinates de, ef, in whatever position they are taken ; hence the series of ordinates, across the semi- circle is a series of proportional means between the several corresponding terms of the increasing and decreasing series. Again, it has been shown (Prop. VII, Sch. B. I,) that the expression for any ordinate ec, drawn across the parabola, CBh, between the curve and the diagonal, parallel to the axis, is equivalent to the rectangle de X ef, and since this is true of every parallel position of the ordinate ec ; hence the sum of the series of ordinates, is equivalent to the sum of the series of rectangles of the series of variables, = the sum of the se- ries of squares, of a series of ordinates dn, &c., across the semi-circle. Let DB be greater than CD, or AB and the ordinates dn, will be equivalent to those drawn across a semi-ellipse, whose major axis is BD, and minor axis CD; but if CD is greater than BD, then will BD be the minor axis of the ellipse, &c. Hence the solid ABCD formed by drawing the correspond- ing terms of the increasing and decreasing lines into each | GEOMETRICAL MAGNITUDES. 155 other is equivalent in its expression to the area CBh of the parabola, and the semi-circle DBn is equivalent to.the sum of a series of square roots of an infinite series of parallel or- dinates ec across the parabola, orof parallel sections defi through the solid. 26. The elements contained in the forego- ing propositions, may be applied also to solids of revolution ; for instead of a prism and its inscribed pyramids in Art. 16, we may substitute a cylinder, and its inscrib- ed cones, &c. ; and if planes HPNQ be passed through the cylinder, the cones and their complements, these will be cut in the relation of their magnitudes, respectively. in each section; which will be in all cases, the same as that of the prism, and the inscribed pyramids, of equal base and altitude. If the semi-circle or semi-ellipse CIN be made to revolve about the axis CI, and by its revolution to produce a sphere, or spheroid, then, because every section described by the pa- rallel ordinate ON, would be proportional to the squares of their circumscribing lines respectively, the sum of the sections de- scribed, would be proportional to the sum of the squares of the generating lines ; hence the sum ofan infinite series of those sections, and consequently the solid generated by the semi- circle would be equal to four times a middle section, multiplied by 1 of the series. And any segment or zone of the sphere or spheroid, will be equal to the sum of the bases + four times a middle section X } its altitude. 27. If every section defithrough the pyramid ABCB (Art. 25) should be contracted in one of its dimensions tillit becomes a square, and if the edge DB should continue to be a right line, then will the sides BCD, ABD, be plane surfaces, and the side ACD, ABC will become curved, and the whole solid will be equal and similar to a quadrant of a revoloid ; and four of such pyramids, would constitute a perfect right revoloid ; moreover the sides ABC, and ACD would become semi-cir- cles. : Forit has been shown that the square root of every section defi, through the solid, is equal to an ordinate drawn across the semi-circle DnB through the same _pa- rallel ; but the square root of the section, is the side of a square equivalent to the section; hence, if the sides ef the sections are severally the ordinates belonging to a | 156 PRODUCTION AND RESOLUTION OF semi-circle, then the solid must acquire a form, similar to that of a quadrant of a rectangular revoloid, such as would be formed by passing planes through the axis, bisecting its oppo- sides. ' If, when the sections of the solid become squares, its vertices are conceived to be at the extremities of an axis, passing through its centre ; then the solid would become an elliptical revoloid ; its conjugate axis being = to } its vertical or trans- verse axis. The pyramid ABCD is equal to } of the prism, whose base is the square of AB, and altitude BD ; four such solids, or the whole right revoloid is = 4 = 2 the circumscribing prism. Also, the prism, circumscribing the elliptical revoloid, whose base = defi, and altitude BD, being = 1 of the former prism, is the prism circumscribing the elliptical revoloid ; hence, the elliptical revoloid is = % its circumscribing prism. If from a cylinder of equal base and altitude, two equal cones be taken, one on either base, and of an altitude equal to that of the cylinder, the two remaining portions would be each equivalent to the cylinder; and every section through each of these portions, by planes parallel to the cylinder’s base, would be equivalent to a corresponding section through the quad- rant of the sphere ; each of these portions are = 2 of the cyl- inder ; hence, four of these portions = the sphere, are = 2 the circumscribed prism as found in the Elements of Geometry. Scholium. We may, from the preceding investigations, draw the following deductions and conclusions. First, that any series of quantities in arithmetical pro- gression, varying from z to 0, drawn into any series of constant quantities, will produce a quantity whose value is = 3 the sum of the base or maximum product of the variable + four times the value of the product of half the maximum value drawn into the series. Also, that the value of any multiple of this series may in like manner be determined. Second, that if any series of numbers varying from 2 to 0, in arithmetical progression be drawn into another similar series, direct or reciprocal, the value of the product is = 3 the sum of the two bases produced + four times a middle base formed by drawing 4} the maximum values of the terms of the series into each other. And that this is true for any multiple, or power of the va- riable series, whose exponent is an integer or when, any number of variable quantities are drawn into each other, whe- ther direct or reciprocal, and this is the basis of the Integeral Calculus, as will appear in the subsequent pages. GEOMETRICAL MAGNITUDES. 157 CHAPTER ITI. ¥. ON THE CONSTRTCTION OF QUANTITIES WHOSE ELEMENTS ARE A SERIES OF CONSTANT OR VARIABLE QUANTITIES. Art. 1. Having proceeded thus far, in analyzing the pro- ductien of geometrical magnitudes, showing the manner and Jaw of their generation, we are enabled, by having the ele- ments and the law of the production of any magnitudes, to give a geometrical construction of such magnitudes. We were taught, in the application of algebra to geometry, the mode of constructing integral algebraic quantities or ex- pressions geometrically ; we are now to represent a series of quantities, under asingle construction; or to construct quanti- ties whose elements are a series, either of constant, or variable quantities. | 2. In considering the relations which exist between different quantities, those which, during the whole of any investigation are supposed to retain the same value, are called constant quantities ; those to which different values are assigned, are called variable quantities : constant quantities are usually re- presented by the former letters of the alphabet, as a, b,c, &c., and variable quantities by the latter, as uxyz, &c. 3. When two or more variable quantities are connected in such a manner, that the value of one of them is determined by the value assigned to the other, the former is said to be a function of the other variables. Thus, in the equation y=azr+bz’?+c, where the value of y depends on the value assigned to 2; y is said to be a function of x, which is usually expressed by f’(z), 9(x), (2), or similar abbreviations. Also, if an infinite series of equidistant ordinates are drawn across a surface, the sum of those ordinates is a function of the surface. So, also, the sum of an infinite series of planes through a solid may be regarded as a function of the solid, or the solid or surface a function of the planes or ordinates. 4, When any quantity or magnitude as the element of other quantities or magnitudes is variable, the sum of a series of the variable quantity, within its variable limits, may be represented by a dash drawn below or above the letter representing the variable quantity. Thus z or 2 may represent a series of the variable quantity, z. If the incipient value of the variable is z, and the series is decreasing to s, or 0, the dash must be placed below the let- 158 PRODUCTION AND RESOLUTION OF ter ; but if its incipient value is 0 or s, and its terminate value is z, being an increasing series, the dash must be placed above. Thus z iudicates a series decreasing from z to s or 0; and Zz indicates the series increasing from 0 or s to z 5. If it is required to express a series of quantities in arith- metical progression from z to s, or from s to 2, it may be thus written—z2: -- -s, or Z++s3 or which is the same,@hst! OT Lesa, x being the incipient and 2’ the terminate value of the series. ‘The condition of questions involving these variables, gene- rally indicate the incipient and terminate values of the increas- ing or decreasing variables. 6. The expression z is equivalent to that of Z, when consi- dered independent of other variables; but the product arising: from drawing z into Z is not equivalent to that ‘of drawing z into z; for zz indicates that the greatest value of z is drawn into its least value, and consecutively ; and zz indicates that the greatest value of z is drawn into the greatest, and so on through the series. A series of the squares of z or 2, is represented by z’ or 2? ; a series of roots by /z or V2. 7. When it is designed to express, a series either of con- stant or variable quantities, without regard’to their progres- sion or law of variation, a small capital, of the letter denoting a single term of the series may be used. Thus in the equation y= /dz, if we would express a series of y, it may be written y, and the equation will be y= Ydz 5 this, unless otherwise restricted, expresses an infinite series of the quantity represented by y. If y= v(dz) is the equation: of any figure, then y, is a function of the surface, or y= /(dz)x is the equation to the surface. Hence the equation to a surface consists of the equation of the figure considered as a series, drawn into the axis or ab- scissa. 8. If a be the magnitude AD of any series whose number is n, of lines AB, and whose length is the constant quantity z, then the sum of the lines will be nz; and their magnitude a made by drawing them into a surface, will be qth a | i: a Tor we have shown that if an infinite series of lines are drawn into their respective dis- tance, the product is a surface ; hence, we have the following construction, viz: a rectangle. ABDC. a GEOMETRICAL MAGNITUDES. ~ 159 9. Let a be a series of lines z, con- stantly decreasing in arithmetrical pro- gression from z to 0, or from the line AB to the point D, while the line BD repre- sents the number or magnitude of the se- ries; and the series of z drawn into the quantity a, will be equivalent to the tri- angle ABD. If a be a series of lines, AB decreasing from z to s, or from AB to EC, then if AD=a = the magnitude of the series, the construction will be the trapezium ABCK. 10. If a be a series of z*, where z decreases uniformly from AB, to a point D; or from z to 0, then az? may be constructed by the exterior space of a parabola AeDB, for while az ge- nerates the triangle ABD, a.z’ will generate the parabola AeDBA, (Prop. VII, Sch., B. 1,) whose axis is CD. C D If a be a series of Vz, then may az be put under the construction of a semi-parabola, AcDB, whose axis is DB=a, and whose base AB=z, the series of z being a series of ordi- nates across the triangle ABD, parallel to AB. A B 12. Let it be required to construct a quantity dV (z.z), or a se- ries of mean proportionals between the corresponding terms of two equal increasing and decreasing series drawn into d, the number of the series. If d = the line AB, and if z = the same line, the surface generated by drawing d into /(z.z), would be the semicircle ABD. D For the equation to the circle is y’= / (dx—z’) Gr me DC af AGL CL. AC B But if z be greater or less than d, then the construction will be a semi-ellipse ; which if z is less than d, will have AB for its major axis; butif z is greater than d, AB will be the minor axis. For the ellipse by its equation, d?: ¢:: 2(d—z) :.y’ or d*y?’ =c’(dx — x’) is only the circle expanded, or contracted, in the ratio of the major and minor axes. The equation for the surface of a circle or an ellipse, may more properly be expressed by d./(xx) for a semi-circle or semi-ellipse, and if x = d, then the equation will be that of a 160 PRODUCTION AND RESOLUTION OF circle; but if d and x are unequal, the equation becomes that of an ellipse, and d will be the major or minor axis, according as it is greater or less than 2, and 2 will be its conjugate ; and its equivalent in either case is 1}dxX+, or for the whole circle or ellipse 1dzxX: hence 7 isa function of ./ (xz) or fv (Er) Hence we have a finite expression for the circles quadrature in algebraic terms, viz: 2d./(x.) or 2x /(z.2)=r'x, where x is equal to the diameter. Andr=(2r+r") / (z.%)=the circum- ference of any circle whose diameter is x ; z and x being series of increasing and decreasing quantities. Hence, t=2d/ (xx) =1d'=(8~d) y (a2), or (82) V (22). 13. Let ABD be the segment of a circle, the height of the segment DE being equal z; and if the diameter = 2, then will the chord AB= (zz 5) and if EG = x—z = 2’, then will the equation to the surface be 2yz=2z Bice 2)2). 4 ach erpmnvenlte: minh 64) 34 Also, let CE = 1x—z=u, then if AB in a decreasing series is drawn into CH, or if (u./(xz)) be constructed, it will be equal to the triangle ABC ; and the sum of the segment and tri- angle, = 2z/((z-:x')z) + u(/(xz)) = the sector ACBD. (2) In the triangle ABF the side AF is equal 2CK=xz—2z ; and BF:BA:: AF: AS, or 2: xz: :a2—-2z: AS the sine of the angle, BCA, or sine of the arc of the segment =f (zr) — 2zV (zz) +r. ee ee ee (8) Let AS=s, and let the are ADB=7’ and the area of the segment, (Prop. XVIII B. IV) will be (#’—s) xX{z, hence the expression 22./((@-- 2’)z)=1a0' — jars: - - - + = (4) Therefore the arc ADB of the segment may be expressed 2 af ((2-°2')z)-+s % Jf (LX of = VIE ETS = ary (eat VC) 6) 4 14. If a series of variables consisting of two or more fac- tors, as its elements, are drawn into the number denoting the series, the construction can more conveniently be represented by a solid. Surfaces properly consist only of factors, equivalent to the second power ; cubes of factors equivalent to the third power, or the product of three factors. a But we may, under certain conditions, construct quantities representing solids as surfaces ; those whose factors are con- stant, by any surface arbitrarily chosen, and those which are variable by such curves as yield to the conditions of the ex- pression ; and we may thence proceed to construct such solids as would result from drawing such surfaces into a series ; or GEOMETRICAL MAGNITUDES. ~ 161 such as would be the result of drawing the variable series of elements of such surface into a constant, or a variable quan- tity ; and the quantity so constructed may, by being put under a superficial construction, become the base of another quantity or magnitude, made by drawing this quantity into another given quantity or into a series; and we may proceed, in this manner, to construct quantities geometrically, converting one construction into a base for the next, and so on. This may be performed algebraically, and with greater fa- cility, inasmuch as a quantity involving any power of the va- riable may be assumed as a base for a higher power of the series, and thus the powers and roots of variables may be extended at pleasure; and the laws of variation or their in- crease or decrease in value, may be determined by geometri- cal construction and analytical deductions. 15. If it were required to construct a series, a, of the quantity bc, drawn into each other, b and c being constant quantities ; here it is evident that, for the purposes of construction, be may represent either a line or a surface; if the series of be be re- presented by a line, then bc, drawn into a, or abc, will re- present a rectangle. But, if bc represents a surface, then abc will represent a prism. And for the purposes of investigation, the conditions of the quantities would be similar in either case; for the line would have the same relation to the rectangle ge- nerated by drawing the line into the series, as the surface to the prism, generated by drawing the surface into the series. 16. Let it be required to construct a series a, of bz or a.bz, where z is a series decreasing uniformly to o. This would evidently be represented under the form of a wedge ABCDEF, whose base ABCD is represented by bz, and the perpendi- ,, cular ED represents the value of a. A If z decrease only to s, then the construction would be the prismoid ABCD abcd, whose greater base, ABCD, is repre- sented by 6z, and whose lesser base, abcd, is represented by bs. 17. If « be also a quantity decreasing uniformly with z, in such ratio as to reach o at the same time, then the ex- pression a.uz would generate a pyramid ABCJ; which will be its proper construc- tion; and if w=z, or if the . expression can be but under | the form a.z’*, then the pyramid will have a square base. 162 PRODUCTION AND RESOLUTION OF 18. Let a series a.zz, be constructed. This quantity, Sheet senting the products” of the corresponding terms of two series, one increasing from o to z, while the other decreases from z to 0, drawn into a, the quantity denoting the number of the series, generates the double wedge ABGD, where AB, or GD, is equal to z, and the per- pendicular AD=a. If uw represent a series of ordinates drawn across the triangle ADG, then az will re- present the triangle; and if these ordinates are severally drawn into a series of uw, the result would be the same as before. 19. Let it be required to construct y°=pz, which is the equation to the vertical parabolic revoloid, y= y(pz,) being the equation to the parabola, p being the parameter, is a constant quantity, we have z variable. First, we may draw p into a series of 2, which gives us the triangle DEA (see diagram to art. 16,)=pz, p being equal to DA, x=the axis of the revoloid=DE the altitude ; Satie if this triangle, as a base, is also drawn into z, the axis of the revo- loid, we shall have the solid ABCDEF. Or, more properly, if we first draw the axis z into the se- ries of the variables, we shall have the triangle as before, with a base AD; and if we multiply this by p=DC, we shall have the solid described as before, then will any section abed of this solid parallel to the base ABCD, be equivalent to a similar section through a parabolic revoloid, or pyramoid, which it is designed to represent. The expression for a cube may at all times be constructed on a surface by means of curves; but every different species of solid requires some peculiar construction, according to the equations for the ordinates or sections of the solid, the expres- sions for which may be transferred to expressions for ordi- nates to a surface, or equations to some curve; and since any multiple, or power of a series x, whose exponent Is an integer, is known, when the series or root is known; we can hence dis- cover an innumerable variety of curves, which are quadrable ; but, in general, in descending powers, it is not certain that the series of roots of a given series may be so. Thus, the circle is equivalent to a series of square roots, of the ordinates to a pa- rabola; the series of the parabala is quadrable, but not the circle, except in certain functions of given quantities; we may get an expression for its value, but it will be under an incom- mensurable form. Let ABD represent the vertical segment of a rectangular spherical revoloid ; and its equation considered in relatinn to its figure, will be y* = (z.. 2’)z - - - - (1) GEOMETRICAL MAGNITUDES. 163 the equation considered as a solid, will y°z = z(z.. a')z (2) which from its organization will readily be discovered to be cubable, for its value is a (x'z + 4 (Lx + 42’) 42). = 222? nA RMR i 2s ty ing tths esha ®) itsconvex surface will be 4xz - thawen Shes pe (4) Hence, the solidity of a spherical segment will be = (2227+ 42'2%)le = Lazo + la'2e - - - - (5) and its surface will be xz - - - . : (6) 2 ; 7 c ; 20. Let it be required to construct v'=7,(dr¢ +2") the equation to an hyperbolic revoloid, vy’ being a series of parallel planes, being the squares of the ordinates to the axis. Because this equation involves powers higher than the Cc square ; that is, because the terms (dz+z’,) multiplyed by 5 = produces a solid, the series represented by these ane drawn into a, would be of a higher power than a cube ; hence, in order that our construction shall be under a cubie form, @ (dx +2”) must be represented onaplane. Let a=CB. Hence, we have a.z*=the exterior pa- G rabolic surface ABC, AB being the axis of the parabola, and ufo dé = CBE; now, if this is drawn into 7 we shall have the solid ABECFDG; for the construc- tion of a solid equivalent to the quadrant p of an hyperbolic pyramoid or revoloid ; every section NIKL through this solid is equivalent to a corresponding section through a hyperbo- ~ lic pyramoid of the same altitude and equivalent base. Hence, it will be perceived that the hyperbolic revoloid is cubable; for this, its representative is equal to one-sixth of the product of the sum of the base AEFD plus four times a mid- dle section NIKL drawn into the altitude BC. This-is also true in relation to the parabolic revoloid: And since a paraboloid or hyperboloid is to its respective circumscribed revoloid as a circle to its circumscribed square, or as r°x : 4r’, it follows that the paraboloid and hyperboloid, are also cubable in terms of =. 21. It may be perceived that in constructing the quantity az, where x is a series decreasing to 0, we have a triangle ; hence. the value of az is tax; or the value of ex may, in like manner? be shown to be ryt, ’ Also, since 2°a is equivalent to a pyramid whose base is 2”, and whose altitude is x; henee, z°x is equal to 32°. Ss 164 PRODUCTION AND RESOLUTION OF And if we have a series of z* to be drawn into 2 or a, we may also construct 2° by art. 20, and we shall here have 2°z =127'; alsoz*z= “ihas My Hence, the law of the progression of the powers of the va- riable quantities is manifest, for the value of the continued pro- duct of any series of power, of a series of variable quantities drawn into any constant quantity, is equal to the same power of the maximum value of the variable drawn into the same quantity, divided by a number denoted by the index of the power plus 1. If the variable has a fractional exponent, the same consi- 1 Moy deration will also apply. Thus, a xis equivalent to 3a and yee 4 x*x is equivalent to 2z m} anda" x equivalent to cana &c. Where the product of any power of a variable quantity is made the base of another construction, according to its deter- minate value, the magnitude constructed on this base, is no longer considered as a function of the variables, entering into the former constructions as elements of this base; but this new construction is subject to the laws of its own organiza- tion; and if raised to any power, or multiplied by any number of factors, either as constants or variables, the different powers depend on, or are a function of the base assumed, and not of its original constituents. Thus, if we have the quantity 2°z, its value is }z*; now, if 1z* is assumed as the base of another construction, as (12“)z, or 1z‘z the product is not =;2°, but is =1z°. Ifa series uxz are drawn into a, the product is equivalent to }uzza; the co-fficient being the same as that for the pro- duct of z*z, which gives }2* as its equivalent. When the terms of the variable series denoted by x or z, are drawn into each other according to their reciprocal value, or where an increasing series is drawn into a decreasing se- ries, the value of some power of the product of the series may be obtained ; and the other powers and roots of this value are subject to peculiar laws for theirdevelopment. For the contin- ued products of these series, are not subject to the same ratio, or the same order in the variation in their successive changes from one power to another, as is observed in the products of such as are either all increasing or all decreasing. ‘Thus, if we have eV 22, let this quantity be involved till its radical is removed, and we have z°zz, the value of which may be de- termined by a construction. Thus, x%.x2, which is equivalent to a double wedge, (Art. 24, Chap. I,) or which will be more available for our present purpose, it may be put under the GEOMETRICAL MAGNITUDES, 165 construction of the parabola cut off by a diagonal, (Art. 25, Chap. 1,) the value of which is evidently equal 12°x4x1a =12°, which is equal to one-half the value of G xx; andif this value is again put into a series, "i < | and drawn into z, we shall have 42°.2, or \ x’z.x, whose proper geometrical construction will be the solid CELI, which is the solid of a prism circumscribing a parabolic complemen- tal ungula, which (Prop. X VIII, Cor. 2, B. II.) is equal to one-fifth of the prism CANELI —17* Tow Let there be a series of 2’z, a series of xx, and two series zz, and the sum of the series will be equivalent to 2°, as may be represented by the prism AH. For 2x equals the pyramid ABCDE, xx equals the pyramid EFHGD, and 2¢%zrzx 4) = equals two wedges CDEG completing the : cr ae Tj D prism whose value is 2’. 4 Wj YY A B Let there be a series a of z drawn into a reciprocating Se- ries of Z, the terminate value of z being z’, and the incipient value of x being 0. Since the product of these series a.(z ++ 2')% gives us for a construction a solid hinoCD, whose value is equal to jax(2z'+z); and this also represents a segment of a quadrant of a revoloid where z and z increase and decrease in the same arithmetical ratio, equal to a segment of a cylin- drical ungula. Also, if we have a series of the products of reciprocating variables, or zz drawn in- to a series of A or z, its value has been found equal to tAz* or 32°, let this be drawn into another series of Z, and we have the following construction, which is a parabo- la whose vertice is D; and we have 222? equal the parabolic ungula ADBC equal 2ABXHAXIBC=2(zX (42X12) X12z)=347'; hence, the series zzz’ is equivalent to ;;2°. Let this be drawn into another series of Z, and we shall have 22z°=12(2?+-4(42° x $z))=12'+.12°; if this is drawn into another series of z, we shall. have 22zz*=12z(z°+4(74-2'X12)) =}2'+,5;z"; hence the law of progression is manifest. If we have a series of zzz’, its value will be 42(2z?+2*) =12°+,';2°. Hence, the law of the progession of this series will easily be discovered. 166 DIFFERENTIAL AND Scholium. The results obtained by the preceding notations and geometrical constructions, are similar to those obtained by the integral calculus, the same principles as here used serv- ing as the foundation of that science; and in order that a com- parison between the two modes of notation may be instituted, we will present some of the elementary principles of the cal- culus, as the subject of the next chapter. CHAPTER III. DIFFERENTIAL AND INTEGRAL CALCULUS. Art. 1. Asa basis of the deferential calculus, we may premise that if y be a function of 2, and if a change takes place in the value of f (x) so that x becomes x+h, x being quite indetermi- nate, and h any quantity whatever, either positive or negative, a corresponding change must take place in the value of y, which may then be represented by y’. If the quantity f (ec +h) be now developed in a series of the form TAL) TAR BiETCh Re... oe which is always practicable, in whlch the first term is the ori- ginal function f (x), and the other terms ascend regularly by positive and integral powers of h, and A, B, C, &c., are inde- pendent of h; then the co-efficient of the simple power of h in this series is called the first differential co-efficient of y or f (x). This is the fundamental definition of the differential cal- culus. 2. Let us now examine the change which takes place in the function for any change that may be made in the value of the variable on which it depends. Let us take, as a first example, Uy (ALi, and suppose x to be increased by any quantity 2. Designate by wu’ the new value which w assumes, under this supposition, and we shall have | u’ = a(x-+h)’, or by developing wu! = ax*+2azrh-+-ah’. If we subtract the first equation from the last, we shall have u’ — u=2arh+ah’; hence, if the variable x be increased by h, the function will be increased by 2axh-+ah’. If both members of the last equation be divided by h, we shall have u'—U ici eon 2ax-+ah, (1) INTEGRAL CALCULUS. 167 which expresses the ratio of the increment of the function to that of the variable. The value of the ratio of the increment of the function to that of the variable is composed of two parts, 2axz and ah. If now, we suppose / to diminish continually, the value of the ratio will approach to that of 2az, to which it will become equal when h=0. The part 2az, which is independent of h, is therefore the limit of the ratio of the increment of the func- tion to that of the variable. The term, limit of the ratio de- signates the ratio at the time becomes equal to 0. This ra- tio is called the differential co-efficient of u regarded as a func- tion of x. 3. Then let y be a function of z, such that y=ax* Let x become x+A and y become y’ y'=a(x+h)* expanding =az>+2az..h-+-ah This, it will be perceived, is a series of the required form, the first term az’ is the original function y, and the other terms ascend by integral and positive powers of h; hence, according to our definition, 2az the co-efhicient of the simple power of h in this series, is the first differential co-efficient of y or f (x) 7 4. Again, let Ue Let z become x+h and y become y' y'=(xz+h)* expanding =2°+32? .h+3z .h?+h* Here, therefore, 3z’, the co-efficient of the simple power of h is the first defferential co-efficient of x’. 5. Again, let yar’ + ba’ +ca-+td Let z become (x+A) and y become y/ Re af dia) Ober icc tb) expanding = axv*+3axr*h+ 3arh’+h?+b2’+2bxh+-bh?+cx+ch +d, arranging according to powers of h, =(ax*+bx?+cx+d)4+(8ac?+2bz2+c)h+(Bar+b)h?+h, a series of the required form, for the first term is az*+b2’?+cx +d, the original function, and the succeeding terms ascend regularly by powers of h. Hence, 8ax*+2bx-+c the co-efficient of the simple power of h in the development of y’ is the first differential co-efficient of yf or ax’ ba -+ or + as ‘ 6. We have now to introduce a notation by which this ratio may be expressed. For this purpose we represent by dz the last value of h, that is, the value of h which cannot be dimin- ished according to the law of change to which h is subjected, r 168 DIFFERENTIAL AND ay without becoming 0; and let us also represent by du the cor- responding value of «~: we then have du ett 2az., (2) The letter d is used merely as a characteristic, and the ex- pressions du, dx, are read, differential of u, differential of x. It may be difficult to understand why the value which A as- sumes in passing from equation (1) to equation (2), is repre- sented by dz in he first member, and made equal to 0 in the second. We have represented by dz the /ast value of h, and this value forms no appreciable part of h or x. For, if it did, it might be diminished without becoming 0, and therefore would not be the last value of hk. By designating this last value by dz. we preserve a trace of the letter 2, and express at the same time the last change which takes place in 4, as it becomes equal to 0. 7. Let us take as a second example, u = ax’, If we give to z an increment h, we shall have, u'=a(a+h)*=ax'+8ahz’+3ah*c+ah*, hence, u! —u=8ahz’+3al’xz+ah', and the ratio of the increments will be u'— uu Rap sax’ +3ahz+ah’, and the limit of the ratio, or differential co-efficient, Ha 3ax* dz 7 In the function ’ du = nz*, we have— = 4nz*. dx And if y=f (2) the first differential co-efficient of y is denoted by the symbol d oa thus in the above examples. (Arts, 4. & 5.) y = ax’ hy dp ee yee dy de Oe y = ax+bz'?+crz+d dopant se 3ax?+2ba+c we INTEGRAL CALCULUS. 169 in like manner if w=f (z) the first differential co-efficient of xy or f (z) will be represented by = Note.—Since a constant quantity is not susceptible of change, it is manifest that it can have no differential co-efficient, or if dy U feeat 2 pill 0. 3. To find the first differential co-efficient of any power of a simple algebruic quantity. Let y=2" Let z become x+h os Sfi=a(e-8)2 Expanding by the binomial Ee? A Caga! asi) aon AE my pli—l n—-2 },2 n—3],3 =2*% +na"At+ hei eh? + ae ze shs + & eo, dy ° Siete n—l ; qe From this it is manifest that The first differential co-efficient of any power of a simple Algebraic quantity is found by multiplying the quantity by the index of the power, and then diminishing the exponent by unity. Ez. 1. y= x" ay =a" {a 2. ya arte a = @ (p+q) zPti! 3. y = at+x-4 ie == — gat) 4. y= Gis _ = eG) INTEGRAL CALCULUS. The object of the Integral Calculus is to discover the pri- mitive function from which a given defferential co-efficient has been derived. : This primitive function is called the integral ofthe proposed differential co-efficient, and is obtained by the application of the different principles established in finding differential co- efficients and by various transformations. & 12 _ 170 DIFFERENTIAL AND When we wish to indicate that we are to take the integral of a function, we prefix the symbol §, Thus if y = ax’ We know that dy = 4aa‘dx If then, the quantity 4az* dx be given in the course of any calculation, and we are desirous to indicate that the primitive function from which it has been derived is ax‘, we express this by writing 8 4az* dx = azx* The characteristic §$ signifiesintegral or sum. The word sum, was employed by those who first used the differential and and integral calculus, and who regarded the integral of 2™dx as the sum of all the products which arise by multiplying the mth power of z, for all values of z, by the constant dz. When constant quantities are combined with variable quan- tities by the signs + or — they disappear in taking the differ- ential co-efficients, and therefore they must be restored in taking the integral. Thus, if y = az +b or, y = ax* —b or, y = ax’ In each three cases equally dy = sax’ dx Hence in taking the integral of any function itis proper al- ways to add a constant quantity, which is usually represented by the symbol C. Thus, if it be required to find the integral of a quantity such as dy = 3ax’* dx yY = § 8ax' dz =azx'+C where C may be either positive, negative, or 0. We cannot determine the value of C in an abstract example, but when particular problems are submitted to our investigation, they usually contain conditions by which the value of C can be as- certained. By reversing the principles established for finding the dif- ferential co-efficients, or differentials of functions, we shall ob- tain an equal number of rules for ascending to the integrals from the derived functions. Recurring therefore to these we shall perceive that 1. The integral of the sum of any number of functions is equal to the sum of the integrals of the individual terms, each term retaining the sign of its co-efficient. Thus, if INTEGRAL CALCULUS. © 171 dy = 4ax’* dx + 8bx? dx — 2bx dx + dx y =§ 4az* dx + $ 3bx* dr—$ 2 bz dx + §dx+C II. Since, if | y = a™ dy = maz™ az it is manifest that The integral of a function raised to any power is obtained by adding unity to the exponent of the function, and by divid- ing the function by the exponent so increased, and by the differ- ential of the funciton. ee 1. dy = ax" dx _ axl ~ n+l Ex. 2: dy = de = ux dz ' ax er Paps 29) a Re Pinay che Ez. 3. dy = (a+ 2x)" dz _ (a + x)etl n+l dy = seldaie. dx I= (a+a) se (Grit) oar 1 +C +6 7 =~ ~(@—l) (a+2z)-2 This rule applies to all functions of the form dy = (a + ba™)™cx"7! dx for these can all be reduced to the form az™ dz. Scholium. The coincidence of the results obtained by the notation, used in Chap. Hl, with those determined by the Calculus, in- dicates that the same principles serve as the foundations of both ; which, as was observed at the close of Chapter I, are the following, viz., that when any series of quantities vary in arithmetical progression, the sum of the series as well as their powers, or products, with a constant quantity, or with another arithmetical series, may be measured by the sum of the pro- ducts of the maximum values + the minimum values + four times the products of their medium values multiplied by } of the magnitude, representing the number of the series. 172 DIFFERENTIAL AND &c. Any series of numbers which are not subject to these con- ditions, cannot be made the subjects of accurate numerical determination or calculation, by the Calculus ; hence, we may always determine, by construction, whether any geome- trical magnitude, or algebric quantity, is susceptible of accu- rate development numerically in terms of given quantities, One advantage possessed by the investigation of geometrical subjects, by the principles contained in Chapter If, of this subject, is, that the notation there used, is more elementary than that of the Calculus, and expresses the conditions of geometrical subjects in a more obvious and inteligent manner, and is more direct in its results. Hence, instead of pursuing the method of defferentation, we express the conditions of quantities depending on variable factors according to their several conditions, or organization, and proceed to integrate or sum up the series of functions, and determine the magnitude of the production, according to the principles therein contained. Hence S§ x’ expresses definitely the function of the series of squares of a series of variables, and if this series be drawn into a constant quantity x, and integrated, we shalt have according to our notation iz’. " If this is made equal to uw, or if we have u = 12°, its differ- ential is 2 =z", which, being again integrated by the cal- culus, we have again 32°, showing similar results, from differ- ent considerations. The notation by the calculus is arbitrary, but by this inductive ; and hence, by its intimate connection with the principles of the calculus, may serve to render that subject more obvious. There are cases, however, for which the calculus is more particularly adapted than the notation here referred to; such as drawing tangentsto, and rectifying curve-lines. This is accomplished by that science in a manner the most elegant and complete ; but in relation to surfaces and solids, nothing can be more complete or satisfactory, than the discussions which we have introduced. OF THE VIRTUAL CENTRE. 173 CHAPTER IV. On the centres of surfaces and solids, the method of finding them, &c. Article 1. The centre of surfaces and solids is susceptible f two considerations, that of magnitude and distance. Hence, we have the centre of aggregation and the virtual centre. 2. The centre of aggregation or centre of magnitude of any plane surface, is that point through which, if a line is drawn in any direction in the plane, it shall divide the surface equally. 3. The centre of magnitude of any solid is that point, through which if a plane is passed in any direction, the plane shall divide the solid equally. 4. The virtual centre, or centre of gravity of any plane is that point through which, if a line be drawn in any direc- tion in the plane, the sum of all the points, on one side, drawn into their distances from the line, shall be equal to the sum of all the points on the other side drawn into their respective dis- tances from the line. 5. The virtual centre of any solid is that point through which if a plane be passed in any direction, the sum of the points, on one side, drawn into their distances from the plane, shall be equal to the sum of al! the points on the other side, drawn into their distance from the plane. 6. The virtual centre between any two points is evidently in the right line connecting the two points, and equidistant from each. -Hence the virtual centre of a right line, is in the middle of that line, oris in the centre of magnitude of the line. 7. In a system of points, their virtual centre has respect to the magnitude of the lines drawn from the several points to such centre. But the centre of magnitude, has regard only to the number of points. 8. If there be drawn froma system of points, lines perpen- dicular to a given base, the sum of those lines divided by their number, will give their average length ; whichis = to the distanee of the virtual centreof the system of points from the base. 174 _ METHOD OF FINDING Let AB be a given base line, and let a, b, c, d, e, &c., be a system of points ; if perpendiculars al, b2, c3, &c., be drawn from the points to meet the base line AB, and if the sum of these lines be divided by their number, the quotient will give their average length, or the distance AC, from which, if a line CD be drawn, parallel to AB, it shall pass through the virtual centre of the system of points. For if the line CD is drawn, according to the conditions ex- pressed in the proposition, the portions of the lines cut off beyond the line CD, are sufficient to extend those which fall short of that line, to the line CD. Hence, the line CD passes through the virtual centre of the system, for the lines drawn from the points on both sides of the line are equal. Hence, if two base lines 8 AB, GH are drawn, not pa- rallel to each other, and if the system of points are connected to both of the bases, and we proceed, as in the proposition to draw CD parallel to AB, and IL parallell to GH, the inter- section F’, of the lines CD, and IL will be the virtual centre of the system. If the base line AB, should pass through the system of points, so as to leave one portion of them on one side, and another portion on the other, then the lines connecting the points to the base must be estimated on one side positive, and on the other negative ; and the virtual centre will fall on one side or the other, as the positive or negative signs predomi- nate. 9. If, instead ofa system of points, the virtual centre of a system of parallel lines is required. Let a base line be drawn parallel to those lines, and because the lines are to each other, as the number of equidistant points, arbitrarily taken in each ; hence, if we take the product of the several lines into their re- spective distance from the base, the sum of these products divided by the sum of the lengths of the lines, will deter- mine the distance from the base, through which, if a line is drawn parallel to the base, it shall pass through the virtuai centre. KH WNWRraAn tag? THE VIRTUAL CENTRE. 175 10. The virtual centre of a surface may be found by sup- posing parallel ordinates drawn across its surface, and assuming those ordinates, as representing the surface, and computing the distance of the virtual centre of the ordinates from a given base ; and if from the properties of the figure, it cannot be readily discovered what part, of the line passing through the centre, is occupied by the centre, then another base may be assumed, making an angle with the former, and if ordinates are supposed to be drawn across the figure, parallel to this base, another line may be found parallel to this base, passing through the centre ; hence, the intersection of these two lines will be the centre required. 11. If AQ bea base line drawn through B__ Lv dt any point, as suppose the vertex of any ||x ==” body, or figure QBD, and if @ denote any ordinate EF of the figure, d = AG, its dis- tance from the base line AQ, and S = thea sum of all the ordinates, or the whole figure QBD ; then the distance IC of the virtual centre from AQ, 1s denoted by the (sum of all the ad) + S. PROBLEM I. To find the centre of a triangle. Through the vertiee C of the triangle draw a base line CD, parallel to AB, let an indefinite number of ordinates ab, or ’ z, parallel to DC, be conceived to be drawn across the triangle each of which we may conceive to be drawn into their re- spective distances be, &c., from the line CD, which may be represented by a se- ries of x; hence, if the series of z, whose F E magnitude is x, bedrawn intothe series of zx, the product may be represented by a pyramid, whose base ABEF is = zz and whose altitude is AC, or x; hence, } AB X BD X BD= 1 AB xX BD? or xzx,=417z,=the sum of the products, but (Art. 2) the sum of the products + S, the sum of the series of or- dinates, or series of z, whose measure is zz, is equal to the distance of the centre, from the line CD. The sum of the series of ordinates may be expressed by half the rectangle of AB x BD, or}ze. Hence, by (A sista x BD, orize. Hence, by (Art. 11.) AB xX HD &. = 2AC ="2DB or 4232 + $zr = 2x, that is the virtual cen- tre Is some where in a line ab 2 of the distance from the ver- 176 METHOD OF FINDING tice C of the triangle to the base AB, in like manner, we shall find it also in a line GH, 1 of the Retanch from the vertice B toward the side CA ; heater it must be in the intersection of the two lines. Problem 2. Let it be required to find the virtual centre of a trapezium, ABCD. First, let DC, parallel to AB, be taken as Dees FG the base and imagine an indefinite number of equidistant ordinates ab, drawn across the figure, parallel to the base, and if these are severally drawn into their respective dis- tances from the base DC, we shall have con- structed a wedge, whose base is equal to AB, AEE E==—SB drawn into EF, and whose altitude is equal ne to EF, Let z = AB, and z’ = DC, then will the series of ordinates be represented by z..z’ let EF = z, then will the series of de- creasing distances be represented by z. Then, we have the solid generated by the production of E X(Z..2)) asd Be eles iar Lae x(2z +z’), let this be divided by 4 aheras) 3(z+2') tual centre I from the line DC. Now, in order to find in what part of the line Ab is the cen- tre, we may proceed as before to construct a quantity on CB, DG, or any line parallel to CB, consisting of ordinates EF, &c., across the figure, drawn into their respective distances from the bases, and dividing the quantity thus constructed by the sum of all the ordinates, or the area of the trapeziums. Or, we may divide the trapezium into the triangle ADG, and the parallelogram GDCB, and proceed to find the virtual. centres of each of these figures ; and it is then evident, that if the distance between the two points, thus found, is divided in. the alternate ratio of the two figures, the result will determine the centre required. Thas, the distance he of the triangle, from the side DG is } aH, (Prob. 1,) and the distance ne of the centre of the pa- rallelogram from the line DGis evidently = } nb. Let ¢ = the area of the triangle, and p = that of the paral lelogram, then as ce the distance of the two centres, is to t + p, so is p, to the distance el, of the vertical centre of the trapezium from the point e. (z + 2z')z, and we have = the distance of the vir- THE VIRTUAL CENTRE. 177 Any rectilinear figure may, if necessary be divided into triangles, and the virtual centres determined for each, and then finding the common centre of every two of these, till they are all reduced to one only, which will be the virtual centre of the whole. Problem 3. To find the virtual centre of a segment of a circle ABD. Let parallel ordinates ab, &c., be drawn across the segment parallel to its base AB, and let each of these ordinates be con- ceived to be drawn into their respective ¢ distances from AB; then by their proper- A E B ties we shall have generated an ungula ABCF; divide the soli- dity of this ungula by the area of the segment, and the quotient is the distance EF of the virtual centre from the chord AB. D H Or we may let GH be the origin, G passing through the vertice D of the segment, parallel to AB, and if the ordinates ab, &c., are drawn into their distances from this line GH, or the vertice D, the product will be the ungula ABDL, which is the compli- ment of the former ungula, and the distance DI of the centre will be found by dividing the solidity of the ungula by the area ABD. If we make the origin at the centre of the circle, we shall have the ungula ADBRCF, which is a segment of the un- gula GDHF, and the solidity of this seg- ment, divided by the surface of the seg- G O ment ABD, will give the distance OI, of the centre required. Problem 4, Let it be required to find the virtual centre of a sector of a circle CADB. Let C be the origin, and if an inde- finite number of equidistant ordinates db, parallel to the chord AB, are drawn into their respective distances Ce, &c., we shall have produced the sectoral un- gula, FADBGC, which if we divide by the surface, CADB, we shall have the distance Cl of the centre. Let r = CD the radius of the circle, and the chord AB = c then will rc = the convex surface AFDGB of the ungula, and ir’c = the solidity of the ungula. Let 7’ = the arc ADB, 178 METHOD OF FINDING then we have the area CADB = i1’r; hence the distance CI prin bg 2rc = la'y — 37? Resolving this into a proportion, we have, 37 : 2r::c: the distance CI. Hence, generally, the distance of the virtual centre of any sector of a circle is = the fourth proportional to three times the arc of the sector, twice the radius, and the chord of the arc. Let CE = V(r? —1c’,) hence we have !7°c — 1c* = the so- lidity of the pyramid ABGFC ; subtract this from the sectoral ungula, and we have j, c* for the solidity of the segment ABGFD, divide this by a, the area of the segment ABD and we have the distance of the virtual centre of the segment 3 an i T2© from the centre C of the circle = Problem. 5. To find the virtual centre of the parabola. Let its distance be estimated from the vertex. . Phe equation to the parabola is y?= ./pa, if this be drawn into a series denoted by z, the surface of the parabola will be- come y x/dz. Let this be drawn into a series Z, and we have zz/pz, and since the quantity p will not affect the result, it may be omitted, . . : 1 3 and we have by removing the radical sign x12 = 2.22; let this be divided by zzz = 2 23 and we have ZU ertex. Problem 6. To find the distance of the virtual centre of a semi-parabola from the axis. Its equation considering the origin at the extremity of the axis 1s y= (d—z)’+22(d—zx);d=the maximum value of =the axis ; or its equation may be expressed 2? + 2zz. This being put into a series, whose measure is a, represented by the base, we shall have its area, which may be ex- pressed ax’+2ax % = dy, which being integrated, we have 4 ax" + § az’ = 2az’. Let the expression a2” + 2axx, be drawn into z, z being = the base, and we have aza* + 2azzz, let this be inte- grated, and we have 4(;zx7 + 12x) 1a = la(11z2*) = [7 Gea. 2o2 , - j = 3x” = the distance of the virtual centre from the 3 7 == 2a= the distance of the virtual centre from the axis. THE VIRTUAL CENTRE. 179 Prob. 7. Let it be required to find the virtual centre of a solid. If the solid be a pyramid, its solidity may be expressed x?z. If we estimate the centre in relation to its distance from the vertex, we may draw this quantity into another series of z, and we have xz, let this be divided by z?z, which represents the base of the latter construction, and we have ex 4x é —~— = “=z, that is the vertical centre is 2 the distance De LE from the vertex of the pyramid to the base. Problem. 8. To find the virtual centre of a vertical seg- ment, of aspherical revoloid, or the virtual centre of a seg- ment of a sphere. The equation of the reveloid is 4y2 = 4dz—z?, let this be drawn into a series of z,and we have 4rxz (d—z), the segment, x being the altitude of the segment, and d the diameter of the sphere, the value of this is 1(2dx2—2zxr3) + 1(4dx?—2zx*) = dr—?2z?. Let the former series be drawn into another series of z, and its value will be 4%(dx2+Zx? = 2dx3—1 24; 2dxz3—1r4 = 4dr—3x? hence, ——-——- = ——— —— di 47% 6d—4x centre of the segment from the vertex. This construction will serve for a segment of a sphere, a spheroid, a spherical, or an elliptical revoloid. =the distance of the virtual Problem. 9. To find the virtual centre of a vertical parabo- lic revoloid pyramoid, or of a parabolic conoid. The equation to the revoloid is y’=pz, let this be put into a Series, whose measure is the axis, and we have its solidity apx; let this, as a base, be also drawn into z, and we have pxz” equivalent to }pz° ; xpx is equivalent to } p2’ ; 1 3 Me 1 g Pe hence ; = 2x = 2% the distance from the vertex to. the base. Problem. 10. To find the virtual centre of an hyperbolic conoid, pyramoid or vertical revoloid. c2 ; c? Its equation is 4y? = 4 7 (dx + x*) : since 4a stant quantity, and a factor of the whole expression, it may be omitted, without affecting the result ; then the expression will become dx + x?. Let this be drawn into a series of the form z(d+xz)_, which expresses the segment, whose value is, 3(dx+a?2) +} (idx+}2*)=fa(8de+2z?) = 5 (3dz*+2z') is a cone 180 METHOD OF FINDING Let the former series be drawn into another series of x, and there will be produced x(d + x)x? == 12(dx? + x3) + tex (jdz? +123)4=12(2dz*+112°)=1(2dz*+ 112°), hence, we have I 3 Lint 2 4Qde% + Liz), Ade. Bm Riddatp enw tenes the distance of the centre from the vertex. Problem. 11. To find the virtual centre of the arc of a circle. The virtual centre of an are of a circle, is the same in reference to the centre of the circle, as that of the segment of arevoloidal curve, whose conjugate diameter is the same as that of the circle, and the base of whose segment is equal to the given arc of the circle. For, if a_revoloidal curve AEB circumscribe atu | the semi-circle DEH, i Use and if any chord FG, be /| \ produced to fg, soas to | meet the curve, the ordi- nate fg will be equal to the arc FEG. Draw across the seg- ment fEgf, equidistant ordinates, perpendicular to fg, and they will represent the distance of the several points in the arc, from the line fg, since they are supposed to be drawn from points equidistant from each other on the line fg, or the arc FEG; and, since if there is an infinite number of ordinates, they may be regarded as the area of the segment, it there- fore, follows that if this area is divided by the arc, the quotient is the distance of the centre, from the line FG ; also, if the area f EgHD is divided by the arc FEG or line fg, the quo- tient is the virtual centre of the arc f Kg from the axis DH. Let c = the chord FGa = the arc FEG, and r = CE, then will cr = area Df EgH, (Prop. Ill, Cor. 4, B. III.,) hence, we ler have aa the distance EI of the centre. This is also the virtual centre of the segment f Egf of the revoloidal curve, and it may also be shown that if a series of equidistant ordinates to the axis Cl, are drawn through the revoidal surface, or any segment of it parallel to fg, the series of ordinates so constructed, drawn into their several distances from any given line, parallel to such ordinates, will determine the virtual centre of the segment fgef, or of the arc FEG, by proceeding as before. Art. 12. If it be required to find the virtual centre of the arc of an ellipse, a parabola, or an hyperbola, it may be done in 4 similar manner, by taking a portion of the surface of the THE VIRTUAL CENTRE. 181 elliptic, parabolic, or hyperbolic reyoloid, and proceeding as for the arc of the circle, which also gives the virtual centre of the segment of the revoloidal surface pertaining thereto. Problem 3. To find the virtual centre of the surface of a solid. Let the proposed surface be the convex surface of a pyra- mid ABCDYV ; and because any portion of the convex sur- face, included between any two sections, by planes parallel to the base, is proportional to the portion of a vertical triangle through the pyramid included between the same planes; it follows that the vertual centre of the convex surface, is the same as that of the vertual triangle ; hence the same process will determine both. If it is required to find the virtual cen- tre of the whole surface of a pyramid, including its base, we have only to imagine an infinite number of ordinates to be drawn across the several triangular sides parallel to their se- veral bases, and also a similar series of parallel ordinates across the base, and if each of these ordinates are severally drawn into their respective perpendicular distances from the vertex of the given pyramid, we shall have produced, as many new pyramids AEFCQ, whose bases R ACFE, ABHG, &c, are, severally equal to the bases of the sides of the pyramid multiplied by IA, the dis- tance of the base from the vertex, as the given pyramid has sides, and also a prisn ABDCQIKR, formed by drawing every line in the base, or the whole surface of the base into the distance of the base from the vertex. And the sum of the imaginary solids so generated, divided by the whole surface of the pyramid, will give the distance of the virtual centre from the vertex. Let AB=z and if the pyramid is generated from 2’ or hz’, or if the base of the pyramid is a square, the perimeter of the base will be 4x ; and since each pyramid ACF EQ, is equal to 3 the prism ABCDQRKI: hence the four pyramids generated by the series drawn into the four sides of the base are = 4 the prism, and if h = the altitude IA of the pyramid, ha’?+4ha?= thx" = the sum of the four pyramids + the prism ABDCQIKR ; the surface of the given pyramid is = 2’+92/(h?+32’) Hence we have, aha Thx 4+ Qn /(h?+4a%7 242 Vh?+42? equal the distance of the virtual centre from the vertex. 182 ON THE RELATIONS OF MAGNITUDES. BOOK. VII. CHAPTER V. ON THE RELATIONS OF LINES, SURFACES, AND SOLIDS, GENERATED BY MOTION. * Tue capacity of any solid generated by the motion of a surface perpendicular to itself, is measured by the generating surface drawn into the distance moved ; which distance is al- ways equal to the distance passed through by the virtual cen- tre of such surface. If the motion of the generating surface is such, as that it always maintains a parallel position, and moves in a direction perpendicular to itself, the proposition is sufficiently manifest. BS) \G Let now the rectangle ACBD revolve about R the side BC, which remains fixed, and the product will be the cylinder DF’, whose solidity is equal to the surface ACBD drawn into the circumference PK, described by the virtual centre K, of the plane, which centre is in this case also the centre of aggregation. If a right-angled triangle ABC revolve about the perpendi- cular BC, so as to describe the cone ABD, P B N this is also measured by the triangle ABC drawn into the circumference FL, described by the virtual centre of the triangle. The virtual centre of the triangle we have shown to be situated at the point F, on the line BE, from the vertex bisecting the base at a distance from B =? its length. Let 4’ the surface ABC be multiplied by the cir- cumference described by the centre F’; and since the radius FG=2EC, hence the circumference h}G=2 the circumference EC, and because the triangle ABC=1Lits circumscribing rectan- gle ADBC, which generates a cylinder ADNM, the generating surface of the triangle ABC drawn into the circumference, is equal to one-third the cylinder generated by the rectangle, or one-third the rectangle drawn into the circumference EC, as it ought to be. And in general, let any plane figure be revolved about any line or axis without the figure, but always in the same plane, GENERATED BY MOTION. 183 and the solid generated will be measured by the generating surface drawn into the are described by the virtual centre of the surface. Let AFHD be a solid generated by the plane ABD; through C, the virtual centre of which, draw DCAE, perpen- dicular to the axis of rotation, and meet- ing HGFE in E, let an indefinite num- ber of parallel ordinates, ef, 2k, &c., be drawn across the generating surface, parallel to the axis about which it re- volves; and the solid generated is equal to all of those ordinates, drawn into the distances passed through by each; viz., the ordinate drawn across the point Axthe arc AF+the ordinate ab drawn through cXthe arc CG+, &c., through © the whole series. And because EA, EC, ED, &c., are as the arcs AF, CG, DH, &c. Hence EhxXef, and Elxik, &c., are as phXef, and rlxik, and be- cause EC drawn into all the ef, ik, &c., is equal to all, the EhxXef, Elx tk, &c., it follows that CGXall the ef, ik, &c., is equal to all the phxXef, rlxik, &c.; or that the solid ABDHF is equal to the generating surface ABDe drawn into the line described by the virtual centre of the surface. Cor. 1. Hence, if any curve or any line be made to revolve about any axis exterior to such curve, but in the same plane, the surface described by its motion will be equal to the line or curve drawn into the distance passed through by the virtual centre of such line or curve. For, let the perimeter of the figure generating the solid above, be the generating line, and let us suppose its virtual centre the same as before; let every point in this perimeter be reduced to the line AD by means of perpendiculars thereto ; and the figure generated by its revolution about the axis, is equal to all the ph, rl, &c., described by every point; but we have seen that all the pA, rl, &c., are as all the Eh, El, &c. ; and since the sum of all the EA, EJ, &c., is equal to as many times EC, therefore the sum of all the ph, rl, &c., is equal to as many times CG, or equal to ABDeXCG, that is, the surface described by the perimeter ABDe, is equal to ABDe drawn into the line described by its virtual centre C. Cor. 2. From E draw EIKL, cutting the upright prismatic figure erected on the given base ABD, so as that any perpen- 184 ON THE PRODUCTION OF MAGNITUDES. dicular AI may be equal to the corresponding arc AF. Then will the figure AILD be equal to the figure AFHD. For, by similar figures, all the AF, CG, DH, &c., are as all the AI, CK, DL, &c., each to each; and as one of each are equal, therefore they are all equal, each to each; viz., all the Al, CK, DL, &c., equal to all the AF, CG, DH, &c.; that is, the figure AILD equal to the figure AFHD. Cor. 3. Through K draw MKNO; then the figure ANMD will be equal to the figure AIKLD, or equal to the figure AFHD. For, by the last corallary, AFMD is equal to the figure de- scribed by the base AD, revolving about O, till the arc de- scribed by C be equal to CK; which, by the proposition, is equal to ADXCK, or ADXCG. Cor. 4. Hence, all the upright figures AQKRD, AIKL, ANKMD, AKPD, &c., of the same base, and bounded at the top by lines or planes cutting the upright sides, and passing through the extremity K, of the line CK, erected on the vir- tual centre of the base, are equal to one another; and the va- lue of each will be equal to the base drawn into the line CK. Hence, also, all figures described by the rotation of the same line or plane about different centres or axes, will be equal to one another, when the arcs described by the virtual centre are equal. But if those arcs be not equal, the figures gene- rated will be as the arcs. And in general, the figures gene- rated, will be to one another, as the revolving lines or planes drawn into the arcs described by their respective virtual centres, Cor. 5. Moreover, the opposite parts NIK, MLK, of any two of these figures, are equal to each other. Cor. 6. The figure ASPD is to the figure APD, as AS to CK ; for, by similar triangles, they will be as AD to AC. For ASPD is equal to ADX AS, and APD equal to AD xCK. Cor. 7. If the line or plane be supposed to be at an infinite distance from the centre about which it revolves, the figure generated will be an upright surface or prism, the altitude be- ing the line described by the virtual centre; so that the base drawn into the said line will be equal to the base drawn into the altitude, as it ought for all upright figures, whose sections parallel to the base are all equal to each other. BY MOTION. 185 Scholium. If a right line, or parallelogram, revolve about a line perpendicular to the length, there will be described a ring either superficial or solid; and as the virtual centre of the describing line, or parallelogram, is also the centre of magni- tudes, it follows, therefore, that such surfaces or solids, are equal to the generating magnitude drawn into the distance passed through by the centre of magnitude. When the centre of rotation is in the end of the line, the line will describe a circle whose radius is the said describing line, and whose circumference is double the circumference described by the virtual centre; consequently, the radius drawn into half the circumference, will be the area of the circle. If a semi-circle revolve about a diameter, and describe the surface of a sphere, then will the surface of the sphere be equal to the revoloid arc X by the circumference described by the : cr ’ ‘ virtual centre = ee See Xa = 2rev = the circumference into z the diameter. And for the solidity of the sphere, we shall have the dis- 3 tance of the virtual centre equal , where d is the diameter, 12a and a the area of the segment ; let twice this distance be mul- 2d°r a’ h ign tg the so- lidity of the sphere, as before found in the Elements of Ge- ometry. tiplied by +, and also by a; and we have For the solidity of the parabolic spindle: putting b = the base, and a@ = the altitude, or axis of the generating parabola. We have found that 2a is the distance of the centre of gra- vity from the base, and consequently *%a7 = the line describ- ed by the centre of gravity ; but 2ab is = the revolving area ; therefore 1Sar x 2ab + the 22a’br will be the content, which is ;% of the circumscribed cylinder. For the paraboloid. Making the notation as in tne last ex- ample, and making 2 = the area of a circle, whose diameter is 1; 3b will be the distance of the centre of gravity of the semi-parabola from the axis, consequently 3b X 8n X 2ab= 2ab’n = the solidity = half the circumscribed cylinder. 13 MENSURATION. Having, in the elementary parts of the work, introduced such subjects of mensuration as depend on principles therein discussed, it only remains for us now to present the higher branches of the subject, or such subjects in mensuration, as de- pend on the higher branches of geometry. The subject of mensuration admits of three general divi- sions : lines, superficies, and solids ; but since the mensuration of lines is so intimately connected with that of surfaces, we shall make but two general divisions, termed superficies, and solids. PART I. MENSURATION OF SUPERFICIES. PROBLEM I. To find the area of a segment of a circle. CASE I. When the arc, sine, and radius are gwen. Rutz.—Multiply the difference between the are of the segment and its sine by half the radius. (Prop. XII, B. IV.) Let «’ = the arc, s = the sine, 7 = the radius, and A the area of the segment; and A = 1a’r—tsr. Ex. 1. What is the area of a seg- ment AE, whose arc AE is 2,09438, and whose sine ES = 1,73205, the ra- diliss— 22,4 2.094388 the area of the segment AE. D Eix. 2. What is the area of the segment EADB, whose arc EADB = 4.18878 and sine ES = .86602 ? In this example because the sine is considered negative, by Trigonometry, the arc being greater than that of a semi-cir- cle, we shall have by the rule 4.18878 ) . ; + .86602 | 2” = 1.66138 = the area required. MENSURATION, ETC. 187 : Ex. 3. What is the area of the segment whose arc is 6.9813, and whose sine is 6,4278, the radius being 10? Scholium. If the are and radius, or the sine and radius only are given, the other parts may be taken from the table of natural sines, and the area of the segment calculated by the rule. The arc of any segment less than a semi-circle may be found approximately by formula 3, (Prop. IX, B. IV.) 4¢ = J/(4vur-+1s’) —4s. Where v is the versed sine or height of the segment, r the radius, and s the sine of the half arc, or the 1 chord of the segment, x being = the arc of the segment. It will appear that this gives the value of the arc to a great degree of exactness when the segment is small. Let us see how near the truth this comes for a semi-circle. In this case, the sine and versed sine are each equal to the radius, which suppose = 1. Whence we have iz = V41—4 = 1,56155 And x = 3.12310 the true number being 3.14159 The difference of which is .01849, the error in a segment = the semi-circle. Let us assume that the error in any smaller segment, is pro- portional to the sixth power of v or v’, then if we correct this by deducting v’ x.01849 therefrom, the result will, in this case be correct, and if our hypothesis is correct, it will give the proper result for any smaller arc. CASE II. When the arc, cherd, versed sine, and radius are given. Rute.—Multiply the difference between the chord and arc by half the radius, to which add half the product of the chord and versed sine. Investigation. Let ABD be a segment; this is <7 composed of the segments AD + DB + triangle »f/ \ | ADB, but the segment AD = DB = (are AD — AF) linn Xir; segment AD+DB = (are ADB —chord AB) & X2zr and triangle ADB = ;(ABx DF). Ex. What is the area of the segment ABD, whose arc ADB, is 10,4719, and whose chord AB=10, the versed sine DF or height of the segment =1.3398 ? 10,4719 10 X 1.3398 segment. =,9058 = the area of the > 188 MENSURATION CASE Hi. When the radius of the circle, and the degrees of the arc only are given . Rure.—Find by Frigonometry, or by the table of natural sines, the sine of the given are for a circle whose radius is 1, observing that the same sine answers for an are and comple- ment, multiply this by the given radius, which gives the sine of the given arc. Then say, as 180° is to the given arc, so is + to rt’, so is the semi-cireumferenee of a circle whose diameter is 2, to the length of the given arc; then proceed as in Rule 1. Ex. I. What is the area of the segment AK, (see diagram to Case 1,) whose arc AK: = 60°; the radius EC being 1? / (EC? — SC’?)=ES= ¥(r——-ir)= v (4) =.86602=s And 180° : 60°: : 7:4’: : 9.14159 : 1.04719 = the arc AE =, Hence, (1.04719 — .86602) x3=.09058 = the area of the segment AE. Ex. 2. What is the area of the segment EBF, whose arc EFB is 120°, the other quantities remaining the same as be- fore? The sine of the arc AE is also the sine of the arc EFB. The are EFB=2 arc AE=2.09438; hence (2.09438 — 86602) X+=.61418 = the area of the segment EBF. Ex. 3. What is the area of the segment EADB, whose arc EADB is 240°, the other quantities remaining the same ? As 180° : 240: : 3.14159: 4.18878=the arc EADB. Since the segment EADB is greater than a semi-circle, its sine, ES, is considered negative by Trigonometry, we have (4.18878 +-,86602) X $=2.52745 = the area of the segment E ADB. Ex. 4. What is the area of the segment ADBFE, whose arc is 300° 7 A Scholium. The difference of the segments EBF, and the segment AF is equal to the sector ACE ; the segment ADC ~ segment EBI — sector ACE + triangle ECB. OF SUPERFICIES. 189 CASE IV. When the chord of the segment, its height or versed sine and radius are given. Rutrsr.—As the radius is to half the | ‘chord, so is twice the difference of the versed sine and radius, to the sine of / the arc of the segment; divide this by Fr the radius, reducing it to the sine of an arc of a circle, whose radius is 1. Then in the table of natural -sines, take out the arc answering to that sine in degrees, and proceed as in Rule II], to find its length ; then proceed as in rule Ist., to find the area ~~ of the segment. Investigation. Inthe right angled triangle FAB, we have FA—2CH, and because the triangle ASB is similar to FAB or CEB—hence, CB: EB:: FA: AS. Or without finding the sine AS, of the arc ADB, proceed te take out from a table of natural sines the arcs AB, DB, an- swering to AK, BE, the sines of those arcs respectively ; and af- ter finding their lengths as in Rule ILI, proceed by Rule II, to find the area. Ex. 1. What is the area of a:ssegment ABD, whose chord AB=17,3205, and whose height ED=5, the radius being 107 In this example we have FB=20, AB=17,3205 and AF== (10 — 5)2—10, to find AS. Or, we may make the triangle CEB, whose side CB—10, EB—8.6602, and CE = 5, and FA=2CE to find AS, by the rule. Hence, 10 : 8.6602 : : (10 — 5)2 : 8.6602 — the sine; and 8.6602— 10—.86602 = the tabular sine: the arc answering thereto is that of 60°, but this segment being greater than a quadrant, the arc must be the supplement of 60°120°. Then 180° : 120° : : 3.14159 : 2.09439 = the length of the arc of 120° in a circle whose radius is 1. Hence, 2.09439 x 10==20.9489 = the arc of the given seg- ment. ; Then 20.9439 — 8.6602 ABD. Or, taking the same example, having found the length of the arc ADB =20.9439, we have by Rule II. 20.9439 aa +) —17.3205 5+17.8205-+$==61.418 = the area the same as before. 5, = 61.418 = the area of the segment 190 MENSURATION Scholium. If two of the following parts, viz: the chord, versed sine, and radius are given, the other may be found by the formule (in mensuration Hl..Geom. Prop. XIII.) Scholium 2. The triangle ACB is = the triangle AFC. EBxCE=1CBx AS. That is, the product of the sine of an arc X its cosine = } sine of twice. the arc X radius. CASE V.. When the chord and radius only are given. Ruxe.—Divide half the chord by the radius, and the quo- tient will be the sine of half the arc of the segment ; find the arc corresponding thereto in the table of natural sines, in de- grees, and multiply it by 2; then find its length as in Rule 3, and multiply it by the radius. Take the versed sine — ED—CD—vy(CA—AE.) Then proceed as in Rule 2 to find the area of the segment. Ex. Taking the same example as in the last rule, having: found the arc = 20.9439, we have ED — 10 — J/ (100 — 69,899905404) = 5. Hence, 20.9439) . , hs 17.3205 UP tn pee es = 61.418. Examples for Practice. Ex. 1. What is the area of a segment whose arc is 3.14159 and whose sine is .87785, the radius being 10 7 Ex. 2. Required the area of the segment whose chord is == 12, the radius — 10. Ans. 16.35. Ex. 3. What is the area of the segment whose height is 2, the chord being 20 ? Ans. 26,8804, Ex. 4. What is the area of a segment of a circle whose arc is 110° the radius being 1? Ex. 5. Required the area of the segment whose height is 5, the diameter being 8. Ans. 33.0486. PROBLEM If. To find the area of a circular zone AKDB, or the space in- cluded between two parallel chords AB, ED, and two arcs AE, BD. Rute. Multiply the two ares AE, BD, of the zone by } the radius, to which add the sum of the products of the sines of the OF SUPERFICIES. 191 external segments, with ' the radius, if on different sides of the centre, or add the difference of those products, if on the same side. (Prop. XXIII Schol. Formula 3, B. IV.) A = 3(rv'—rs')—}(rz—rs,) L Ex. What is the area of a zone ABDE, the sum of whose arcs AE+BD is=6,28318 , xe the sine of the arc ALB=2,57178, and the sine of the arc EFD = 2,62386, the radius being = 3? A vols tee F 6,28314 -++ 262386 3 = 17,31817 the area required. +257178 Scholium. The same rule will apply to any portion ABEF of the circle includ- ed between two chords, that are not par- allel. Hence, if the sum of the axes AF, AE in this figure, is equal to the sum of AE, BD in the last, and if the arcs AIB, FLE in this are respectively = ALB, EFD in that, then the portion AFEB in this will be = the zone AEDB in that. Ez. 2. What isthe areaof azone AGHB whose two chordsare on the same side of the centre, the sum of the arcs of the zone being 2,09436, the sines of the arcs on each side being 2,59806, and — 2,95440, the radius being = 3? 2,09436 +2,59806 X 2 = 2,60703 the area required. —2,95440 Nore. Other rules have been given for finding the areas of circular segments and zones in Mensuration, El. Geom ; formule are there given for finding such data as are required, for the elements of the area. Examples for Practice. Ex. 1. Required the area of the zone ABEDHGH, the greater chord AB = 186 feet, the less chord HD = 68 feet, and the distance LP = 248 feet. Ans. 55655.1965159 sq. feet. « 192 MENSURATION Ex. 2. Suppose the zone to have its parallel chords equally distant from the centre of the circle O, each chord AB and HD = 12.49 feet and their distance LP = 10 feet ; required the area of the zone. Ans. 148.86672 sq. feet. Ex. 3. Supposing the circular zone ABEDHGA, having its greater parallel chord = 40 yards, being equal to the diame- ter of the circle, the less chord = 20 yards, and their distance LP = 17.310508 yards ; required the area of the zone. Ans. 592.08244 sq. yards. dix. 4. Required the area of the zone ABEDHGA, the pa- rallel chords AB, and HD, being 16 feet and 12 feet, and their distance HP = 14 feet. Ans. 253.0792 sq. feet. fix. 5. The zone, whose parallel chords AB = 40, HD = 30, and the breadth = 35 ; required the area of the zone. Ans. 1581.745. Ex. 6. Suppose the two parallel chords AB and HD = 80 feet and 60 feet, and the perpendicular distance from each other = 70 feet; it is required to find the distance of the greater chord AB from the centre atO ; and also to find the radius of the circle. The distance OP. .... ..... = 30 feet = 1st Ans. The radius of the circle OF = 50 feet = 2nd Ans. Kix. 7. Required the area of the zone ABEDHGA, whose arcs AGH, BED are together = 160°, the arc AOB being 110°, and the radius of the circle 10 feet. Fix. 8. Required the area of the portion ABEF, included between the two oblique chords AB, FE, (see diagram to Scholium above,) whose arc AF= 60°, BE=80°, and AIB = 110°, the radius being 20. PROBLEM III. To find the circumference of a circle approximately. CASE I. When the radius sine, and cosine of any small arc is given. Rue 1. Divide 11 times the product of the radius and sine by the sum of the radius, and half the cosine, which will give the length of the arc, multiply this by the number of such are in the whole circumference, and the product is the circumfer- ence of the circle ; the accuracy of which depends in the small- ness of the arc. Let s — the sine, c = the cosine, OF SUPERFICIES. 193 and ’ — the arc corresponding to these functions, * — being the semi circumference, and r the radius of the circle = 1. | v ‘ «= ih being the number of parts each = «’ that the _ semi-circle is supposed to be divided into. Then (Prop. IX, B. IV. Formula 1.) wae 2rs r+ic T Letia? ==. “™ 20000 Then s = ,000157079632033 c= ,999999987462994 and rs = ,000235619448050 r+le = 1,499999993831497 3rs Hence —-——= _,000157079632676 » rT +46 and, 000157079632676 xX 20000 = 3,1415926535,2 = the circumference of the circle, whose radius is 1, which is true to the last figure, which should be 8 instead of 2. Scholium. Other formula may be found for determining the circumference of a circle at Prop. IX B. XIV, viz., formule 2 and 3, to which the student is referred ; at which place, will also be found some important trigonometrical formule for finding the value of the sines and cosines, and other functions of the circle. Rute 2. Divide 6 times the product of the radius and versed sine of a small arc, by the sum of the sine + 4 times the sine of half the arc, which will give the length of the arc, which, multiplied by 2, the number of times this arc is con- tained in the circle, gives the circumference. 6A’ vow ie . XVII, = .5 +b (Prop I,) which may be expressed 6r.versin.«’ — 6rv sin. t + 4sin.in! — s+4s' Ex. Having the sine and cosine of anarc=to ;1, part ofa quadrant= ,0157073173118 = s and ,9998766324816 = c; it is required from these data_to find the arc 7’ the radius being 1. The versed ‘sine = 1 — ,9998766324816 = ,0001233675183 = v Hence, 67v = 0007402051098 And by Trigonometry we shall find s’ = ,0078539008887. Hence, s + 4s’ = ,04771229208666. 194 © MENSURATION : orv And 0007402051098 — 0471229208666 = 3-4! = ,015707963267,5 = s' which is true to the last figure, which should be 9 instead of 5. | 6A’ peaeay also be put under the — 6rs " 6rs Cos. «/-+4 Cos. x! e-F4e' iz. 2. Let it be required to find the value of 2’ to twenty decimal places ; for this purpose let x’ be an are of yytz— part of a quadrant, or 1 minute, according to the French centesimal divisions of the circle. The sine of this are to 21 decimal places, according to Legendre’s Trigonometry, is = .000157079632033525563=s and its cosine=.999999987662994524005 =c piesa And by the trigonometrical formula cos. 6= <4 1 + cos. ° Scholium. The expression form 2 we have cos. de! = .999999996915748676195. Hence we have 67s=.000157079632033525563, and 6+-4c’ =,5999999975325989228785. 6rs : ; Therefore,- x’ = 0001570796326794896,5, which is c+4e’ A true to the-last figure, which should be 6 instead of 5. Scholium. Since there is no limit tothe smallness of the are, which may be taken, and since its sine and cosine may be cal- culated to any number of decimal places whatever, it there- fore follows, that there is no limit to the accuracy with which the circle’s circumference may be calculated by this method. The circumference of the circle, as found by M. DeLagney, to 128 decimal places, by a method furnished by the calculus, is as follows The circumference of a circle whose diameter is 1, is 3.1415926535897932384626433832795028841971693993751 058209749445923078 1640628620899862803482534211706 798214808651 32723066470938446 + or 7—. ’ The series has more recently been extended to 154 decimal places. We might proceed by this method to verify the re- sults obtained by the calculus, and extend the number of decimals much farther, were it worth the labor ;- but since we have the result already, extended beyond what is prac- tically useful, the labor may be reserved for those who. have leasure and inclination to pursue it. It may be shown that, however far, the circumference should be developed in terms of the diameter, the expression would never terminate, OF SUPERFICIES. 195 or in other words, the circumference and diameter of a circle arc incommensurable in terms of each other, See notes. _ PROBLEM IV. 1. To describe an Ellipse. Let TR be the major axis, CO the Nor minor axis, and c the centre. With the radius T’c and centre C, de- scribe an arc cutting TR in the points F, f; which are called the two foci of the ellipse. “D> Assume any point P in the major axis ; then with the radii PT, PR, and the centres F f, describe two arcs intersecting in I; which will be a point in the curve of the ellipse. And thus, by assuming a number of points P in the major- axis, there will be found as many points in the curve as you please. Then with a steady hand, draw,the curve through all these points. Otherwise with a Thread. Take a thread of the length of the axis-major AB, and fasten its ends with two points in the foci, SH. Then stretch the thread, and it will reach to P in the curve: and by moving a pencil round within the thread, keeping it always stretched, it will trace out the el- lipse. There are various instruments used for the construction of this and the other conic sections. But we have not room, consistantly with our plan, to describe them here. PROBLEM V. In an ellipse having either three of the following parts given, viz., the major or minor-azxis, the ordinate, or abscissa, to find the fourth. CASE, Ix To find the ordinate. When the major axis, the minor axis and abscissa, are given. Rute. As the major-axis is to the minor-axis, so is the square root of the product of the two abscisse to the ordinate. 106 MENSURATION Fix. 1. In the ellipse of ABHD _B the major-axis AH = 70, the minor- axis BD = 50, and the two abscis- se AS = 14, HS = 56, itis requir- ed to find the length of the ordinate 4 Hi AH: BD:: y(AS x HS): LS, viz, 70: 50:: v(14 X 56) : 20, the length or the ordinate required. Ex. 2. If the major, and minor-axes, of an ellipse are 80 and 60, the abscissee AS = 16, what is the length of the ordinate ? Ans. 24. CASE Il. To find the two abscissa. When the major, and minor axes, and ordinate are given. Rute. As the axis-minor is to the axis-major,so is the square root of the difference of the squares of the semi-minor axis and ordinate, to the distance between the ordinate and centre ; which distance, added to and subtracted from, the semi-axis major will give the two abscissz. Ex. 1. The major-axis AH = 10, its conjugate BD = 50, ang the ordinate LS = 20; required the two abscisse AS, S. BD: AH:: y( (£BD)’—LS’) : CS, viz., 50: 70:: ¥ (25*—20’) : 21, the distance from the centre to the ordinate. Hence, 2AK SC "(70 2) 235 220 = "56 and 14 = AS, HS, the two abscisse. 2. What are the two abscisse AS, HS, the ordinate LS = 24, and axes AH BD = 80 and 60? Answer 16 and 64. 3. The major-axis AH = 36, its conjugate BD = 24, and ordinate LS = 8 ; required two abscisse HS, AS. Answer 18 + 3/2= 18 + 4,2426408 = 22,2426408 and 13,7573592. OF SUPERFICIES. 197 CASE UL. To find the major azis. | When the minor axis, ordinate, and abscisses, are given. Rute. From the square of half the minor axis, subtract the square of the ordinate; then extract the square root of the remainder. Next add this root to the semi minor axis, if the less abscissa be given, but subtract it if the greater abscissa is given, reserving the sum or difference. Then say as the square of the ordinate, is to the rectangle of the abscissa and minor axis, so is the reserved sum or difference to the major axis. Ex. 1. In the ellipse ABHD, there are given the minor axis BD = 50, the ordinate LS = 20, and the less abscissa AD = 14 ; required the major axis AH. First. /{ (:BD)’ — LS’) = v(16? — 20’) = v225 = v (5° X 3°) = 5 X 3 = 15, the square root of the difference of the semi-conjugate axis, and the ordinate. Then }3BD + 12 = 25 + 15 = 40, the sum. Secondly. LS’: BD x HS: : 40: AH, viz., 207: 50 x 14:: 40:70 = AH, the major axis required. Ex. 2. If the minor axis BD = 40, the ordinate CS = 16, and the less abscissa AS = 36; what is the length of the ma- jor axis AH. Ans. 180. CASE IV. To find the minor azis. When the major axis, ordinate, and absbiss@, are given. Rute. As the square root of the product of the two abscis- sz is to the ordinate, so is the major axis to the minor axis, Ex. 1. The major axis AH = 180, the ordinate HS = 16, and the greater abscissa HS = 144; required the length of the conjugate axis AD. Here AH — AS = 180 — 144 = 36 = HS, the less ab- scissa. Then /(ASXBS) : LS: : AH: BD, viz., y(144X 36) : 16 : ; 180 : 40, the conjugate axis AH. Ex. 2. The major axis AB = 70, the ordinate LS = 20, and the abscissa HS = 14; required the ipa BD. ns. 50. 198 MENSURATION ' PROBLEM VI. To find the area of an ellipse. CASE I. ¢ When the major and minor axes are given. Ruts. Multiply the product of the semi axes by 7= 3,14159 or the circumference of a circle whose diameter is 1, and this product will be the area. : Fiz. 1. Required the area of an ellipse ABLD, whose axes are AL = "70, ahd BD = 50. PAL X IBD Xx*= 10 X 50 X 3,14150 = 2748. 9, the area of the ,4 = ellipse required. Ex. 2. What is the area of the ellipse whose major axis 1s 23, and the minor axis = 18 ? Ans. 339.2928. Fix. 3. The major and minor axes being 61,6, and 44 re- spectively, required the area of the ellipse. Ans. 2128.7481,6. Ex. 4. What is the area of an ellipse, whose axes are 25 and 19 ? Ans. 373,06381. iz. 5. What is the area of an ellipse whose axes are 23, and 17 respectively ? Ans. 307,09042. Scholium. If there be two or more concentric ellipses FGHK, fghk, the area of the inner one subtracted from that of the outer one, will be the area of the elliptical ring included between them. Hence, also as for a circular ring (Mensuration Hl. Geom.) so with the elliptical ring, its area is equal to the difference of the rectangles of the semi axes of the inner one and outer one, multiplied by v = 3,14159. Let the ellipse be taken, whose axes are 25 and 19 ; 23 and 17, in the last two examples ; 25 19 23 17, . and we have Cora xX Pe cera isy x ares w= (12,5 x 9,5— 11,5 X 8,5) « = (118,75 — 97,75) 7 = Q2le* = 65,97339 = he area of the ring Ff, Gg, Hh, KA. "+ OF SUPERFICIES. __ 199 Let the results be taken from the two examples referred to, and the area of the ring will be found to agree with this, viz., the area of the outer ellipse is there found - - «=373,06381 The area of the inner one- = - =807,09042 The difference is =65,97339 the same as found above CASE Il. When any two conjugate diameters are given. Roxe. Multiply continually together any two semi conjugate diameters, the sine of their included angle, and x. (Prop. IV, Cor. 5, B. I.) Ez. The two conjugate diameters yan wy ty AB, FG, of the ellipse ADFBGEA being 32 and 28, and their included dif_p< A r Wey B angle 77° 344’; required its area. The sine of 77° 34’f is 9765625 ; therefore, 9765625 xX 16 xX 14x 3,14159 = 687.225 = the area. PROBLEM VII. To find the area of the segment of an ellipse, cut off by an ordinate to any diameter. CASE I. When the ordinate is perpendicular to either of the principal axes. Rute. Find the corresponding segment of a circle of the same height, described on the same axis, to which the cutting line or base of the segment is an ordinate. Then, as this axis is to its conjugate, so is the circular seg- ment to the elliptical segment. Or find the area of a circular segment, whose versed sine or height is equal to the quotient of the height of the elliptic seg- y* 200 MENSURATION ment divided by its axis. Then multiply continually together, this segment and the two axes of the ellipse, for the area of the segment required. — Ex. What is the area of an elliptic segment JLA, cut off by the line IL, parallel to, and at the distance of 7 from the minor axis EF’, the G axes being 35 and 25 ? ne H 17: — 74 = 10 the height Ae of the segment. Then 2./(Ae X Be) = GH the corresponding ordinate or chord to a segment of the circumscribing circle=2./(10X 25) = 15,8113883 X 2 =31,6227766. Let the semi chord Ge be divided by the semi diameter, and we shall have the corresponding sine of the arc GA of a circle, whose radius is 1 = 15,8113888 ~~» 17,5 = .903508) corresponding to which, is the arc of 64° 373’; hence the arc GAH = 64° 37)’ X 2= 129° 15’. Then AB X * = 35 X 3,14159 = the cireumference of the circle ACBD = 109, 95565. And 360° : 129° 15’ : : 109,95565: 39,4771 = the areGAH ofthe segment ; and by Problem I, Case IJ, (39,4771 — 31,6227) Sr + 31,6227 X 10 + 2 = 17,8544 X 8,75 + 31,6227 X 5 = 68,726 + 158,1135 = 226,8395 = the area of the circular segment GHA. Then 35 : 25, or7 : 5: : 226,8395 : 162, 171 = the area of . the elliptic segment ILA. Scholium 1. If the area of the segment ILFBE had been required, the circular arc GCBDH should have been taken instead of the arc GAH. 2. If the segment Io, whose base is parailel to the major axis, is required, it may in like manner be found from its re- lation to the segment POE of the inscribed circle. OF SUPERFICIES. 201 CASE Il. When the base of the elliptic segment is oblique to the axes. Rute. Divide the abscissa Pv by its diameter Pp, and find a circular segment whose versed sine or height is the quotient. Then multiply continually together the area thus found, and the two axes, for the elliptic seyrment. Or multiply continually together the circular segment, the diameter Pp, to which the base of the segment is a double ordinate, its conjugate diame- ter Dd, and the sine of their included angle, for the area of the elliptic segment. x. The principal axes of an ellipse being 35 and 25, it is required to find the area of a segment QgP, whose base Qq is an ordinate to the diameter Pp, whose length is 33, it being divid- ed by the ordinate into the two ab- scisse Pv = 7, and pv = 26. FA + AB = 7 ~ 33 = .2121,7,= the versed sine or height of the segment. The area of a segment corresponding to this height in a cir- _ cle, whose diameter is 1, is .12162866. Hence, .12162869 X 25 X 35 = 106.4251 = the segment QgP. PROBLEM VIII. To find the circumference of an ellipse. First, find the area of an elliptic ring included between an interior and exterior concentric ellipse, whose axes are sever- ally the axes of the given ellipse + n, and the same axes — nj; then divide this area by the average distance between the exterior and interior curves, and the quotient will be the circumference of the given ellipse. (Prop. VIII, B. IV.) Ex. Let it berequired to find the circumference AEBDA of an ellipse, whose axes AB, ED are 24 and 18. (See diagram to Scholium, Prob. VI.) Let us assume two other exterior and interior concentric ellipses, whose axes FH, GK are = AB + n, and ED + 2; and fh, gk, are = AB —n and ED — 2; and if n = 2, we shall have the area of the ring, as found in examples under + 14 202 MENSURATION Scholium Prob. VI=65,97339. Let this be divided by the . : Ae 9899+! average distance as found in Proposition VII, B. v= and we have 66,3115 for the elliptical circumference AEBDA. Scholium 1. If great accuracy is not required, the following approximating rules from Hutton’s Mensuration may be used. Ruue 1. Multiply the sum of the semi axes by =, or 3,1416, and the product will be the circumference nearly. Ex. Required the circumference of an ellipse, whose axes are 24 and 18. : (12 + 9) + 314159 = 21 X 3,14159 = 65,9735 equal the cir- cumference nearly. Or Rue 2. Multiply the square root of half the sum of the squares of the two axes by 7, and the product will be nearly = the circumference. Ex. Taking the same example as before, we have 24 + 17° : : 5) xX 3,14159 = 66,6483 = the circumference nearly. It will be observed by comparing the last two results, that the former one is nearly as much in defect, as the latter is in excess ; hence, if we take half the sum of the two we shall have the circumference of the ellipse more accurately. Thus + pes + 2 = 66,3084, which is very neaf the truth. PROBLEM IX. To construct a parabola ; having given any ordinate PQ to the axis, and abscissa VP. First, find the focus F thus ; bisect PQ in A; draw AV, and AB perpendicular to it; take VF = PB, and F will be the focus. Arithmetically. Divide the < square of the ordinate by four times the abscissa, and the quo- B tient will be focal distance VF. * OF SUPERFICIES. 203 Then, in the axis, produced without the vertex V, take VC = VF; draw several double ordinates SRS ; then with the radii CR, and the centre F’, describe arcs cutting the corres- ponding ordinates in the points S. — Draw the curve through all the points of intersection, and it will be the parabola required. PROBLEM X. Of any abscissa X, its ordinate y, and latus rectum, or para- meter p; having two given, to find the third. + CASE 1. To find the latus rectum. Divide the square of the ordinate by its abscissa, and the quotient will be the latus rectum. Or, take a third proportional to the abscissa and ordinate, for the latus rectum. That is, p = y’? + =. EXAMPLE. If the abscissa be 9, and its ordinate 6 Then 6 X 6 + 9 = 36 + 9 = 4 = the latus rectum. CASE IL. To find the abscissa. Divide the square of the ordinate by the latus rectum, and the quotient will be the abscissa. That is,z =y'? > p. EXAMPLE. If the ordinate be 6, and the latus rectum 4. Then 6 X 6 +4=367+4=9= the abscissa. CASE Ill. To find the ordinate. Multiply the latus rectum by the abscissa, and the square root of the product will be the ordinate. That isy = pz. * 204 MENSURATION EXAMPLE. The absciss being 9, and the latus rectum 4. Then v(9 X 4) = 36 = 6 = the ordinate. PROBLEM XI. Of any two abscise A, B, taken upon the same diameter, and their two ordinates a, b ; having any three given, to find the fourth. The abscissz are to one another as the squares of their or- dinates. That is, as any one abscissa is to the square of its ordinate, so is any other abseissa to the square of its ordinate 3; and conversely. Or, as the root of one abscissa is to its ordi- nate, so is the root of any other abscissa, to its ordinate. B «avAB SALnLB >: was Ovigrm_g ae a VA::b:byA _bvAB Pisce 4 ( Hence Bas 3 Ab’ And : a =— so B 2 . B od Ez. 1. If an abscissa = 9 correspond to an ordinate = 6, required the ordinate whose abscissa is 16. Here V9: Y16::6:6 X 4 + 3 = the ordinate. Ex. 2. Required the abscissa corresponding to the ordinate 6, the ordinate belonging to the abscissa 16 ole} S. Here 8’: 6’: : 16: 9 = the abscissa. PROBLEM XII. To find approximately the length of any arc of a parabola, cut off by an ordinate to the axis. | When the abscissa and ordinate are given. Rue. To the square of the ordinate add four thirds of the square of the abscissa, and twice the Pe root of this sum pa be the tea of the curve, nearly.* * See Hutton’s Mensuratien. # OF SUPERFICIES. 205 Fx. The abscissa VH = 2, and the ordinate AB = 6, re- quired the length of the curve EVE. Here 2, (AB? + 4VB’) = 2v [6+ Qi + 3)) = 2v-34 = 2 V3l X 3= 3793 = 9.6436508 X 3 = 12.8582, A the length of the curve AVC, nearly. Examples for Practice. Ex. 1. What is the length of the parabolic curve AVG, whose abscissa VB = 2, and the ordinate AB = 8 ? Ans. 17.4356. Ex. 2. Required the length of the parabolic curve DAVCF, when the abscissa VE = 16, and the ordinate DE = 12. Ans. 42.142615. Ex. 3. Required the length of the parabolic curve DAVCF, when the abscissa VE = 8, and the ordinate DE = 16. Ans. 36.951. PROBLEM XIII. To find the area of a parabola, when the base and height are given. Ruts. Multiply the base by the height, and two-thirds of the product will be the area. Ex. Required the area of the parabola AVCA, the abscissa VB = 2, and the base, or ordinate, AC = 12. Here 2 (AC x VB) = 2(12 X 2) = 16, the area of the parabola AVCA required. Examples for Practice. Ex. 1, What is the area of a parabola DAVCFD, whose abscissa VE = 10, and the double ordinate DF = 16. Ans. 1062. & 206 MENSURATION | Fix. 2. Required the area of a parabola DAVCFD, whose base or ordinate DF — 15, and the abscissa VE — 22? Ans. 220. Ez. 3. What is the area of a parabola AVCA, the base or ordinate AC = 20, and the height or abscissa VB = 6. Ans. 80. PROBLEM XIV. To find the area of parabolic frustum, or zone of a parabola, or of the space included between two parallel ordinates. The two ordinates, and their distance being given. Rute. To the sum of the squares of the two ordinates, add their product, divide the result by the sum of the two ordinates, the quotient multiplied by two-thirds of the altitude of the frustum, will give the area. Hix. Required the area of the parabolic frustum ACFDA, the two parallel ordinates DF’, and AC = 10, and 6, and the distance BE = 4. (DF’ + AC’) + (DF x = pails fet onpE PAG o> x3 BB by vasa ore), 2 1386+60 8 Tt Vie (HIOGE Os 2, surat, waameae a TO Minates 196 8 98 . =syo*3=5 = 322, the area of the frustum ACFDA. Examples for Practice. fix. 1. What is the area of the parabolic frustum ACFDA, whose two ordinates DF and AC — 10 and 6, and the dis- tance BE = 3? Ans, 241, Ex. 2. The greater end of the frustum DF = 30, the less end AC = 20, and their distance BE = 15; required the area. Ans. 380. Ex. 3. The greater end of the frustum DF — 20, the less end AC = 10, and their distance BE = 12. Ans. 1862. OF SUPERFICIES. 207 OF THE HYPERBOLA, PROBLEM XV. To construct or describe a hyperbola. Let o be the centre of the hy- perbola, or the middle of the transverse AB; and BC per- pendicular to AB, and equal to half the conjugate. With the centre o, and radius Co, describe the circle, meeting AB produced in F and f, which a the two foci of the hyper- ola. Then assuming several points vv, &c., in the transverse produced, with the radii Av, Bv, and centres f, F, describe arcs intersecting in the several points g, g, &c., through which points draw the hyperbolic curve. If straight lines oM, oN, be drawn from the point 0, the mid- dle of the transverse diameter, through C, and D, the extrem- ities of the conjugate, they will be the asymptotes of the hy- perbola, the property of which is to approach continually to the curve, but not to meet it, until they be infinitely produced. PROBLEM. XVI. In an hyperbola to find the'transverse axis or conjugate axis, or ordinate or abscissa. CASE I. To find the ordinate. When the transverse axis, conjugate axis, and the abscissa are given. Rute. As the transverse axis is to the conjugate axis, so is the square root of the product of the two abscisse to the ordinate. Nors. In the hyperbola, the less abscissa added to the axis? gives the greater abscissa. Ez. If the transverse axis AB = 24, the conjugate axis CD = 21, and the less abscissa BH = 8, what is the length of the 0 rresponding PH. 208 MENSURATION Here AB: CD:: y[(AB+ BH) x BH]:: PH, viz. 24: 21:: /[(24+ 8) x 8]: 14, the length of the corresponding ordinate PH, required. Examples for Practice. Ex. 1. The transverse axis AB = 60, the conjugate axis CD = 36, and the less abscissa BH — 20, required the cor- responding ordinate PH. Ans. 24. Ez. 2. The transverse diameter AB = 50, the conjugate diameter CD = 40, and the greater abscissa AH = 64; re- quired the ordinate PH. Ans. 32/14. Ex. 3. Required the length of the ordinate MK, whose transverse axis AB = 609, the conjugate axis CD = 588, and the less abscissa BK = 116. Ans. 280. CASE Il. To find the two abscisse. When the transverse axis, the conjugate axis, and the ordinate, are given. Ruts. As the conjugate axis is to the transverse axis, so is the square root of the sum of the squares of the ordinate and semi-conjugate to the distance between the ordinate and cen- tre, or half the sum of the abscisse. Then will the sum of this distance and the semi-transverse be the greater abscissa, and their difference the less. Ex. The transverse axis AB — 24, the conjugate axis CD = 21, and the ordinate PH = 14; required the two abscisse AH, and BH. Here CD: AB:: y[PH? + (}CD)’] : HO, viz. 21: 24:: v(142 + 10.5%): 20. Then the two abscissas AH and BH =HO + 1AB= 20 + 12 — 32 and 8. Examples for Practice. Hix. 1. The transverse axis AB — 60, the conjugate axis CD — 36, required the two abscisse AH, and BH, corres- ponding to the ordinate PH = 24. Ans. AH = 80, and BH = 20. Ex. 2. The transverse axis AB = 120, the conjugate axis CD = 72, and the ordinate MK = 48, required the two abscisse AK and BK. Ans. AK = 160,and BK = 40. OF SUPERFICIES. " "209 PROBLEM XVII. To find the length of any arc of an hyperbola approximately beginning at the verlex. When the transverse and conjugate axis, the ordinate, and abscissa, are given. Ruts. First. Add 21 times the the square of the conjugate to 19 times the square of the transverse, and multiply this sum by the abscissa; to this product add 15 times the transverse, multiplied by the-square of the conjugate, and call this quan- tity the dividend. Secondly. Add 21 times the square of the conjugate to 9 times the square of the transverse, and multiply this sum by the abscissa ; to ‘this product add 15. times the transverse, multiplied by the square of the conjugate, and call this quan- tity the divisor. Thirdly. Then divide the dividend by the divisor, and mul- tiply the quotient by the ordinate for half the length of the curve, or multiply the quotient by twice the ordinate for the length of the whole curve, nearly.* Ex. 1. Required the length of the hyperbolic curve PLBRG to.the abscissa BH=2.1637, and the ordinate PH=10 ; the two axes AB and CD=80 and 60. Here (2ICD'H19AB’)x BH +(15ABXCD") oy Mere (91CD'+ 9AB') xX BH+(15ABXOD*)*~ __[(21 x 60’) + (19 X 80") ] x 2.1637-+(15 X 80 X 60") ~ [(21X 60°) +( 9X80") |X 2.1637 + (15 X 80 X 60) __ (75600-+121600) x 2.1637+4320000 ~ (75600-+ 57600) X2.1637-+4320000 *7° 426681.64+4320000 4746681.64 = "288204.84+4320000 * 7° —4608204.84 * 20 = 1.03005 x 20 =20.601, the length of the whole curve PLBRG required. *2X10 —_ * Hutton’s Mensuration. 210 MENSURATION Ex. 2 Required the length of the hyperbolic are PLBRG, the abscissa BH = 20, the ordinate PH = 24, and the two axes AB and CD =60 and 36. Ans. 62.652. PROBLEM XVIII. To find the area of an hyperbola. When the transverse axis, conjugate axis, and the abseissa, are td given. Rute. To the product of the transverse axis and ab- scissa, add } of the square of the abscissa, and multiply the square root of the sum by 21 ; to this product add 4 times the square root of the product of the transverse axis and abscissa ; ;- then multiply this sum by 4 times the product of the conjugate axis and abscissa, and divide this last product by 75 times the transverse axis, the quotient will give the area of the hyper- bola, nearly. * Ex. 1. Required the area of the hyperbola PBGP, whose ab- scissa BH = 10, the transverse and conjugate axis AB and CD = 30 and 18. {21 v[ (ABx BH)+BH’]+4v(ABXRH)} x4CDx BH ae 75 AB _ f21 v[(80x 10) +(§ x 10*)]+-4 v(30x 10)}x4x 18x10 ie 75 X 30 1 J/371¢+4/300)X8 (21 J 2529 +-40/3)X8__ ea DS ak Wi tithe ik veae he 8 10 siz X 5 \/ (267) +40 8) 55 X (30 v182+-40 3) = 8x10 a x (8182448) = = 18, (40.4722128-+-6.9282082)= 151.681328, the area of te hyperbola PBGP required. Ex. 2. What is the area of the hyperbola MBNM, the ab- scissa BK = 25, DP transverse and conjugate axis AB and CD = 50 and 30? Ans. 805.090844. PROBLEM XIX. To find the area of any mizxtilineal figure by means of equidis- tant ordinates, terminated by a curve on one side, and aright line as a base on the other. Rute. To the sum of the first and last ordinates add 4 times. the sum of all the evem ordinates, and twice the sum of all the * Hutton’s Mensuration. OF SUPERFICIES. 211 odd ordinates, rejecting the first and last; and 1 of this result, multiplied by the common distance of the ordinates, will give the area, very nearly. Prop. XV, B. IV. Scholium. This rule is absolutely true for a parabola, if the ordinates are parallel to its axis. And if the distances be- tween the ordinates is small, it is approximately true for any other curve. Ex. 1. Required the area of an irregular figure, bounded on one side by a curve line at five equidistant ordinates, the breadths being AD=8.2, mp = 7.4 nq=9.2, or = 10.2, BC = 8.6; the length of the base AB = 39, and the common dis- tance of the ordinates Am, mo, no, oB, each = 9.75. Here + [((AD+BC)+4 (mp p q + or) + 2ng}) X° R75 = | | c 11(8.2 + 8.6) +4(7.4 + 10.2) + (9.2X2)] X 9.75 = 4(16.8 + 70.4-+ 18.4) X9.75=(105.6 5-7 EN OTT +3) X 9.75=343.2, the area of the space ADCBA required. fix. 2. Required the area of an irregular space ADqnA bounded on one side by acurve line, and divided by three equidistant ordinates perpendicular to the base An, the or- dinates being AD=8, mp=6, and nq=10, the length of the base An=14, and the common distance Am, mn, each equal 7. Ans. 98, Ex. 3. The abscissa of a parabola being 2, and the base or ordinate 12, required the area of the parabola. Here, by taking three ordinates, of which the first and last are each nothing, the middle one being the abscissa=2, and the common distance=6 ; hence the area of the parabola=16 = Ans. 212 MENSURATION. MENSURATION OF SOLIDS. PROBLEM IT. To find the solidity of a sphere, spheroid, a spherical or an elliptical revolord. Rute.—Multiply a central conjugate section by the vertical axis, and take two-thirds of the product for the solidity. Ex. 1. What is the solidity of a sphere, whose Zs diameter is 10 feet ? if «o 31.4159 xX $=78.5397 = toa central sec- 4 \iC >», major and minor axes of the middle Al& | = A or greater elliptic section AB are 50 yy att, 30, and ay major and minor “Es a 4 axis'at one’ end EF are 40 and;24,° @ “**-~_4. the height IK=9 ; required the solid content of we middle frustum. Here (50X30X2) + (40X24) x 9 X .2618= (300+960) x 2.3562 = 9330.552, the solidity of the frustum EFHGE re- quired. Ex. 2. In the middle frustum EFHGE of an oblate spheroid, the two axes of the middle ellipse are 50 and 80, and those of each end are 30 and 18, the height of the frustum IK = 40 ; required the solid content of the frustum KFHGE. Ans. 37070.88. PROBLEM IX. To find the solidity of a paraboloid or a vertical parabolic revoloid. Rute. Multiply the area of the base by half the height. Ex. 1. Ifthe diameter of the base of a paraboloid be 12 feet, and height 22 feet, what is the solidity ? Ans. 1243. OF SOLIDS. 219 Ex. 2. If the sides AB, CB of the rectangu- lar base of a parabolic semi-revoloid or pyra- moid ABCD are each = 10 inches, and the al- titude I-D=18 inches, required its solidity. ; : Ans. 1800 cubic inches. =a PROBLEM X. To find the solidity of the frustum of a parabolic conoid, or or paraboloid, or of a vertical parabolic revoloid. Roce. Multiply the sum of the areas of the two ends by half their distance. Ez. 1. What is the solid content of the frustum of a paraboloid, the greater diameter DC = 30, the least diameter AB=24, and the altitude EF=9 ? Ans. 5216.6268. Ex. 2. What is the content in wine gallons of a cask in the form of two equal frustums of a paraboloid ; the length=2EF =40 inches, the bung diameter DC=32 inches, and the head diameter AB=24 inches ; the gallon containing 231 cubic inches ? Ans. 108.768 gals. PROBLEM XI. To find the solidity of a hyperbolic conoid, or otherwise called a hyperboloid. Rute. To the square of the radius of the base, add the square of the diameter in the middle, between the base and top; multi- ply this sum by the altitude, and the product again by .5236, for the solidity of the hyperboloid. (Art. 22, Chap. I, B. V.) iz. 1. What is the solidity of an hyperboloid MBNM, whose altitude KB=10, the radius of its base MK=12, and the middle diameter PG=6 7? 2230 MENSURA TION: Here (MK? + PG’) x KB x .5236 = [122 + (6v7)] X 10 xX 5.236 = [144 + ( X 7)] X .5236 = (144 + 252) x 5.236 = 2073.456, the solidity of the hyperboloid MBNM re- quired. RE og Ex. 2. Required the solidity of the hyperboloid MBNM, whose altitude KB=50, the radius: of its base MK=52, and the middle diameter PG=68. Ans. 19184'7.04., PROBLEM, XII. To find the solidity of the frustum of a hyperbolic conoid, or hyperboloid. Rots. To four times the square of the middle diameter, add the sum of the squares of the greatest and least diameters ¢ then multiply this sum by the altitude of the frustum, and that product again by .1309, (et being the sixth part of .7854,) for the solidity.. Ex. 1. Required the solidity of the & frustum of the hyperbolic conoid PGDCP, the height EH=7, the great- est diameter CD=48, the middle diam- eter MN=38, and the least diameter Bi PG=27. Here (4MN?+CD? + PG’) x EHx mn .1309=[ (38° 4)+48°+277] X .1309= 4, A? _\\\\\w (5776+ 2304+-729) X7X .1309 = 8809 x 7X.1309==8071.6867, the solid com- cg tent of the frustum PGDCP required. Lz. 2. Required the solidity of the frustum ‘of a hyperbo- lic conoid PGDCP, whose greatest diameter CD = 10, the least diameter PG=6, the diameter MN=8}, and the altitude BEH=12. Ans. 667.59. fiz, 3. A cask, in the form of two equal frustums of a hy- perbolic conoid, having its bung diameter CD=32 inches, its head diameter PG = 24 inches, and the diameter in the mid- dle, between the bung and head MN=£,/310, the length of the cask 2 EH = 40 inches; required the content in wine gallons. Ans. 24998.7584 cubic inches= 108.219, &c. gallons. OF SOLIDS. 224 PROBLEM XII. Po find, by a general rule, the solidity of any-solid, frustum, or segment, produced by the revolution of any conic section, or of any revoloid circumseribing such solids. GeneraL Rute. To the sum of the ends, ‘add four times ‘a section equidistant therefrom, and multiply this sum by one- ‘sixth of the length. Scholium. This rule is true for any solid or segment, which is generated by any multiple or power of a-series of numbers in arithmetical progression. (See Beok Y, Chap. 1 & 2.) Ezx.1. What is the solidity of a-sphere, whose diameter is 2? The area ef its central section, or of its great circle is 3,14159. Hence 3,14159x 4x 2=4,1887%8 the solidity. ix, 2. What isthe solidity of a zone of a spheroid, whose two bases are10 and 5 square inches, and whose central section parallel to the bases is 9 square inches, the height ef the zone being 18 inches ? Ans. 1538 cubic inches. ‘PROBLEM XIV. To find the solidity of & circular spindle, produced by the revo- lution of a@ circular segment about its base er chord as an avis. Rots. From tof the cube ofhalf the axis, subtract the pro- duct of the central distance into half the revolving circular segment, and multiply the remainder by four times 3.14159. If a=the area of the revolving circular segment, /=half the length-or axis of the spindle, [c=the distance of the axis from the centre of the circle . to which the revolving segment belongs ; The solidity = (1’—1ae)X 4X 3.14159. Cc Ez, Wet a circular spindle ACBD be pro- duced by the revolution of the segment ABC, about AB. If the axis AB be 140, and CP half the middle diameter of the spindle be 38.4 ; what is the solidity ? The area of the revolving segment is 3791 The central distance OF 44,6 The solidity of the spindle 374402 . 222 MENSURATION PROBLEM XV. To find the surface of a circular spindle. Rute. Subtract the product of the revolving arc, multiphed by the central distance, from the product of the length of the spindle into the radius, and multiply the remainder by twice 3.1416, and this product will be the surface. Ex. Required the surface of a circular spindle ACBDA, whose length AB = 40, and middle diameter CD. = 30, the length of the arc ACB being 531. The height of the revolving segment=15, the radius of the circle==202, the central distance OM =55. (AB x OC—ACB x OE) X 3,1416 2 = (40 X 203—531 X 5) 6.2832= 3281. ta ll the surface of. the spindle ACBDA required. PROBLEM XVI. To find the solidity of the middle frustum of a circular spindle. Rute. From the square of half the axis of the whole spin- dle subtract $ of the square of half the length of the frustum ; multiply the remainder by this:half length ; from the product subtract the product of the revolving area into the central dis- tance ; and multiply the remainder by twice 3.14159. If L=half the length or axis of the whole spindie, J=half the length of the middle frustum, c=the distance of the axis from the centre of the circle, a=the area of the figure which, by revoling, produces the frustum ; The solidity =((L’—}?) x ac) X2X3.14159. Ex. Hf the diameter of each end of a frustum of a circuiar spindle be 21.6, the middle diam- eter 60, and the length 70; what is the solidity ? C The length of the whole spindle is 79.75 The central distance 11.5 The revoling area 1703.8 The solidity 136751.5 Scholium. The middle frustum of a circular spindle may be resolved into a cylinder, whose two bases are AD, CB, and a OF SOLIDS: 223 ring described by the revolution of the segment CDE about the axis KL; this ring may also be resolved into a cylindric seg- ment, whose base is the segment DCE, and whose altitude is =the inner diameter of the ring, and a circular spindle formed by revolving the segment DCE about its chord DC. (Prop. XJ, Cor. 3, B. Il.) Hence its content may be calculated ac- cordingly. PROBLEM XVII. To find the solidity of an elliptic spindle. Rute. First find the solidity of a circular spindle, generated from a segment, whose height CG, is the same as that of the elliptical. segment,.generating the elliptic spindle, the radius of the circle being CO. Then, as the length of the circular spindle is to that of the elliptic spindle, so is the solidity of the circular spindle to that of the elliptic spindle. Fiz. If half the middle diameter CD of anelliptic spindle is 38.4, and its axis AB=200, its central distance OG being 44.6, what isits solidity ? 1 The solidity of a circular spindle having the same middle diameter, | and the same central distance, we H- have found (Prob. XIV,)=874402, but its length is 140, there- fore by the rule. 140 : 200 : 374402 : 534860 the solidity required. PROBLEM XVIII. To. find the solidity of a parabolic spindle, produced by the re- volution of a parabola about a double ordinate or base. Roe. Multiply the square of the middle diameter by ;4; of the axis, and the product by .7854., Ex. If the axis of a parabolic spindle be 30, and the middle diameter 17, what is the solidity ? Ans. 3631.7 PROBLEM XIX. To find the solidity of the middle frustum of a parabolic spindle. Ruts. Add together the square of the end diameter, and twice the square of the middle diameter ; from the sum sub- 224 MENSURATION tract 2 of the square of the difference of the diameters, and multiply the remainder by + of the length, and the product by .7854. If D and d=the two diameters, and /=the length ; The solidity = (2D’+d?—2 (D—d)’) x j/X.7854. Ex. If the end diameters of a frustum ofa parabolic spindle be each 12 inches, the middle diam- eter 16, and the length 30; what is the solidity ? Ans. 5102 inches. PROBLEM XX. To find the convex surface of a cylindric ungula. Ruz. From the product of the diameter and sine, subtract the product of the arc and cosine, and multiply the difference by the altitude divided by the versed sine. Let h=the altitude AD, v=the versed sine AF, d=the diameter AB a=the arc EAG, s=the right sine FG, c=the cosine of the half are. ds—ac Then—-— x h=the convex surface. Scholium 1. When F is the centre of the base ; then v=s= d,c=o ; and then the rule becomes dh, viz., the surface is= the product of the diameter into the height. 2. When AF excedes + AB, then ac must be added, and the ds-+-ac expression becomes — xh=the surface. Fix. 1. What is the curve surface of an ungula EGDA, whose base is half the base of the cylinder and height, AD= 10, the radius FG=10? Ans. 100. Hz. 2. Given the diameter AB = 100 the height AD=140, and the versed sine AF=10, required the curve surface. Ans. 5962,738. Scholium. The same considerations will apply to cylindric ungulas, as for circular spindles, taking GE as the axis of the spindle, and AD the circumference of a middle section. (Prop. IV, Cor. 1, B. III.) OF SOLIDS. 225 . XXL, To find the solidity of a cylindric ungula. CASE I. When the base of the ungula is = half the base of the cylinder Ruts. Multiply the square of the radius of the base by the altitude of the ungula, and take 2 the product for the solidity. Scholium. This rule is absolute, without reference to the circle’s quadrature, (Prop. VI, Cor. 5, B. IIL) (s=27°h, Formula 6, Page 92.) iz. What is the solidity of the cylindric ungula AE FC, whose base EFC is half that of the cylinder, the diameter EF being 6, and the altitude CA of the ungula being 16, 3°X 16X2=96 the solidity required. CASE II. When the base of the ungula is greater or less than half that of the cylinder. Subtract the product of the area of the base by the differ- ence between the radius and the versed sine or height, from one-twelfth of the cube of the chord of the base, if the versed sine be less less than the radius, otherwise add this product, multiplying this result by the altitude ot the ungula, and divide this product by the versed sine. If a=the area FEC of the segment forming the base of the ungula, r=the radius, FI v=the versed sine CI,c = the chord EP, and A=the altitude AC. Then will the solidity of any cylindric ungula= (ge ur N v) a) i. Ex. Given the diameter HC 50 inches, the altitude AC of the ungula=120 inches the versed sine IF =10 inches, requir- ed the solidity of the ungula. The chord EF will be found=40 inches. The area of the base 279.56. Whence, by the rule, AF being less than } HC, we have (40° (rv) a) =136709} cubic inches, the solidity of the ungula EFCA. 226 MENSURATION Scholium. When the section passes ob- liquely through the opposite sides of the cylinder, the content of the ungula may be found by multiplying the sum of the greatest and least heights of the ungula by the area of the base, and its surface may be found by multiplying ithe sum of the greatest and least heights by the perimeter of the base. Hence, the ungula ABD is equal to half pg the cylinder ABCD, both in its surface and | solidity. 2. The complement LMHKD, DKHCB | of any ungula LMPD, BGAED, may be ? found by subtracting the solidity of the un- gula from a portion of the cylinder of equal base and altitude. | oy il Nh IA PROBLEM XXII. To find the solidity of a conical ungula. cut from the cone, or frustum, by a plane parallel to-the side of the cone. Rute. Multiply the area of the base by the diameter of the. base of the frustum, and divide the product by the difference of the diameters.of the two bases ; fom this quotient subtract four-thirds of the product of the less diameter by the square root of the product.of the less diameter, and difference of the diameters. Multiply the remainder by one-third of the height, and the product will be the content. Let a=the area.of.the base cmB, D=AB the diameter of the base of the frustum, d=KD the diameter at the top, h=the height. Then will the solidity of the ungula= aD (Fy-tex vd(D— a) th Ex. If the diameter AB=30 inches, the diameter ED=19.2 inches, and the height od = 18 inches, what is the content of the ungula ? Ans. 1606.41. OF SOLIDS. 227 Scholium. Other formule may be found in the general scholia, at pages 191, &c., for cylindric and conie ungulas; it is unnecessary to extend the problems here. PROBLEM XXIII. To find the solidity of a cylindrical ring. Rute.. Multiply half the sum of the inner and outer circum- ferences by the area of a section of the ring. Scholium. This rule answers for all rings, the virtual centres of whose sections are in the centre of magnitudes of such sections. For all other rings see Prop. XI. and Corollaries B. IIL. What is the solidity of the cylindric ring AD, whose inner diameter BC | is 10 inches. and whose outer diam- 4 eter AD is 20 inches ? Ans. 925.436 inches. Scholium. This ring is equivalent to a segment: ABGFA ef a cylinder, whose section DE is equal to that of the ring, and whose length FG = the inner circumference, and AB = the outer circumference. ——S—SS= SaaS K i) ot ee Be Ex. 2. What is the solidity of the cyl- indric ring EF, whose inner diameter is 0, and its outer diamer EF 10 inches? Ans. 308,478 inches. SS g Scholium. This ring is equivalent to the segment ABC of a cylinder, whose length AB is=the outer circumference of the ring. =o, = = SI = f 41 Lat GAUGING OF CASKS. Art. 1. Gauging of casks is a practical art; and since casks are not commonly constructed in exact conformity with any regular mathematical figure, the subject does not admit of being treated in a very scientific manner ; by most writers on the subject, however, they are considered as nearly coin- ciding with one of the following forms : het) ig of a spheroid, 2: | Llsiisetra tenes adlee, of a parabolic spindle. of a paraboloid, The equal frustums of a cone. 4. __The second of these varieties agrees more nearly than any of the others, with the forms of casks, as they are commonly made. The first is too much curved, the third too little, and the fourth not at all, from the head to the bung. 2. Rules have already been given, for finding the capacity of each of the four varieties of casks. As the dimensions are taken in inches, these rules will give the contents in cubic inches. ‘To abridge the computation, and adapt it to the par- ticular measures used in gauging, the factor .7854 is divided by 282 or 321; and the quotient is used instead of .7853, for finding the capacity in ale gallons or wine gallons. , 7854 Now Ja5 = 002785, or .0028 nearly ; 7854 And 93] 70084 If then .0028 and 0084 be substituted for .7854, in the rules referred to above ; the contents of the cask will be given in ale gallons and wine gallons. These numbers are to each other nearly as 9 to 11. PROBLEM I. To calculate the contents of acask, in the form of the middle frustum of a spheroid, being a cask of the first variety. Rute. Add together the square of the head diameter, and twice the square of the bung diameter; multiply the sum by 1 of the length, and the product by .0028 for ale gallons, or by .0034 for wine gallons. GAUGING. 229 If D and d= the two diameter, and /=the length ; The capacity in inches= (2D? +-d?) X 1X .7854. And by substituting .0028 or .0034 for 7854, we have the capacity in ale gallons or wine gallons. Ex, What is the capacity of a cask of the first form, whose Jength AB is 80 inches, its head diameter EF 18, and its bung diameter CD 24 ? | Ans. 41.3 ale gallons, or 50.2 wine gallons, PROBLEM Ii. To calculate the contents of a cask, in the form of the middle frustum of a parabolic spindle, being a cask of the second variety. Rute. Add together the square of the head diameter, and twice the square of the bung diameter, and from the sum sub- tract 2 of the square of the difference of the diameters ; mul- tiply the remainder by 1 of the length, and product by .0028 for ale gallons, or .0034 for wine gallons. The capacity in inches= C (2D?-+-d'—2 (D—d)") x 11x.7854. ZZ | Ex, What is the capacity of a 4) cask of the second form, whose i length AB is 30 nches, its head diam- . eter BF = 18, and its bung diameter CD=24 7 Answer 40.9 ale gallons, or 49.7 wine gallons. PROBLEM IiIl. To calculate the contents of a cask, in the form of two equal frus- tums of a paraboloid, being a cask of the third variety. Ruxe. Add together the square of the head diameter, and the square of the bung diameter ; multiply the sum by half the length, and the product by .0028 for ale gallons, or .0084 for wine gallons. 230 GAUGING. The capacity in inches= (D?+d?) X11x.7854. Ex. What is the capacity of a cask of the third form, whose dimensions are,as before, 30, 18, and 24? “D Ans. 37.8 ale gallons, or 45.9 wine gallons. SS gs PROBLEM IV. To calculate the contents of a cask, in the form of two equal Srustums of a cone. | ‘Rute. Add together the square of the head diameter, the square of the bung diameter ; and the product of the two diam- eters ; multiply the sum by 3 of the length, and the product by 0028 for ale gallons, or .0034 for wine gallons. Tne capacity in inches= (D?+d?+ Dd) x 11X.7854. Fix. What is the capacity of a cask 4{)) of the fourth form, whose length AB is Aji 30, and its diameters EF and CD = 18 \b and 24? Ans. 37.3 ale gallons, or 45,3 wine gallons. Scholium. In the preceding rules, it is supposed that the cask corresponds to the different varieties, whereas, it is seldom that a cask perfectly coincides with either ; but for the greater certainty of the truth, when accuracy is required, the follow- ing rules, the demonstration of which will be found in Hut- ton’s Mensuration, are to be preferred. PROBLEM V. To calculate the contents of any common cask from three dimensions. Ruue. Add together 25 times the square of the head diameter, 39 times the square of the bung diameter, and 26 times the product of the two diameters. Multiply the sum by the length, divide the product by 90, and multiply the quotient by .0028 for ale gallons, or .0034 for wine gallons. GAUGING. 231 The capacity in inches= l (39D +25d" + 26Dd) x — x.7854. Ex. What is the capacity of a cask whose length is 30 inches, the head diameter 18, and the bung diameter 247 Ans. 39 ale gallons, or 471 wine gallons. PROBLEM VI. To calculate the contents of a cask from four dimensions, the length, the head and bung diameters, and a diameter taken in the middle between the head and the bung. Rus. Add together the squares of the head diameter, of the bung diameter, and of double the middle diameter ; multi- ply the sum by } of the length, and the product by .0028 for ale gallons, or .0034 for wine gallons. If D = the bung diameter, d = the head diameter, m = the middle diameter, and /=the length ; The capacity in inches= (D?+d?+2m’) X4/X.7854. A 1 Ex. What is the capacity of a qj cask, whose length cr is 30 in- ™ ches, the head diameter EF=18, the bung diameter CD = 24, and the middle diameter IK 22}? Ans. 41 ale gallons, or 492 wine gallons. Scholium. In making the calculations in gauging, accord- ing to the preceding rules, multiplications and divisions are frequently performed by means of a Sliding Rule, on which are placed a number of logarithmic lines, similar to those on Gunter’s Scale. Another instrument commonly used in gauging is the Diag- onal Rod. By this, the capacity of a cask is very expeditiously found, from a single dimension, the distance from the bung to the intersectiou of the opposite stave with the head. The measure is taken by extending the rod through the cask, from the bung to the most distant part of the head. The number of gallons corresponding to the length of the line thus found, is marked on the rod. The logarithmic lines on the gauging ae are to be usad in the same manner, as on the sliding rule. 232 GAUGING. ULLAGE OF CASKS. Art. 2. When a cask is partly filled, the whole capacity is divided, by the surface of the liquor, into two portions ; the least of which, whether full or empty, is called the udlage. In finding the ullage, the cask is supposed to be in one of two positions; either standing, with its axis perpendicular to the horizon ; or dying, with its axis parallel to the horizon. The rules for ullage which are exact, particularly those for lying casks, are too complicated for common use. The following are sufficiently near approximations. See Hutton’s Mensura- tion. PROBLEM VII. To calculate the ullage of a standing cash. Ruts. Add together the squares of the diameter at the sur- face of the liquor, of the diameter of the nearest end, and of double the diameter in the middle between the other two; multiply the sum by } of the distance between the surface and the nearest end, and the product by .0028 for ale gallons, or .0034 for wine gallons. If D=the diameter of the surface of the liquor, d=the diameter of the nearest end, m=the middle diameter, and l=the distance between the surface and the nearest end; The ullage in inches= (D*+d?+-2m’) X3/1X.7854. Ex. If the diameter at the surface of the liquor, in a stand- ing cask, be 32 inches, the diameter of the nearest end 24, the middle diameter 29, and the distance between the surface of the liquor and the nearest end 12; what is the ullage ? | Ans. 274 ale gallons, or 33? wine gallons. PROBLEM VIII. To calculate the ullage of u lying cask. Rue. Divide the distance from the bung to the surface of the liquor, by the whole bung diameter, find the area of a cir- cular segment, whose versed sine is the quotient in a circle, whose diameter is 1, and multiply it by the whole capacity of the cask, and the product by 14 for the part which is empty. If the cask be not half full, divide the depth of the liquor by the whole bung diameter, and find the area of the segment, multiply, &c., for the contents of the part which is full. Ez. If the whole capacity of a lying cask be 41 ale gallons, or 492 wine gallons, the bung diameter 24 inches, and the dis- tance from the bung to the surface of the liquor 6 inches, what is the ullage ? Ans. 7? ale gallons, or 9} wine gallons. OF THE SPECIFIC GRAVITY OF SOLIDS AND FLUIDS. Tue specific gravities of bodies are their relative weights, contained under the same given magnitude as a cubic foot, or a cubic inch, &e. A table of the Specific Gravities of Bodies, and the weight of a cubic foot of each, in ounces, avoirdupois. Platinum, Common stone, . . 2520 Rolled, . ‘ 22666 Loom, . ’ & 2160 Hammered, . 20335 | Clay, .. : : 2160 Pure gold Brick, . ; j 2000 Hammered, . 19360 Ivory, °. . - 1825 Cast, : 19256 Sand, . : : 1520 Gold 22 car. fine cast, Coal, ; : ET ‘17484 | Sulphuric acid, . 1840 Mercury, ; 13596 | Nitrous acid, . 1550 Lead; . A a Nitric acid, . : 12s Silver, cast, : 10474 Human blood, : 1054 Copper, 4 . ° 8788 | Cow’smilk, . - 1081 Soft steel, Box-wood, . F 1030 Hammered, . 7839 Sea-water, . “ 1028 MOaste. ©. ; 7832 Vinegar, . : 2 lO0eg Hard steel, ‘are’ wi ; : 1015 Hammered, , 7817 Common water, . 1000 Case ee : 7815 Red wine, . : 990 Barwon... et T7987 1.’ Linseed oil, *. blah! ci 3e- Tin, : : 7290 Proof spirits at 510, 923 Cast iron, | % : 7208 Olive oil, . ; 913 Zinc, . , 6860 Alcohol, pure, . Toe Granite, . 3500 to 4000 fither, ~’. s TOG Flint glass, . ‘ 3329 Wir, ‘ : 12 Note. The several sorts of wood are supposed to be dry. Also as a cubic foot of water weighs just 1000 ounces, avoir- dupois, the numbers in this table express not only the specific gravities of the several bodies, but also the weight of a cubic foot of each, in avoirdupois ounces ; and hence, by proportion, the weight of any other quantity, or the quantity of any other weight, may be known as in the following problems. 16 234 MAGNITUDES AND PROBLEM. I. To find the magnitude of any body from its weight Rute.—As the tabular specific gravity of the body is to its: weight in avoirdupois ounces; so is’ one cubic foot, or 1728 cubic inches, to its content in feet, or inches respectively. Ex. 1. Required the solid content of an irregular block of common stone, which weighs 1 cwt. or 1792 ounces. Here, as 2520 oz: 1792 ox: : 1728 cubic inches: to its so« lid. content. Or, As 5 oz. : 256 0z:: 24 cubic inches : 12284 cubic inches, the solid content required. Ex. 2. How many cubic feet are there in a ton. weight of dry oak ? ' Ans. 38138 cubic feet. Ex. 38. What is the solid content, and diameter of a cast iron ball, that weighs 42 pounds, its specific gravity being 7208 ? aes Solidity = 161.095 cubic inches, " / Diameter = 6.75 inches.. PROBLEM II. To find the weight of a body from its:magnitude. Rurs.—As one cubic foot, or 1728 cubic inches, is to th solid content of the body, so is its tabular specific gravity to» the weight of the body. Ex. 1. Required the weight of a block of marble, whose specific gravity being 2700, the length.= 63 feet, the breadth and thickness each = 12 feet; this block being the dimensions» of one of the stones in the walls of Balbeck. Here, as 1 cubic foot : 68 X 12 X 12 (= 9072 cubic feet) : : 2700 oz. : 683,74 tons. weight, almost equal to the burthen of an East India ship. Ex. 2. What is the weight of a block of dry oak, which, measures 10 feet long, 3 feet broad, and 2} feet deep? Ans. 43351 pounds. Ex. 8. What is the weight of a leaden ball, 41 inches in diameter, its specific gravity being 11351 ? Ans. 16 Ibs. 7 oz. Ex. 4. Required the weight of a cast iron shell, 3 inches thick, its external diameter being 16 inches, and its specific gravity 7208. ‘A Solidity = 1621.0656 cubic inches. '( Weight = 2 cwt. 3 qr. 9 lb. .5 oz. SPECIFIC GRAVITY. | 235 PROBLEM II. To find the specific gravity of a body heavier than water. Rute.—Weigh the body both in water and out of water, by a hydrostatic balance, and take the difference of these results, which will be the weight lost in water. Then say, as the weight lost in water, is to the weight of the body in air, so is the specific gravity of water, to the spe- cific gravity of the body. Ex. 1. A piece of stone weighed ten pounds in air; but, in water, only 62 pounds ; required the-specific gravity. Here, as 10 — 62 (= 31) : 10: :.1000: to the specific gra- vity of the body. Or As 13 lbs. : 40 Ibs. : ; 1000 oz. : 3077 oz. = Ans. Ex. 2. A piece of copper weighs 36 oz. in air, and only 31.904 oz. in water ; required the specific gravity of copper. . Ans. 8788 ounces. Eq. 2. Required the specific gravity of a piece of granite stone which weighs 7 lbs. in air, and 5 Ibs. in water. Ans. 3500 ounces. ante f PROBLEM IV. To find the specific gravity of a body lighter than water. Rutz. —Fasten to the lighter body, by a slender thread, ano- ther body heavier than water, so that the mass compounded of the two may sink together. Weigh the heavier body, and the compound mass, separately, both in water and out of it, then find how much each loses in water, by subtracting its weight in water from its weight in air. Then say, as the difference of these remainders is to the weight of the lighter body iz air, so is the specific gravity of water to the specific gravity of the lighter body. xx. 1. Suppose a piece of elm weighs 12 lbs. in air, and that a piece of metal, which weighs 18 Ibs. 7m azr, and 16 Ibs. in water, is affixed to it, and that the compound weight is 6 Ibs. in water ; required the specific gravity of the elm. Here 18 — 16=2 pounds, the metal lost 7m water ; and (18+15) —6 = 33 —6=27 pounds, the compound lost in water. Then 27 —- 2=25 pounds, the elm lost in water. As 27 — 2 (=25 Ibs.) : 15 Ibs. :.: 1000 oz. : to the specific gravity of the elm. x “sith 236 MAGNITUDES, &c. | On, As 1 lb. : 3 lbs. : : 200 oz. : 600 oz. = Ans. Ex. 1. A piece of ash weighs 20 lbs. in air, to which is affixed a piece of metal, which weighs 15 lbs. in air, and in water, 131 lbs. ;,and the compound, in water, weighs only 84 Ibs ; required the specific gravity of the ash. Ans. 800 ounces. Ex. 2. Suppose a piece of fir weighs 11 Ibs. in air, and a piece of steel being affixed which weighed 16 lbs. in air, and an water 14 lbs.; and the compound iz water weighs only 5 lbs. ; what is the specific gravity of the fir? Ans. 550 ounces. Ex. 4. A piece of cork weighing 20 lbs. in air, had a piece of granite fixed to it, that weighed 120 lbs. in air, and 80 lbs. an water; the compound mass weighed 162 Ibs. in water ; what was the specific gravity of the cork ? Ans. 240 ounces. QUESTIONS FOR EXERCISE. 1. Having a rectangular marble slab, 58 inches by 27, I would have a square foot cut off parallel to the shorter edge ; I would then have the like quantity divided from the remain- der, parallel to the longer side ; and this alternately repeated, till there shall not be the quantity of a foot left; what will be the dimensions of the remaining piece ? Ans. 20.7 inches by 6.086. 2. Given two sides of an obtuse angled triangle, which are 20 and 40 poles; required the third side, that the triangle may contain just an acre of land? Ans. 58.876 or 23.099. 3. The ellipse in Grosvenor-square measures 840 links across the longest way, and 612 the shortest, within the rails ; now the walls being 14 inches thick, what ground do they in- close, and what do they stand upon ? A inclose 4a. Or. 6p. stand on 17601 sq. feet. 4, What is the length of a chord, which cuts off one-third of the area, from a circle whose diameter is 289 ? . Ans. 278.6716. QUESTIONS FOR EXERCISE. 237 A A st a ¥en a f ! \ \ 5. What is the area of the heart CABDFEC, the axis CD = 10 inches. (See article on spirals, page 140.) \ ‘ Ans. 104.7198 inches. SE oo 6. There are two pul- leys AHD, BFE, the di- ameters AD, and BE are each = 20 inches, and the distance Cc is 4 feet; the pulley BFE is put in motion around its axis by a belt ABFE-* DH, passing round AHD. Now if the pulley ILK on the same axis with AHD, is 40 inches in diameter, what must te the diameter hi of a corresponding pulley hft, around which the belt I/feK L may pass, so as to be of the same length and tension as that of the belt ABFEDH, and what will be the ra- tio of the angular velocity of the two pulleys. Draw Ca, cb perpendicular to al, and from c draw cn per- pendicular to Ca; el and cn will be parallel to al, and hence will be = al: with the radius Cn describe a circle nrqst, and the tangent nc will be = the tangent al, = y(Cc?— (IC — hic)’). The arc rn is = arc la — arc Al. It is required from these data to find the diameter 7h. Scholium. In the solution of this problem, it will be meces- sary to express the arc rn in terms of its functions; the mode of conducting the solution will be left for the aiidentay 7. Required an expression for the super- ficies, and also for the solidity of a tetraedron ABCD, in terms of its linear edge, AB=A. Ans. A’./3 = the surface. qzA’ V2 = the solidity. 8. Required expressions for the surface J\ and solidity of a regular hexzedron or cube in terms of its edge. Ans. 6A’,= its surface. A*,= its solidity. 238 QUESTIONS FOR EXERCISE. 2 9. How will you express the surface and solidity of an Octedron in terms of itsedge? Ans. 2A’*VW3 = the surface. sA° V2 = the solidity. 10. Let the Surface and solidity of a do- decedron be expressed in terms of its edge. Ans. 15A’/(1+2./5)= the surface. 47+21 7/5 par = the solidity 11. Express in terms of its edge, the sur- face and solidity of an Icoseedron. Ans. 5A’./3 = the surface, 5A°/7+3/5 = the solidity. 2 ches, and its depth 8 inches: what must the diameter of a bushel be when its depth is 71 inches ? Ans. 19.1067. 13. Of what diameter must the bore of a cannon be, which is cast for a ball of 24 lbs. weight, so that the diameter of the bore may be 1-10 of an inch more than that of the ball, and supposing a 9 lb. ball to measure 4 inches in diameter ? Ans. 5.646 inches. 14. Suppose the ball on the top of St. Paul’s church is 6 feet in diameter, what did the gilding of it cost, at 3d. per square inch ? Ans. £237, 10s. 1d. 15. What will the gilding of a right rectangular revoloid, whose axis is 4 feet, come to at 5 cents per square inch ? Ans. $720. 16. A silver cup, in form of the frustum of a cone, whose top diameter is 3 inches, its bottom diameter 4, and its altitude G6 inches, being filled with beer, a person drank out of it till he could see the middle of the bottom; it is required to find how much he drank ? | Ans. 42.899844 cubic inches = .152127 ale gallons, or 1 gill and 4 nearly, the quantity required. QUESTIONS FOR EXERCISE. 239 17. ‘Two persons would divide between them, by a plane perpendicular to the base, a hay rick, in the form of a para- boloid, whose altitude is 40, and the diameter of its base 30 feet ; it is required to find the difference between the solidities of the parts, supposing the’ altitude of the section to be 28 feet. Ans. 11265.758038, the difference required. 18. In the construction ef a railroad, having contracted for the sum of $1000 to excavate acertain section, the area of whose conjugate:sections in 11 different places, taken at equal distances of 3 reds each, including the ends, are as follows, viz: the first, 150 square feet,the second, 160. the third, 165, the fourth, 172, the fifth, 190, the: sixth, 210, the seventh, 224, the eighth, 240, the ninth, 202, the tenth, 108, and the eleventh 0. After having disposed of the materials to be excavated at 10 cents per cubic yard, to be delivered on an adjoining «sec- tion, I afterward received an offer to have the whole labor of excavation and delivery performed for 30 cents per yard ; shall I gain or lose by my contract if I accept of the offer, and how much ? Ars. I shall gain $355,64. 19. To determine the weight of a hollow spherical iron shell, 5 inches in diameter, the thickness of the metal being one inch ? Ans. 11.79]b. 20. It is propesed to determine the proportional quantities of matter in the earth and moon; the density of the former being to that of the latter, as 10 to 7, and their diameters as 7930 to 2160. Ans. as 71 to 1 nearly. 21. ‘What difference is there, in point of weight, between 4 . ~ block of marble containing 1 cubic foot and a half, and ano- ther of brass of the same dimensions, whose specific gravity is 8000? Ans. 496lb. 140z. 22. What position in the line between the earth and moon, is their common centre of gravity ; supposing the earth’s di- ameter to be 7920 miles, and the moon’s 2160; also'the density of the former to that of the latter, as 99 to 68, or as 10 to 7 nearly, and their mean distance 80 of the earth’s diameters? Ans. 633.65 miles below the surface of the earth. 23. How deep will a cube of oak sink in common water ; each side of the cube being 1 foot ? Ans. 11,3; inches. 24. How deep will a globe of oak sink in water; the dia- meter being | foot ? Ans. 9.9867 inches. 25. If a cube of wood, floating in common water, have three inches of it dry above the water, and 4;$5 inches dry when in sea-water ; it is proposed to determine the magnitude of the cube, and what sort of wood it is made of ? Ans. the wood is oak, and each side 40 inches. 240 QUESTIONS FOR EXERCISE. 26. Hiero, king of Sicily, ordered his jeweller to make him a crown, containing 63 ounces of gold. The workmen thought that substituting part silver was only a proper perqui- site; but Hiero, suspecting that fraud had been practised, Archimides was appointed to examine it; who on putting it into a vessel of water, found it raised the fluid 8.2245 cubic inches ; and having discovered that the inch of gold more cri- tically weighed 10.86 ounces, he found by calculation what part of the king’s gold had been changed. And you are de- sired to repeat the process. Ans. 28.8 ounces. 27. Supposing the cubic inch of common glass weigh: 1.4921 ounces troy, the same of sea-water .59542, and of brandy .5368 ; then a seaman having a gallon of this liquor in a glass bottle, which weighs 3.84lb out of water, and, to conceal it from the officers of the customs, throws it over- board. It is proposed to determine, if it will sink, how much force will just buoy it up? Ans. 14.1496 ounces|] 28. Suppose, by measurement, it be found that a man of war, with its ordinance, rigging, and appointments, sinks so deep as to displace 50000 cubic feet of fresh water ; what is the whole weight of the vessel ? Ans. 1395,, tons.. 29. It is required to determine what would be the height of the atmosphere, if it were every where of the same density as at the surface of the earth, when: the quicksilver in the ba- rometer stands at 89 inches; and also, what would be the height of a water barometer at the same time? Ans. height of the air 291662 feet, or 5.5240 miles. height of water 35 feet. 30. If the inner axis of a hollow globe of copper, exhausted of air, be 100 feet; what thickness must it be of, that it may just float in the air? Ans. .02624 of an inch thick. 31. Ifa spherical baloon of copper, of ;4; of an ineh thick, have its cavity of 100 feet diameter, and be filled with inflam- mable air. of +4, of the gravity of common air, what weight will just balance it, and prevent it from rising up into the at- mosphere ? Ans. 21'785l|b. 32. Construct a quantity == and show its value. (See Book IV., Chap. IL.) is 33. I:xpress in a series of variables, the sym- etrical cylindrical ungulas, DBAECAB. where, AB=2I1D=z, and IC=z. QUESTIONS FOR EXERCISE. 241 33. Required the solidity of the vacuity of a gothic roof, and also the solidity of the materials of the the roof ; the span AB = 30 feet, the chords of each are An and Bn =40 feet, the versed sine mo = 6 feet, the thickness of the pear KA or BH = 17 feet, at the spring of the arc, the thickness of the crown ¥¥ of the arch Dn=4 feet of the roof 60 feet. ™ ( 52896.89319 cubic feet, the solidity er of the vacuity. " ) 104854.11776 cubic feet, the soli- i dity of the materials. 34. Required the superficies of a dome in the form of a right hexagonal revoloid, each side of the base being 10 feet, and height 10 feet. Ans. 519.61524 square feet. 35. The circumference of the base of a circular dome is 150 feet, and its height 23.873 feet ; required the superficies. Ans. 3581.1 square feet. 36. If the height BC of a saloon be 3.2 feet, the BoD, of its front 4.5 feet, the distance or, of its middle part, from the arc 9 inches, and the mean circumference at m=50 feet ; required the solidity of the saloon. Ans. 138.26489 cubic feet. 37. What is the whole surface of a saloon round a rectan- gular room, the mean compass at r=67.3187 feet, the girt DrB 3.1416 feet, and the ceiling measures 16 feet long, and 12 feet broad ? Ans. 403.4727 square feet. 38. Required the concave surface of a cireular arch, raised on arectangular base, whose sides are 27 feet 3 inches by 20 feet 4 inches. | | Ans. 632.5415} square feet. 242 DESCRIPTION OF A Description of an instrument constructed by the author, for measuring distances and heights, by a single observation, and without changing the position, or measuring any base line. The principle on which this instrument is constructed, and by which the result is produced, consists in arranging mirrors, _or reflectors, in such manner as to convey two distinct images of any distant object to the eye of the observer, as seen from two positions which are indicated by two mirrors, placed at any given distance from each other on the instrument, and causing the’ object to appear in two positions at the same time, and then measuring the apparent angle under which the two images appear. For this purpose the following diagram represents one form of the construction of the instrument. aa: |Z le AZ a | AGA x \|\GAA4 ty AZ ge , g 1 i La 4a . h o> Bs AB, represents the stock or base; at the extremities of which let two mirrors, I, wu, be placed in the manner of the index glass to a common quadrant, and set perpendicular to the plane of the instrument, and at an angle of 45° with its axis or one of its edges; these we will call the object glasses, one of which, viz: 1, is placed on the centre of motion of an imdex, Mm, which is moveable about a centre at e, ‘and withit the mirror, which for distinction is called also the index glass. There are two other reflectors, n, 7. placed near the middle of the stock of the instrument, one above the other, their edges being in contact, the planes of which cross each other at right angles; one of these reflectors is parallel to one of the object glasses n, the other to the index glass, I. In using this instrument it must be placed or held so that TRIGONOMETRICAL INSTRUMENT. 243 jts axis shall be perpendicular to a line from the object EF, to the fixed object glass u, and images of the object will be formed in the two object glasses I, K, and reflected into the two glasses n andi, and hence, to the eye of the observer, forming two distinct images, ef, ph, one of which appears higher than the other; and the angle under which the two ob- jects appear, varies according to the distance of the object. In order to measure the angle contained by the apparent po- sitions as indicated by the two images, the index is moved, and with it the index glass 7, till the object shall appear in the same position in both glasses, so that there would appear to ‘be but one image formed in both, or till the images formed in both would appear identical. Nowif a vernier scale, v, is at- tached to the index and graduated, it will indicate the apparent angle under which the two images appear, or the angle under which the distance between the two object glasses would ap- pear if placed at the distance of the object. But by the law of reflection, the index would be moved only through half the angular distance of the two images in order to produce an ap- parent coincidence ; and hence the scale should be graduated with double the ordinary divisions, for the angle would be in- dicated by twice the angular motion of the index. In order that the index may be adjusted with accuracy, a tangent screw s,is provided,:by which it can be adjusted with any precision required. When the observation and adjustment is completed, we have a right angled triang'e whose base is the distance of the two mirrors, and whose altitude is the aistance of the fixed object glass to the object, and since we have one of the acute angles, the other side or distance of the object ‘becomes known. When the object EF, is at any considerable distance, the ‘angle uFI, or ull, becomes smaller, and the object will ap- spear to come to the mirror I, from the position GH ; so that HI and Fu produced, shall make an angle with each other equal to the apparent distance of the two images. But when the object is at an infinite distance, then the lines Fu, HI, be- come parallel, and the two images coincide, so’ that but one image appears to the observer; and hence, in this ‘case, the distance cannot be measured. In order to save the trouble of calculation in each case, a table may be constructed to accompany the instrument, which shall contain the distance corresponding to any given angle - pointed out on the scale, or, the scale itself may be graduated to specific distances, which may be read off instead of the angles. | Q44 DESCRIPTION OF A Another construction of the instrument, using but two glasses is as follows : Let two mirrors, a and }, be placed parallel to each other at the opposite extremities of the instrument, making an angle of 45° with its axis as before, and let the mirror }, be move- able about a centre by means of an index as described above. SCSOCC SRA USRs Masseaasnye “+ SEO) ARS CORP SSSane ° . ® ® ° 1 X o . 4 ® ‘ When any distance is to be measured by: this instrument, 18 is placed so that a line from the object P, to the mirror a, shall be perpendicular to the axis 7m, of the instrument, the eye of the observer being at any point e, in the axis produced ; and the image of the object seen in the mirror a, will, by the law of reflection, appear in the direction of this axis ; and the im- age P, in the other mirror, if seen at all, will appear at 0, not coinciding with m. But by turning the mirror b about its axis, the image will advance toward v, in the line of the axis of the imstrument, and will ultimately coincide with it; whem the two images will be seen at e, in the same line, ev. It may be easily shown that the angular motion of the mir- ror b, necessary to bring it into the position b’ n’ so that the two images shall appear to coincide, will be half the angle at P, hence the distance is determined as before. TRIGONOMETRICAL INSTRUMENT. 245 In this form of construction, I place the mirror in a wooden case or tube, leaving openings in its side in front of each mir- ror, and a small hole at the end for the eye of the observer ; the index and scale is on the out-side of the case: the mirror a, must occupy but half a section of the tube, so that the mir- ror ) may be seen over the edge of the mirror a; and the ob- ject is attained by bringing the iniages in both mirrors to co- incide, or so as to appear as one image. In order to determine the powers of this instrument, it is only necessary to observe that the distance mP, is the cotan- gent of the angle P to the radius mr, and that when mr is given, the value of mP may be calculated for any assumed value of the angle P, which is half the angle measured by the index and scale. Assuming the distance between the mirrors to be five feet, the angular motion of the mirror 6 from its po- ‘sition parallel to a, the zero of the scale, will be for 1000 feet nearly 8'35”, and for 1100 feet '7'50”, a difference of 45” or three-fourths of a minute for a difference of 100 feet, or ten per cent of the first distance. If we assume that by the divisions on the scale attached to the index, the motion of the mirror may be correctly found to_ half minutes, then the. distance between the mirrors being taken at five feet, a change of half a minute would correspond at 10 feet to .007 of a foot, at 100 feet to .62 of a foot, at 1000 feet to 63 feet, and at 10000 feet or 1.9 mile the whole angle is but 51”, and considerable variations would entirely escape detection; but by the application of a telescope to the mstrument in making the observations, its powers and accu- racy may be considerably extended. After having measured the distance to any object, as a house, or a tree, its altitude may be easily found by moving the index so that the top of the object in one mirror shall co- incidé with the bottom in the other, when the angle indicated on the scale, less the angle first found, corresponding to, the to the distance, is the angle under which the object appears ; whence having the angle and distance of the object, its alti- tude becomes known. The following investigation of the powers of the instru- ment, showing its limits of practical accuracy, is taken from a report furnished by a committee of the Franklin Institute of Pennsylvania, to whom the two instruments designated above were submitted by the author in 1833. (Published in Vol. XI., No. 8, Journal of the Franklin Institute.) 246 DESCRIPTION OF A Call the variation in the angle P, y, the distance m P a, and mr, b. Suppose the angle P, to become P — y, and that then a+x, x denoting the increase of length of a, corresponding to a decrease, y, of the angle P. By trigonometry, fan,. Pb ys= Q and a tan. (P—y = Fee ar or by substituting for tan. P and tan. (P — y) their values found above. 6 ® ds tan. y, GGhisc ov: cA aM» BHR +g” tan. y bs 2) — a> tan. h | atx a+b tan.y’ else joe (a Bibs be okt b—a X tan. y’ a’ +0° b eee eine Geen tan. ¥ If, as assumed above, b = 5, and y = 1, the general. equa- tion becomes i et 25+a v= 17241.4—a 125 ¢ Wh 244 — Spa eS vi en a — 10,2 172314 007.- 10.025 Kor a= 100, z = aaa For a = 1000, xz = 61.6, and for a = 10,000, z = 13,809 ; which is greater than the distance a. By assuming a limit to the accuracy required, calculation will show how the instrument may be adapted to this limit when possible. Tor example, let the greatest inaccuracy al- lowed be one foot in 100, then b and y must be so adjusted that at the greatest distance for which the instrument is to be used z = a0" Calling this value of a, a’, we shall have, TRIGONOMETRICAL INSTRUMENT. 247 qa? b? .O1 a’ = b sas or ) —. q! tan. y O1 a’ ier re x 6 =— 1.01 a”, an equation which must exist in order that the required accuracy may be attain- able. ‘To examine by it the instrument already supposed, let us ascertain whether at 1000 feet, as the greatest distance at which it is to be used, the accuracy will come within the li- mit of one foot variation, in 100. In this case a’ = 1000, 8, as before, = 5, and tan. y = tan. 1’. Whence 2? = 25, = 17241.4, 0la' = 10, and 1.01 a? =1.010.000. Sub- tan. y stituting these values in the equation above, it requires 25 — 172,414 = — 1,010,000, the equation is not fulfilled, and the instrument does not come up to the re- quirement. It would be easy to determine values of y and 6 required for all possible degrees of accuracy, and thus by the possibility of making the half divisions accurate, and by the length which convenience might limit, to ascertain whether the instrument could be constructed to give the required de- gree of accuracy. : he B . a The investigation may be made more general, thus ; let 3 express the required limit of accuracy at the greatest distance for which the instrument is to be used, then at that distance a g tiie HOT calling, as before, the value of a, a’, Gd ao o- thee rewnence n b tan.y / 12 ] a pa MRE dy ta Wat 1 gti(T, hy nm tan. y Tes nN NAT’ AS AB FRO OH BITE tan. y 4 n* tan. TP Gi Oa ot 1 b=a(itV1- 4 n? tan. *y (1 + = 2n tan. y This equation is possible when 4n tan. *y (n+1) < 1. Nata. a ee BOOK I. Havine in the preceding volume treated of the properties of the Paraboza, Ellipse, and Hyperbola, we show in this, that these curves are the sections of the cone, and are each formed by a plane passing through a cone, accord- ing to certain conditions. Of these sections the quadrature of the parabola iseasily attained, by the principles embraced in Prop. IV., in relation to the sectional divisions of a prism. Props. VI. and VII., show the application of the principles* to the quadrature of the parabola. The quadrature of the Milipse, though not so accurately determined as that of the parabola, is nevertheless easily expressed in terms of the circle’s quadrature. The Hyperbola is of more difficult determination, in relation to its quadra- ture, than the other conic sections, but, is nevertheless susceptible of being approximately determined to any extent required. BOOK II. ON SOLID SECTIONS OR SEGMENTS. Since the term section, though originally applied only to surfaces, made by cutting a solid, is extensively used in the arts, as expressing a definite por- tion of a solid, I have taken the liberty of making solid sections, as synoni- mous with segments of solids. This book, consists mostly of the compari- son of cylindric, and conical ungulas. BOOK IMI. OF REVOLOIDS. Revoloids are a class of bodies not usually treated of in works on Geom- etry ; but, from the important considerations connected with their organiza- tion, and from the relations which they bear, both to rectilineal and curvili- neal solids, it is of the highest importance to Geometry, that their properties should be discussed, and that they should receive a conspicuous place ~ among geometrical solids ; more especially, as they are almost the only cur- vilinear solids, that are absolutely cubable. For, we have shown, in the progress of the work, that, not only the right or spherical revoloid, and the elliptical revoloid, Props. VI. and VU. are cubable, is absolute terms, but also Parabolic and Hyperbolic Revoloids ; moreover, we have shown (Prop. Ill.,) the quadrature of the surface of a right revoloid, without regard to the circle’s quadrature, although it is bounded by cylindric surfaces. Some important principles are derived from the subject of mathematical transformation, as in Prop. IV., V., XI., and Corollaries. * In Prop. VI, the parallel lines AK &c., should be parallel to the axis EF of the parabola, instead of the position as there expressed, otherwise, the argument is not correct; but since the principle to be established, is not affected by the errorin that diagram and argument, it is deemed best not to alter the demonstration in this edition. The Scholium to Prop. VII, may be corrected by making AB or CD the axis of the parabola, instead of AC or BD. In all re- ferences to these propositions in the succeeding parts of the work, it will be perceived that the conditions of their application is expressed. ‘ a” NOTES. ' a 249 BOOK IV. The quadrature of the revoloidal surface treated of in Book III., furnishes us with data for the quadrature of the circle, through the medium of the re- voloidal curve ; this subjectis amply discussed in this book ; through the properties of this curve, we are aiso enabled to deduce some important trig- nometrical functions. In Prop. Ifi., it is shown that the revoloidal curve may be by transformation derived from an elliptical curve. expressions are obtained in Props. IX. and XII., for the length of the cir- ele’s circumterence ; it is there shown that these approximations may be ex- tended indefinitely, so that the circumference may be obtained to any degree of exactness required. By applying the principles of the parabola to those of the revoloidal curve in Prop. XVIL., a remarkable approximation is obtained, so that if the sine and cosine of a small arc, and sine of half the given arc is obtained, the arc itself may be expressed in terms of those functions ; and it is shown that these functions may be so taken, that the arc shall be truly expressed to the same number of decimal places, that those functions are truly expressed ; in pur- suance of this, will be found, an example in Mensuration, page 194, where the are is correctly caleulated to 20 decimal places, by a simple process, having the sines and cosine given, as data, to 21 decimal places. Other modes of approximation, for the arc of the circumference may be pointed out, depending on the same principles, derived both from the quadrature of the revoloidal surface, and cubature of the revoloid; bat, by pursuing the course pointed out, page 125, the arc of the circumference may be found, with certainty, by this process, to any number of decimal places we have patience to pursue it. Props. X VIII and XIX, prepare us for the construction of a curve, described in Prop. XXI., termed the curve of the circle’s quadrature, the properties of which are, that a line drawn from any point in this curve, perpendicular to its conjugate diameter, will be equal to the are-of the inscribed circle cut off by a secant drawn from the centre of the circle to this point, and if another line be drawn from the same point in the curve, to the extremity of the conjugate diameter, the area of the space, intercepted by the two lines with- out the circle, will be equa! to the area of the segment of the circle cut off by the latter line, From these properties, we are enabled to deduce an im- portant theorem, in relation te segments of the circle, viz., that the area of any segment of a circle is equa] to the difference between the arc of the segment, and its sine, multiplied by half the radius. BOOK V. In this book, all geometrical magnitudes are discussed from their princi- ples of organization, from elementary magnitudes, without referring them to any specific forms or relations. And after introducing, and explaining the principles of the production of geometrical magnitudes, from variable ele- ments, in Chap. L., in order to render this science subject to analytical, and algebraic consideration, a peculiar notation has been introduced as the sub- ject of the second Chapter; and the mode of application, of this notation, to such subjects is there explained. By this mode of conducting geometrical investigations, and by this nota~ tion, we are enabled in a manner, somewhat more obvious than that of the calculus, to arrive at the same results as are obtained by that science, And although this forms but an introduction to the subject, yet i furnishes evidence of its adaptation to the investigation of the properties o all geometrical magnitudes : and may, perhaps, be rendered of equal univer sality with the calculus, with which it is most intimately allied, s 250 NOTES. Here instead of the infinitely small momentary increments of variable mag- nitudes, or instead of the differentials of variables, this notation recognizes only the conditions of the variables themselves, in their associated capacity, which, from the principles of the science, may be integrated, as certainly, and with more obvious rationality, than those performed by the calculus from their differentials. By this notation, we are enabled to get a positive expression for the cir- cle’s quarature, in known functions of the diameter ; which, since all Geom- etricians are satisfied of the incommensurability of the circumference In direct terms of the diameter, should be received as the quadrature itself. For, from this expression, means may be devised of developing, decimally, the quadrature to any desirable extent. Chapter III: is an introduction to the differential and integral calculus, de- signed to show the first principles of that science, and to show its connec- tion with that of Chapter II., both of which are evidently in their essential particulars, based on the principles contained in Chapter I. Chapter IV. is devoted to the application of the. principles previously dis- cussed, to determine the position of the virtual centre, or centre of gravity of geometrical magnitudes. This has universally been denominated by all authors, hitherto as the centre of gravity. I doubt not, I shall have the ap- probation of most Mathematicians in discarding that term, and supplying in its place that of the virtual centre. The term centre of gravity, though per- fectly proper in works on Mechanics and Natural Philosophy, is highly incongruous in a work on pure mathematics, where the physical properties of bodies is not a subject of investigation, and perhaps not intelligibly un- derstood. MENSURATION. Such subjects in the mensuration of surfaces and solids, as could not con¢ sistently be introduced into the elementary part of the work, is introduced here; most of the subjects embraced in this, have been discussed in the geom- etrical part of the work, and the principles demonstrated ; where this is not the case, a reference has been made to the author, where such demonstra- tion may be found ; on this subject free use has been made of Hatton’s Men- suration ; limited quotations have also been made from other authors. The article on Guaging is mostly taken from Day’s Mathematics, though originally derived from Hutton. TABLE OF NATURAL SINES. In the following table of Natural Sines, the sines are exhibited to every de- gree and minute of the quadrant, and so arranged, that the degrees corres- ponding to the sines will be found on the top of the page in the same column with the sine ; the minutes in the left hand column opposite their sines: and the degrees answering to the cosine will be found in the same manner, at the bottom, with their minutes in the right hand column. The sine or cosine of an arc greater than 90°, is the same as the sine or cosine of its supplement; or the same as those of the difference of 180° and the given arc. The sines and cosines of any arcs greater than 180°, are the same as those for ares less than 186°, but with contrary signs, being by trigonometry considered negative ; they may be found in the table as above, by deducting 180° from the given arcs. If the sine or cosine is required to degrees, minutes, and seconds. Take out the sines or cosines answering to the next less, and the next greater arc expressed in the table; multiply their difference by the given number of seconds and divide the product by 60; then the quotient added to the sine of the next less arc will be the sine or cosine required. Example. Required the sine of 32° 21' 45”, or its supplement 147° 38’ 15”. The sine of 32° 21’ is, .535090 The sine of 32° 32’ is, 585335 The difference is, .000245. Hence, 245 x 45+-60 == .000245X 3 = .000184, the quantity to be added to .535090. Therefore, .535090 +.000184 ° The sine required. = .535274. If the arc of a given sine or cosine is required in degrees, minutes, and seconds : Take out the arcs answering to the next less and the next greater sine and cosine, and multiply their difference by 60, and that product by the dif- ference of the next less, and the given sine or cosine divided by the differ- ence between the next less and next greater sine, and add or subtract as be. fore for the arc of the sine or cosine. Example. Required the degrees, minutes, and seconds corresponding to the sine .495994. The sine next less than that given is, -495964 The next greater sine 1s, 496217 The difference, .000253, The difference of the next less and the given sine is, 000030. The arc corresponding to the next less sine is 29° 44’, , Hence, 000030 x 60+-000253 = 7 = the number of seconds to add to 29° 44’, Therefore, the required arc is 29° 44’ 7”, oh 2 NATURAL SINES. ia m| 0° is Nl 8°.) Sb Re of OP fer? 'p 6? 9° |™ 000000 017452 034899 052336069756 087 156|104528) 121869|139173/156434)6 - 000291/017743'035 190/052626/070047 087446) 104818) 122158) 139461)156722'59 000582|018034) 03548 1|05291 7070337 087735| 105107) 122447 139749'157009'58 000873 018325 035772 053207070627 088025 | 105396) 122735) 140037|157296 57 001164/018616/036062 053498 070917 088315} 105686) 123024! 140325|157584/56 001454|018907 036353053788 071207 088605) 105975/ 123313 140613/157871/55 001745/019197|036644/054079 071497 088894| 106264) 123601 140901/158158/54 002036 019 188]036934 054369 071788 089 184/ 106553) 123890 141189|158445)53} 002327 |019779}037225|054 660/07 2078 (89474 | 106843) 124179) 141477|158732)52} 0026 18)020070) 137516054950 072368,089763} 107132) 124467, 141765) 15902059 10,0029119|/029361| 137806 055241072658 090053) 10742 1| 124756) 142053) 159307/50 11/0032U0| 020652! 138097055531) J72948 090343) 107710, 125045 149341 |159594 49 1200349 10209421)38388]055822|073238/090633/ 107999 125333 143629/159881!48 13/003782/92 1233 033678)056112'073528)090922'1 08289) 125622 142917/160168147 14/004072/021524/038969 056402073818,091212) 10x57 125910/143205)160455/46 15)004363/021815/039260,056693 074108 091502) 108867 126199/ 143403 160743/45 16)004654/022106 039550)056983) 374399 091791)109156 126485/143780, 61030)44 17/0049-45/022397/039841/057274/0746¢ 9 092081/109445) 126776) 144068 161317/43} 18|005236 022687 040132 057564/074979 09237 1|109734/127065/144356/161604\42 t9|005527|022978 040422 057854 075269) 092660] 110023| 127353144644) 161891141 20/005818)023269 0407 13053145 075559 092950) 110313127642) 144932 162178|49 21|006109/023560 041004/058435/075849 093239110602) 127930 145220162465 39 22/006399/023851)/041294 058726 076139) 93529) 1 10891 )128219/145507|162752/38 23 006690)024141/041585/059016 076429 093819) 111180) 128507|145795) 163039 37 24/006931|924432/041876 059306/076719|J9410-|111469) 128796) 146083) 16332636 25/007272\124723 0421 66\059597|077 009) 094395 |112758/129084/146371/163613 3A 96|007563) 125014 042457/059887/077299/094687) 1 12047| 129373) 146659 163900/34 27|007854} 125305 042748 060177/0775891094977) 112336/ 129661) 146946) 164787/33 98|008145|025595|043938|060468/07 7879095267) 112625) 129949|147234/164474 32) 29/008436) )25886) 043329 0607580781 69 095546) 1 12914)130238]147522)16476131} 30008727 /)26177 043619 061049078459 095846) 113203 130526) 147809 165048 30 31/0090 17} 326468)043910 061339 078749 096135/1 13492 130815/145097 165334 '29} 32/0093.)8|026759 044201 064629)973039 096425} 113781 131103) 148385 165621 28 33|009599| 127049)044491 061920079329 096714) 114070 131391) 148673165908 27 34/00989)|027340/044782 062210 079619/097004| 114359) 131680 148960166195 26 35,010181|27631|045072 062500/079909/097293) | 14648) 131968) 149248) 166452)25 36|010472/027922|045363|06279 1/030 199/097583} 114937] 132256) 149535|166769)24 37|010763|0282 12|045654/06308 1/080489 097872) 1 15226) 132545) 149823) 167056)23 38)011054) 128503045944 063371 080779 098162 1 15515 /132833 150111 167342 22 39|011344]02879 4 046235 063661/081069'09845 1/115804|133121 150398) 167629 21 40/011635| 129085046525 063952081359 098741) 116093 133410|150686 167916 20/ 41|01192+} )29375)04681 6) 64242/081649/09903U]1 16382) 133698 150973 16820319 42\012217) 129666|947 106;064532)181939/099320)116671 133986 151261 168489 18 43))12505)29957|047397) 064823 082228/099609) 116960) 134274/ 151548 168776 17) 4.4/012799|)39218}947688) 065113 082518)099899) 117249134563 151836 169063 16) 45/013990)| 80539/04797#|065403 082898) 100188) 117537) 134851) 152723 169350115 46/013380)030829|048269\ 065693 083098) 100477} 127826) 135139|152411/169636 14 47/01367 1) 131120/048559|065954 083383) 100767|118115,135427/ 152698 169923 13) 48}013962|031 411|/048850,066274/083678) 101056)13844/135716 152986 170209 12 49 ,014253)931702 049 140\066564 083965) 1013.46] 18693, 136004)153273 170490 11 50,014544/ 131992/04943 1066854 84258 101635)/118982/136292)/153561 170783) 10 51}014835|)32283 049721) 067145084547) 101924/119270 136580) 153848 171169) 9 52\015126\032574 050012)067435)084837) 102214] 119559 136868) 154136 171356) 8 53)015416 032864 050302) 067725085127) 102503!119848, 137156/154423 171643) 7 54/015707/033155,059593) 068015 085417 102793) 120137 137445/154710,171929 6 551015998|033146 050883) 068306085707) 103082/120426/137733 154998 172316) 5¢ 56 016289) 033737 051174|068596 085997 10337 1)120714) 138021|155285 172502) 4 57016580) 134027 051464) 068886 086286) 103661) 121003 138309 155572 172789) 3 58016871] 134318 051755}069176 086576 103950) 121292 138597, 55860,173075| 2} i 0 COHAOMUEWNH SCS 591017162) 134609 052045069466 086866) 104239) 121581) 138885) 56147) 173362 a 34899 052336|069756 087156) 104528) 121869 139173) | 56434 173648 89° | 88° | 87° | 86° | 85° | 84° | 83° | 82°} 81° | 80° |m Natural Co-sines. M 173643 173935 174221 (174508 174794 175080 6|175367 71175653 81175939, 9/176226 10|176512 11}176798 12}177085 13|177371 14|177657 15|177944 16|178230 178516 178802 179988 179375) 179661 91179947 180233 41180519 5/180805 61181091 181377 181663 29/181950 30) 182236 31) 182522 32| 182808 33183094 34/183379 35\183665 36/183951 37| 184237, 38) 184523, 39) 184899) 40}185095 41/1853: 42185667 43185952 44)186238 451185524 46/186810 om ONO re Cc 19@8u9 191095 191380 191666 191951 192237 192522 192807 193093 193370 193664 m| 10° | 11° | 12° 193949 NATURAL SINES. 13° 207912) 224951 208196/225234 208480) 225518 208765) 225801 20950) 226085 29334) 226368 209619|226651 209903) 226935 210187/227218 210472)/227501 210756) 227784 211040/228068 | 14° 241922) 242204 242486 242769 243051 243333 243615 243897 244179 244461 244743 245025 15° 252819 259 100 259381 259662 259943 260224 260505 260785 261066 261347 261628 261908 Bic 292372 292650 292928 2)3296 293484 277035)293762 277315/294040 277594/294318 277874)291596 278153/294874 278432|295152 278712|295430 16° 275637, 275917 276197 276476 276756 | 18° 309017 309294 309570 309847 310123 310400 310676/327218 310953/327493 311229|32776x 311506/328042 19° 425568 325843 326118 326393 326668 326943 311782)/328317 M 60 59 58 o7 56 95 54 53 52 5] 30 312059 194234/211325 194520/211609 194805/211893 195990/212178 195376)212462 195661/212746 195946/213930 196231/213315 196517/213599 196802)213 83 L97087|214167 197372/214451 197657/214735 197942)215919 198228) 215303 198513)/215588 198798)215872 199983'216156 19936x%)/ 216440 199653)2 16724 199938/217008; 200223) 217292 200598)217575 200793)217859 201078)218143 201363)218427 2)1648/218711 201933218995 202218)219279 202502/219562 202787/2 19846 2113 972)22013) 293357/220414 203642/220697 228351 22863 4 223917] 229200 229484 299767 230050 230333 230616 230899 231182 231465 231748 232031 232314 932597 232880 233163 233 145250330 23372“| 250662 2340111250943 2342941251225 234577/25 1506 234859/251788 935142)252069 9354251252351 935708252632 2359901252914 236273}253195 9365561253477 2368381253758 937121/254039 237403/254321 237686/254602 245307 245589 215871 246153 246435 246717 246999 247281 247563 247845 248126 248408 2148690 248972 249253 249535 249817 | 250098; 203927 47 48 49 50 5] 52 53 54 55 56 o7 58 59 60 M 3 187996) 294211 187381 187667 187953 188233 188524 188-10 189095 189331 20478] 25065 205350 295635 25920 206204 206489 189667 206773 189952 207058 190238207343 190523 207627 24496, 190809 207912 oer 18° 22098 1/237968; 221265)238251 221548)/2338533 221832)/238316 222116/239098 222399 /239381 222683 239663 222967/239946 223250 223534 223317 224101 224384 224668241649), 22495 1/241922 77° | 76° | 240793 254883) 255165 255446 255727 256008 256289 25657 1 256852 257695 252538 258819 ————$__— 15° 262189 262470 262751 253931 263392 263592 2633873 264154 261434 264715 264095 (5276 265556 255837 266117 26397/283179 266678) 283457 266958/283736 267232) 284015 267519) 284294 267799/284573 268079) 284852 268359/285131 268640) 25410 268920) 285688 269200285967 269480) 286246 269760/286525 270040) 286803 270320) 287082 270600\287361 270880287639 271160/28791& 271440/2838196 271720/288475 272000) 288753 |272280/289032 272560)289310 (272841129589; 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ERRATA. BOOK I. In the diagram to Prop. II, a right line should be drawn fron Fito A. | In Prop. VI, the parallel lines kK, sl-ould be drawn parallel to the axis EF. In Scholium to Prop. VII, AB or CD, should be the axis of the parabola instead of AC or BD. On wes 31, eighth line from bottom, read (dz—z’) instead of BOOK II. Page 60, second line from bottom, IN should be changed to LN. Page 61, third line from the top, MH should be changed to GLMH. BOOK III. On page 101, eleventh line from the bottom, for fluction, read fluxion. On page 110. eight lines from the bottom, for rz, read ars, and for 2is read 3rs. Page 112, sixteenth line from bottom, for rs read rs. yee thirteenth line from bottom, supply 2 to 3rs, to make it read 3 ors. Page 113 Formula 2, for 15978 read }sz.; and in the next line for #rs read °rs. On pages 143 and 144, CQ should be changed to AQ CP. to AP, and fifth line from bottom, page 144, change C to A; also in the third line from top for »n/ar—z’ “Pts rA/ 2az—z’. BOOK Y. On 158 page Article 6th fourth line, for zz read zz; and on the fifth line, for z= read z~; also in article “7th on the ninth line, for J (dx) x, tead a/(dx)z. | Page 158, tenth line from top, for AD, read BC; alsoin article 10th, ‘supply the inferior dash to z, in first, second, third and fourth lines, so that they may read z’, az’, az and az’; also on the sixth line, article 10th, for an/z, ead nite Page 152, on tenth line, for au, read au. y Ay vie Fo ad . UNIVERSITY OF ILLINOIS-URBANA 516SCH6H C001 HIGHER GEOMETRY AND TRIGONOMETRY NEW YO < ib “ Ps teed / et ee Loe PS Ss ie talk Ss me = ie 3 uth f ber ; ae 2 >! = 4 1 r i j i a = ad r a ¥ eee ee ee ee ae eee 4 ; x Me ghee 3 Be oe eee ol ei pease at