220 CONSTRUCTION OF GEOMETRICAL PROBLEMS. consequently m. ED=n. PD; hence PD: ED::m:n. But PD: ED: (by sim. tria.) PB: BF; therefore FB: BF :: m : n. Now PB is given, therefore BF is given in magnitude, and, being at right angles to PB, is also given in position; therefore the point F is given, and consequently PF given in position; and therefore the point E, its intersection with the circumference of the circle AEE'B, or the arc BE is given. Hence the following Const. From в, the extremity of any diameter AB of the given circle, draw BM at right angles to AB; in AB (produced if necessary) take BP an nth part of the given sum; and on BM take BF so that BF: BPN m. Join PF, meeting the circumference of the circle in E and E', and BE or BE' is the arc required. Demon. From the points E and E' draw ED and E'D' at right angles to AB. Then, since BF: BP: nm, and (by sim. tri.) BF: BP :: DE: DP; therefore DE DP :: n : m. Hence m.DE=n. DP; add to each n. BD, then will M. DE + n. BD = N. BD + n . DP = N PB, or the given sum. · Ex. 3. In a given triangle ABH, to inscribe another triangle abc, similar to a given one, having one of its sides parallel to a line mun given by position, and the angular points a, b, c, situate in the sides AB, BH, AH, of the triangle ABH respectively. Analysis. Suppose the thing done, and that abc is inscribed as required. Through any point c in BH draw CD parallel to men or to ab, and cutting AB in D; draw CE parallel to bc, and DE to uc, intersecting each other in E. The triangles DEC, acb, are similar, and DC: ab :: CE: bc; also BDC, Bab, are similar, and DC: ab :: BC: Bb. Therefore BC: CE :: Bb: be; and they are about equal angles, consequently B, E, C, are in a right line. m A H a B Construc. From any point c in BH, draw CD parallel to nm; on cn constitute a triangle CDE similar to the given one; and through its angle E draw BE, which produce till it cuts AH in c: through c draw ca parallel to ED and cb parallel to EC; join ab, then abc is the triangle required, having its side ab parallel to mn, and being similar to the given triangle. Demon. For, because of the parallel lines ac, DE, and cb, EC, the quadrilaterals BDEC and Bacb, are similar; and therefore the proportional lines DC, ab, cutting off equal angles BDC, BDC, Bab; BCD, Bba; must make the angles EDC, ECD, respectively equal to the angles cab, cba; while ab is parallel to DC, which is parallel to mвn, by construction. Ex. 4. Given, in a plane triangle, the vertical angle, the perpendicular, and the rectangle of the segments of the base, made by that perpendicular; to construct the triangle. Anal. Suppose ABC the triangle required, BD the given perpendicular to the base AC, produce it to meet the periphery of the circumscribing circle ABCH, whose centre is o, in H; then, by th. 61 Geom. the rectangle BD. DHAD. DC, the given rectangle hence, since BD is given, DH and BH are given; therefore BIHI is given; as also ID = OE: and the angle EOC is = ABC the given one, because EOC is measured by the arc KC, and ABC by half the arc AKC or by KC. Consequently ic and AC = 2EC are given. Whence this E G AD. DC BD Construction. Find DH such, that DB. DH the given rectangle, or find DH = ; then on any right line GF take FE the given perpendicular, and EG = DH; bisect FG in o, and make Eoc = the given vertical angle; then will oc cut Ec, drawn perpendicular to OE, in c. With centre o and radius oc, describe a circle, cutting CE produced in a through F parallel to AC draw FB, to cut the circle in B; join AB, CB, and ABC is the triangle required. Remark. In a similar manner we may proceed, when it is required to divide a given angle into two parts, the rectangle of whose tangents may be of a given magnitude. See prob. 40, Simpson's Select Exercises. Note. For other exercises, the student may construct all the problems except the 24th, in the Application of Algebra to Geometry, at the end of vol. 1. And that he may be the better able to trace the relative advantages of the ancient and the modern analysis, it will be adviseable that he solve those problems both geometrically and algebraically. CHAPTER CHAPTER X. OF FLUXIONS AND FLUENTS. Art. 1. In the 2d volume of this Course has been giveni a compendious and easy treatise on Fluxions and Fluents ; and what follows is a further and more general extension of the same subject, chiefly on the transformation and on the inverse method of Auxions; as the rules for the direct method, given in that volume, will be found quite sufficient for finding the fluxions of the ordinary forms of quantities. From art. 32, to art. 48, of that volume, have been given a collection of the most common and obvious rules for finding the fluents of given fluxions; and which require no further proof or consideration, as they are self-evident, being simply the reverse of the preceding rules for finding fluxions. But, in art. 42 &c, is given also a compendious table of various other forms of fluxions and fluents, the truth of which it may be proper here in the first place to prove. 2. As to most of those forms indeed, they will be easily proved, by only taking the fluxions of the forms of fluents, in the last column, by means of the rules before given in art. 30 of the direct method; by which they will be found to produce the corresponding fluxions in the 2d column of the table. Thus, the 1st and 2d forms of fluents will be proved by the 1st of the said rules for fluxions: the 3d and 4th forms of fluents by the 4th rule for Auxions; the 5th and 6th forms, by the 3d rule of fluxions : the 7th, 8th, 9th, : 10th, 12th, 14th forms, by the 6th rule of fluxions: the 17th form, by the 7th rule of fluxions: the 18th form, by the 8th rule of fluxions. So that there remains only to prove the Ilth, 13th, 15th, and 16th forms. 3. Now, as to the 16th form, that is proved by the 2d example in art. 63, where it appears that *V (dx - x2) is the fluxion of the circular segment, whose diameter is d, and versed sine r. And the remaining three forms, viz, the 11th, 13th, and 15th, will be proved by means of the rectifications of circular arcs, in art. 61. 4. Thus, for the 11th form, it appears by that art. that the fluxion of the circular arc %, whose radius is r and tangent t, gai „n Now put t = x , or t = x and a = p; then is i = inzłn-14, and 72 + t = a + .x", and = = g2 + 12 I Lan.r is 3 = 28 +12 resi Janın-1; a+an 2 = or ; hence x arc to radius · an-1 x a+an 2 2 x arc na na to radius 1 and tang. √, which is the first form of the 5. And, for the latter form of the fluent in the same n°; 2 because the coefficient of the former of these, viz, is na' 1 double of the coefficient of the latter, therefore the arc nfa in the latter case, must be double the arc in the former. But the cosine of double an arc, to radius 1 and tangent t, is 1-12 In ; ť and because by the former case, this substi1412 1 a 1-t2 tuted for f in the cosine it becomes + a In ・a+an 1+ 129 6. Again, for the first case of the fluent in the 13th form. By art. 61 vol. 2, the fluxion of the circular arc z, to radius and sine y, is ry น = to the radius 1. 2 W(12-y2) √(1—y2) hence น Now put y=, or y2 n = X 2 2 Lan an In = (1-y2) 1 a a √(1 − = ) = √ = × √√ (a−x”), and j = √ — xnxn−1; then these two being substituted in the value of ż, give ż ; consequently the given fluxion n-1 √(a-x) 2 2, and the fluent is 1 n 2, and therefore its fluent is Z, that is n 22 x arc to sine, as in the table of forms, for the first n case of form XIII. 7. And, as the coefficient in the latter case of the said 2 form, is the half of the coefficient in the former case, 92 therefore the arc in the latter case must be double of the arc in the former. But, by trigonometry, the versed sine of double an arc, to sine y and radius I, is 2ÿ2, and, by the former case, 2y3· 2xr 1 == ; therefore > arc to the versed sine a 22 2x" is the fluent, as in the 2d case of form XIII. 8. Again 8. Again, for the first case of fluent in the 15th form. By art. 61 vol. 2, the fluxion of the circular arc z, to radius r r28 ¿ and secant s, is &= to radius 1. 5√(52—~9)3 S√(5a — 1) n = Now, put s= √. xzn in √✓ (~7 — 1) — 217 √✓/ (x” — a), and 's = √1⁄2-1⁄2 × ±na‡n−1, — = ja a a x; then these two being substituted in the value of ż, give ż or 1√/4 x = 2 former case, $2 2a arc to cosine 2 a Na s√(s2 — 1) 2 ż, 2 2 na, and theref. its fluent is x arc to secant, as in the table of forms, for the first case of form xv. nva Ich n/a = or = or s2==; hence s√ (s2 — 1 ) = a 9. And, as the coefficient in the latter case of the 2 na said form, is the half of the coefficient of the former case, therefore the arc in the latter case must be double the arc in the former. But, by trigonometry, the cosine of the double arc, to secant s and radius 1, is - 1; and, by the 2 $2 1= 2a xn 1 = 2a-xn ; therefore n√/ax is the fluent, as in the 2d case of form xv. be ут mn x-1x √(x-a); consequently the given fluxion = 1 na 2 Or, the same fluent will be x arc to cosine n√a cause the cosine of an arc, is the reciprocal of its secant. 22 10. It has been just above remarked, that several of the tabular forms of fluents are easily shown to be true, by taking the fluxions of those forms, and finding they come out the same as the given fluxions. But they may also be determined in a more direct manner, by the transformation of the given fluxions to another form. Thus, omitting the first form, as too evident to need any explanation, the 2d form is %= (a + xn) m −1 xn−1, where the exponent (n-1) of the unknown quantity without the vinculum, is 1 less than (7) that under the same. Here, putting y = the compound quantity a+": then is y = n, and = j ; hence, I ym-ly * π by art. 36, z= %, that is (a+x") m mn ", as in the table. |