tween the side of the cube and twice that side, and the first of those mean proportionals will be the side of the double cube. Now these may be readily found by means of two 'parabolas; thus: Let the right lines AR, AS, be joined at right angles; and a parabola AмH be described about the axis AR, with the parameter a; and another parabola AMI about the axis As, with the parameter b; P cutting the former in M. Then AP=x, PMy, are the two mean proportionals, R of which y is the side of the double cube required. = M H For, in the parabola AMH the equation is y2 = ax, and in the parabola AMI it is r2 = by. Consequently a: y::y: x, and y: x:x : b. Whence yx ab; or, by substitution, yby ab, or, by squaring, ybab2; or lastly, y3 = a2b 2a3, as it ought to be. Note. For other exercises of the construction of equations, take some of the examples at the end of chap. viii. GENERAL SCHOLIUM. On the Construction of Geometrical Problems. Problems in Plane Geometry are solved either by means of the modern or algebraical analysis, or of the ancient or geometrical analysis. Of the former, some specimens are given in the Application of Algebra to Geometry, in the first volume of this Course. Of the latter, we here present a few examples, premising a brief account of this kind of analysis. Geometrical analysis is the way by which we proceed from the thing demanded, granted for the moment, till we have connected it by a series of consequences with something anteriorly known, or placed it among the number of principles known to be true. Analysis may be distinguished into two kinds. In the one, which is named by Pappus contemplative, it is-proposed to ascertain the truth or the falsehood of a proposition advanced; the other is referred to the solution of problems, or to the investigation of unknown truths. In the first we assume as true, or as previously existing, the subject of the proposition advanced, and proceed by the consequences of the hypothesis to something known; and if the result be thus found true, the proposition advanced is likewise true. The direct demonstration is afterwards formed, by taking up again, in an inverted order, the several parts of the analysis. If the consequence at which we arrive in the last place is found false, we we thence conclude that the proposition analysed is also false. When a problem is under consideration, we first suppose it resolved, and then pursue the consequences thence derived till we come to something known. If the ultimate result thus obtained be comprised in what the geometers call data, the question proposed may be resolved: the demonstration (or rather the construction), is also constituted by taking the parts of the analysis in an inverted order. The impossibility of the last result of the analysis, will prove evidently, in this case as well as in the former, that of the thing required. In illustration of these remarks take the following examples. Ex. 1. It is required to draw, in a given segment of a circle, from the extremes of the base A and B, two lines AC, BC, meeting at a point c in the circumference, such that they shall have to each other a given ratio, viz, that of м to N. Analysis. Suppose that the thing is affected, that is to say, that AC: CB:: M: N, and let the base AB of the segment be cut in the same ratio in the point E. Then EC, being drawn, will bisect the angle ACB (by th. 83 Geom.); consequently, if the circle A E be completed, and CE be produced to meet it in F, the remaining circumference will also be bisected in F, or have FA FB, because those arcs are the double measures of equal angles: therefore the point F, as well as E, being given, the point c is also given. Construction. Let the given base of the segment AB be cut in the point E in the assigned ratio of M to N, and complete the circle; bisect the remaining circumference in г; join FE, and produce it till it meet the circumference in c: then drawing CA, CB, the thing is done. Demonstration. Since the arc FA the arc FB, the angle ACF = angle BCF, by theor. 49 Geom.; therefore AC: CB :: AE: EB, by th. 83. But AE: EB:: M: N, by construction; therefore AC: CB:: M: N. Q. E. D. Ex. 2. From a given circle to cut off an arc such, that the sum of m times the sine, and n times the versed sine, may be equal to a given line. Ana Suppose it done, and that AEE'B is the given circle, BEE the required arc, ED its sine, BD its versed sine; in DA (produced if necessary) take BP an nth part of the given sum; join PE, and produce it to meet BF to AB, or to ED, in the point F. Then, since m. ED+n. BDn. BP = n. PD + n. BD; AL D consequently 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 B, 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: BP :: N: 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 :: n: m, 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 sum. PB, or the given 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 mân given by position, and the angular points a, b, c, situate in the sides AB, BH, AH, of the triangle ABH respectively. A H C E B a Analysis. Suppose the thing done, and that abc is inscribed as required. Through any point c in BH draw CD parallel to mвn or to ab, and cutting AB in D; draw CE parallel to be, 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. 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, вba; must make the angles EDC, ECD, respectively equal to the angles cab, cha; while ab is parallel to DC, which is parallel to mân, 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. ED G K 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. DH = AD. DC, the given rectangle hence, since BD is given, DH and BH are given; therefore BI = HI 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 EC and AC = 2EC are given. Whence this 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 given 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 fluxions; 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 fluxions; 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 11th, 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 (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, is &= Now put t = x -1 then is in-1%, and 72 + 12 = a+r", and i = |