tion. And thtis exactly it was answered by; Jtfr'i Nesbit and Rylando. Another answer, by Mr. Gawtkorp. Assume .r = y + v, and z = 7 + 3 v; then the three formulae become y' + 4j/t> + 4 —4_y p -r- 11 p% and j/2 — 6 p; the first & a square, and by equating the second with (y—'we 8et V — -J"' which sub 529 o' stituted fory in the third it become—j^— = a square; let v = 4 a, then y = 25 a, x =z 29 a, and z — 37 a. Exam. When a = 1, y is = 25, i = 29, and I = 37; if a = 2, y = 50, i = 58, and 2 = 74; &c. Very good answers were sent by Messrs. Hine, Jones, Maffett, Wainman (of Armlej Mills), and Winward. 10. Qu. (84) Answered by Mr. J. Whitley, Masbro'., Let AB represent the given rod, revolving about the point A, in the vertical plane VPN, and B W the string fastened to the rod at B, having the given weight W attached to its lower extremity. Let G be the centre of gravity of the rod; then, the segments AG, GB, will be given; join GW, and fromg, the common centre of gravity of the rod and weight, drawgP parallel to W B, meeting AB in P. Put to for the weight of the rod, and, by Art. 37, Marrat's Mechanics, and sim. triangles, we have, W-f»: W::GW:G^::WB :gP :: GB ; G P, and since W, w, WB, and GB, are given, g P, G P, will be given magnitudes; hence A P is given, and the locut of P is evidently the given circle PVN. Draw the vertical diam. VN parallel togP, or WB, and complete the parallelograms Pg E V, and Pg FN; then, because VN is given in length and position, and V E, N F, are each equal to the given line Pg, the points E, F, are given, and since Eg, gF, are respectively parallel to VP, PN, the angle EgF is equal to'th« right angle V PN; consequently, the locus of the point g required, is the given circle EgFE. Cor. The circles EgFE, VPNV are equal. Very nearly as above were the solutions by Messrs. Brooke, Butterworth, Eyres, Gawthorp, Hine, Jones, Kay, Maffett, Nesbit, Rylando, and WinWord. 11. Qv. (85) Answered by Messrs. Eyres, Hine, Maffett, and Winward. Put AB = a, A D = x, and DP ~y; then, by the property of the circle, AC =s a x, and D C ss J {ax — jt') also, by the question, jf x (ax — x') = axt therefore,y = — °X —is the equa y'lffr X') * tion of the curve. When x — o,y o also, therefore the curve begins at A; when x = a, then y is infinite; consequently, B X is an assymptote to the curve. .2 The fluxion of the area is y x zz -,, . — a X ,, to integrate which put x1 = z, and by taking the fluxions, and substituting the latter expression is transformed to 2o X —; rr, the fluent of which by Art. 279, Simpson's. Fluxions, is found = 2a x | —;t-"~ ^V(a— 2') | — 2ax{~£~ — a */(a* — i') j-2 Where Aisa cfr. arc. torad.unity, and sine —-. «i When x = a this expression becomes a2 x 1,5708; therefore 2 X 1,5708 a2 is the whole area included between the assymptote and curve. Remark. The above curve is a line of the second order, having two equal branches lying on contrary sides of AB. ,: Mes therefore, the whole area on each side of the diameter BF is = to the circle whose rad. is a. W. W. R.'': Exactly as above, the question was answered by 3fr. S. Jones. A mistake in the question prevented several gentlemen from answering it; it ought to have been "FG parallel to DE meets AD in P." Observations, sup ying the defect one way or other, were sent by lessrs. Gawthorp, Eyres, Nesbit, Rylando, Sfc. 13 Qu. (87) Answered by Mr. J. Whitley, Masbro'. Construction. Divide the radius O C (l) i'n G, so that T : t:: OC : OG (T and t being the tangents of the given hours arches 75" and 60°). On G C describe the semicircle o G s c jr F G P C; to meet which in P, P, draw OB, making the angle BOC = 21° the given angle; join PC, and perpendicular to it draw the radius O A; then having drawn AF, the tangent of the arc A C, find a fourth proportional to T, AF, and O C : which will be the sine of the required latitude. Demonstration. Let CP, OP, meet OA, AF, in D, E; and join PG. The angle BOC is equal to the given angle (21°) by construction. And by the principles of Dialling rad. : S. lat. :: tan. hour arch : tan. hour angle : therefore T : t :: tan. of the greater hour angle : tan. of the lesser :: (by const, and parallel lines) OC:OG:: CD:PD:: AF= tan. £. AOC: AE = tan. z. AOB, hence by equality T : AF::rad. S. lat. S. E. D. Compilation. Dra*v P S to the centre S of the semicircle. Then tan. 75° : tan. 60° :: rad. 1 : OG =r ,46+1; hence G C = ,5359, GS zz S P = 4 GC = ,26795, O S = ,73205. And bv trigonometry P S : S. Z. POS = 21° :: OS : sine of'78* 16' or 101° 44' =S /. OPS; ergo i. OSP = 80" 44' or 57c 16', and L. jt)Cb = ii'OSP= 40° 22' or 28° 38', the corapliment of which is 49°,38' or 6l» 22*= iDOCj then as tan, 75° : tan. z. DOC :: rad. : sine of 18° 22' 2*" or 29° 23' 29" the required latitude. Vide Prob. I5i, Emmerson's Algebra. ",„ Remark. The angle B O C will be a maximum when O P is a tangent to the semicircle at P; in which case the sine of the latitude will be equal'to . r—-; that is, equal to unity divided by the mean proportional between the natural tapgents of the given hour arches. As the demonstration cannot be well effected without drawing another figure, I have therefore omitted it. Other true answers were sent by Messrs. Cattrall, Gaiuthorp, Hine' Jones, Kay, Maffett, Nesbit, Rylando, and Winwurd. 14. Q,u. (88) Answered by Messrs. Hine, Jones, Maffet, and Winviard. Construction. On AB the given base, describe the segment of a circle to contain the given angle; complete the circle, and draw the diameter D E perp. to A B, bisecting it in L; join EB, and in LD take any point r, and draw rs parallel to A B, take rs such, that Lr may be to 4 rs in the given ratio; from L through a, draw L F meeting EB produced in F; draw FH parallel to A B cutting the circle in C, join A,C; B, C; and A C B is the A required. Demonstration. Join AE and AD; then hy the similar triangles ALD, HEF, HF : H E :: LD: LA:* LB, and alternately, H F : LD HE : L B :: 4 HE: 4 LB; therefore HF X 4LB = 4LD x HE = AC + CBV; (Prop.7.M.G.Student). Aijain,by similar triangles Lrj, LHF, Lr : rs :: LH : H F, and (by Leslie's Geom. book v. Prop. 13,) Lr: 4r* :; LH : 4HF :: LH'. LB : 4 LB . HF = AC + CBV in the.given ratio by construction, and the base and vertical angle are of the given magnitude. 2. E. D. |