tion. And thus exactly it was' answered by Mr. Nesbit and Rylando. Another answer, by Mr. Gawthorp. 3* Assume r = y to, and x=7+30; then the three formulæ become ya + 4 yo + $0", y: 470 110, and y: -60; the first is a square, and by equating the second with(y ), we get y = 2; which sub 529 02 stituted for y in the third it become = a square; let v = 4 a, then y = 25 a, x = 29 a, and x = 37 a., Exam. When a = 1, y is = 25, x = 29,'and % = 37; if a = 2, y = 50, x = 58, and 2 = 74; &c. - Very good answers were sent by Messrs. Hine, Jones, Maffett, Wainman (of Armle; Mills), and Winward. 10. Qu. (84) Answered by Mr. J. Whitley, Masbro'., Let A B represent the given rod, revolving about the point A, in the vertical plane VP VPN, and B W the string fastened to the rod at B, having the given weight Wat. tached 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 from g, the common centre of gravity of the rod and weight, draw g P parallel to W B, meeting AB in Put w for the weight of the rod, and, by Art. 37, Marrat's Mechanics, and sim, triangles, we have, W + w:W::GW:Gg:: WB:gP :: GB : GP, and since W, w, WB, and GB, are given, g P, GP, will be given magnitudes; hence A Pis given, and the locus of P is evidently the given circle PVN.. ¿? Draw the vertical diam. VN parallel to g P, or WB, and coniplete the parallelograms Pg EV, and Pg FN; 'then, because V Ň is given in length and position, and VE, 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 Eg F is equal to the right angle V PN; consequently, the locus of the point g required, is the given circle Eg F E. Cor. The circles Eg FE, VPNV are equal. Very nearly as above were the solutions by Messrs. Brooke, Butterworth, Eyres, Gawthorp, Hine, Jones, Kay, Maffett, Nesbit, Rylando, and Winward. . 11. Qu. (85) Answered by Messrs. Eyres, Hine, Maffett, and Winward. Put AB = a, AD=x, and DP=y; then, by the property of the circle, A C* = a x, and DC=> (a x - x?) also, by the question, y x ♡ (a I -22). = a, .. ax . . therefore, j = Vantay is the equa-, tion of the curve.. . When x = 0, y is = o also, therefore the curve begins at A ; when x= a, then y is infinite; consequently, B X is. an assymptote to the curve. ' in, The fluxion of the area is y* : x Jar-22) = ax vän); to integrate which put * = 2, and by taking the fluxions, and substituting the Sa? A found = 20 x 30 -iva-2) = 2a - i vlas—x)}; where A is a'cir. arc. toʻrad. unity, and sine . a rin... ., When x = a this expression becomes a? x 1,5708; therefore 2 x 1,5708 a? 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. Mr. S. Jones finds the quadrature thus.--The fluxion of the area's . 2 a x j . of the area is y = = mm (2a = AB) = 20 x . a = – x ; the fluents of which W(24 x x ) (2 a x—2)? give the area A PD=2a x (arc A C-CD), and when D coincides with B, the whole infinite space A BXP = 2a x semicircle ACB = 1,5708 X A B. It was answered also by Messrs. Gawthorp, Kay, Nesbit, Rylando, and Whitley. 12. Qu. (86) Answered by Messrs. Hine, Maffet, and Winward. Complete the semicircle BCF, Y O and draw the lines as per question : Put AB = a, AF = x, and FP sy; then by similar As PF:FA:: DE ra =FA: EA, that is, y: x::x: = AE;". EB=a- and F'É= , the equation of the curve.. If x = 0, then y = 0, and the vertex of the curve is at A, and when x = a, then y is infinite, :. if B X be drawn i to B A it will be an assymptote to the curve. The curve is a line of the fourth order having two infinite branches on each side of BF.- Quadrature The fluxion of the area AFP = y*=-* of which, by Simpson's Flurions, Art. 279, is found az A- va? - x?, where A is a circular arch, whose rad. is unity and sine, and when x = a, this becomes = az x,7854 = the area of the quadrant ABC, , the fluent 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 Mr. S. Jones. A mistake in the question prevented several gentlemen from answering it; it ought to have been “ FG parallel to D E meets A D in P.” Observations, supplying the defect one way or other, were sent by Messrs. Gawthorp, Eyres, Nesbit, Rylando, &c. 13 Qu. (87) Answered by Mr.J. Whitley, Masbro', Construction. Divide the radius OC (1) in G, so that T:t:: OÇ :OG (T.and t . being the tangents of the given hours arches 750 and 609). On GC describe the semicircle o GPC; to meet which in P, P, draw O B, 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 OC: which will be the sine of the required latitude. Demonstration. Let CP, OP, meet O A, AF, in D, E; and join PG. The angle BOC is equal to the given angle (219) 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. 2 AOC: A E = tan. < AOB, hence by. equality T: AF:: rad. S. lat. 2. E. D. Computation. Draw PS to the centre S of the semicircle. Then tan. 750 : tan. 600 :: rad. 1:0G ,4641; hence G.C= ,5359, GS = SPEGC = ,26795, OS = ,73205. And by trigonometry PS:S. LPOS = 21° :: 0 S: sine of 78° 16' or 101° 44' ŁOPS; ergo <OSP = 80' 44' or 57° 16', and a OCD=IZOSP = 40° 22' or 28° 38', the compli. ment of which is 499,38' or 61° 22' = 4DQC; then as tan. 75o : tan. Ź DOC :: rad. : sine of 18° 22' 24" or 29° 23' 29" the required latitude. Vide Prob. 151, * Emmerson's Algebra. · Remark. The angle BOC 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 7 ; that is, equal to unity divided by the mean proportional be tween the natural tangents 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, Gawthorp, Hine, Jones, Kay, Maffett, Nesbit, Rylando, and Winward. - 14. Qu. (88) Answered by Messrs. Hine, Jones, Maffet, and Winward. Construction. On AB the given base, describe the segment of a circle VW to contain the given angle; complete the circle, and draw the diameter A DE perp. to A B, bisecting it in L; join EB, and in LD take any point r, and draw r s parallel to A B, take r s such, that Lr may be to 4 rs in the given ratio; from L through s, draw LF meeting EB produced in F; draw FH parallel to A B cutting the circle in C, join A,C; B, C; and AC B is the A required. Demonstration. Join AE and A D; then by the similar triangles ALD, HEF, HF:HE :: LD: LA LB, and alternately, HF:LD:: HE:LB :: 4 HE: 4 LB; therefore HF X 4LB = 4 LD x HE = AC +CB)'; (Prop. 7. M. G. Student). Again, by similar triangles Lrs, LHF, Lr:rs:: LH:HF, and (by Leslie's Geom. book v. Prop. 13,) Lr:4rs::LH:4HF :: LHLB : 4 LB. HF =ĀC+CB2 in the given ratio by construction, and the base and vertical angle are of the given magnitude. 2. E. D... |