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tion. And thus exactly it was answered by Mr. Nesbit and Rylando.

Another answer, by Mr. Gawthorp. Assume x = y to, and x=7+30; then the three formulæ become ya + 4yu + $0*, y?

- 4yo 110", and ye - 60; the first is a square, and by equating the 9

250 se

which sub. 4

529 02 stituted for y in the third it become- = a square;

16 leto = 4 a, then y = 25 a, r = 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, Maffote, Wainman (of Armle, Mills), and Winward.

10., Qu. (84) Answered by Mr. J. Whitley, Masbro'. Let A B represent the given rod, revolv

B ing about the point A, in the vertical plane VPN, and B W the string fastened to the rod at B, having the given weight Wattached 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 A B in P.

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 :P :: 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. V N parallel to g P, or WB, and complete the parallelograms Pg EV, and Pg FN; then, because V N is given in length and position, and VE, NF, 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

W

right angle V PN; consequently, the locus of the point g required, is the given circle Eg FE.

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.

ar

is the equa

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, AC = a x, and DC=(a x - **) also, by the question, y xv (a x ----2-) = ax, therefore, y =

Var-) 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. The fluxion of the area is ..arx = ax

to integrate which put vlax - x2) var) = 2, and by taking the fluxions, and substituting the

? Ź latter expression is transformed to 2a * Vazvythe fluent of which by Art. 279, Simpson's. Flusions, is found = 20 x

áv

= 2a X

zax { $ - v(ar—-1)}; where A is a cir. arc. to rad. unity,

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and sine

When x = a this expression becomes a? * 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.

•x2

Mr. S. Jones finds the quadrature thus.-The fluxion

2 a x si of the area is y = =

(2a= AB) = 24 x ✓(2 ara

a 2 X

i

the fluents of which v (2 a x) v (2 a x- -xo) give the area APD=2u x (arc A C-CD), and when D coincides with B, the whole infinite space A B XP = 2 a x semicircle ACB=1,5708 X AB?.

It was answered alsó by Messrs. Gawthorp, Kay, Nesbit, Rylando, and Whitley. 12. Qu. (86) Answered by Messrs. Hine, Maffet, and

Winward. Complete the semicircle BCF, Y| and draw the lines as per question : Put A B = a, AF = x, and FP =y; then by similar As PF:FA:: DE

x2 =FA:EA, that is, y:8:: x:

y AE;..EB=a.

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and F'É= F
a+, and by the circle F’E. EB
= E D", that is (+3)(-3)

Y

= x, or az ya - 22 y = 4*, and y

the

va? - 82 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, .-. jf 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 AFPyx

the fluent

ta
of which, by Simpson's Flucions, Art. 279, is found

ar A ir vaz - j?, where A is a circular arch, whose rad. is unity and sine, and when x = a, this becomes = 22x,7854 = the area of the quadrant ABC,

x2

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W.

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B

therefore, the whole area on each side of the diameter BF is = to the circle whose rad. is a. 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 0 C (1) in G, so that T:t:: OÇ :OG (T and t being the tangents of the given hours arches 750 and 60°). On GC describe the semicircle GPC; to meet which in P, P, draw OB, making the angle BOC = 21° the given angle; join P C, and perpendicular to it draw the radius O A; then having drawn A F, 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 (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:0G :: 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. 75o : tan. 600 :: rad. 1:0G ,4641 ; hence GC = ,5359, GS = SP = {GC = ,26795, OS = 573205. And by trigonometry PS:S. LPOS = 21° :: 0 S: sine of 780 16' or 101° 44' == 40PS; ergo <OSP = 80' 44' or 57° 16', and a OCD=IZOSP = 40° 22' or 28° 38', the compli. ment of which is 490,38' or 61° 22' = 4DQC; then

as tan. 750 : tan. “ DOC:: rad.: sine of 18° 22' 24" or 29° 23' 29" the required latitude. Vide Prob. 151, 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

that

T.t is, equal to unity divided by the mean proportional between 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.

B

E

Construction. On AB the given base, describe the segment of a circle to contain the given angle; complete the circle, and draw the diameter DE perp. to A B, bisecting it in L; join EB, and in LD take any point r, and draw r s parallel to AB, take rs such, that Lr may be to 4 rs in the given ratio; from L through s, 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 AC B is the 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 Lra, LHF, Lr:rs :: LH: HF, and (by Leslie's Geom. book v. Prop. 13,) Lr:4rs::LH:4HF :: LH.LB : 4 LB. HF = AC+CB) in the given ratio by construction, and the base and vertical angle are of the given magnitude. 2. E. D.

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