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

Another answer, by Mr. Gawthorp.

Assume ry + v, and z = 7+30; then the three formulæ become y2 + 4y v + 4 v2, y2 — 4 y v

110,

and y2 — 60; the first is a square, and by equating the

9

second with (y-22),

250

we get y=

; which sub

4

529 v2

16

= a square;

stituted for y in the third it become

leto 4 a, then

25 a, x = 29 a, and z = 37 a. Exam. When a 1, y is = 25, x = 29,'and z = 37; ,y if a = 2, y = 50, x58, and 274; &c.

Very good answers were sent by Messrs. Hine, Jones, Maffett, Wainman (of Armley 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 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 from g, the common centre of gravity of the rod and weight, draw g P parallel to W B, meeting A B in

P.

B

W

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:: WBgP:: GB: GP, and since W, w, WB, and GB, are given, g P, GP, will be given magnitudes; hence AP is 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

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.

11. Qu. (85) Answered by Messrs. Eyres, Hine, Maffett, and Winward.

Put AB a, AD=r, and DP=y; then, by the property of the circle, AC1

= ax, and DC = the question, y ×

therefore, y =

√ (a x — - x2) also, by √ (α x — x2) = ax ̧

ax

√ (α x − x2)

tion of the curve.

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When xo, y iso also, therefore the curve begins at A; when x = a, then y is infinite; consequently, BX is an assymptote to the curve.

The fluxion of the area is y * =

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P

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z, and by taking the fluxions, and substituting the latter expression is transformed to 2a ×

√(a), the fluent of which by Art. 279, Simpson's Fluxions, is

found 2 ax

2

A 2

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Sa2 A

— ± √√(as —x2)}, where A is a cir. arc. to rad. unity,

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When a this expression becomes a2 × 1,5708; therefore 2 × 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.

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

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√(2 a x — x2) √(2 ax- - x2)

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give the area APD=2a x (arc A C-CD), and when D coincides with B, the whole infinite space ABX P 2ax semicircle A CB 1,5708 × A B2.

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 and draw the lines as per question: Put A Ba, AF r, and FP=y; then by similar As PF: FA:: DE

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=ED, that is (a+)x(a).

= x2, or a2 y2 — x2 y2 = x2, and y

equation of the curve.

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If xo, then y = o, and the vertex of the curve is at A, and when xa, then y is infinite, .. if BX be drawn 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.

fluxion of the area AFPyx =

Quadrature-The

x2 x

the fluent

of which, by Simpson's Fluxions, Art. 279, is found — a2 ́A — — x √ a22, where A is a circular arch,

whose rad. is unity and sine

[ocr errors]

a

,

and when xa, this

becomes a2x,7854 the area of the quadrant ABC,

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 DE meets AD 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',

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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 OA; 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 OA, AF, in D, E; and join PG. The angle BOC is equal to the given angle (210) by construction. And by the principles of Dialling rad. S. lat. :: tan. hour arch: tan. hour angle therefore 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. AOB, hence by equality T : A F :: rad. S. lat. 2. E. D.

Computation. Draw PS to the centre S of the semicircle. Then tan. 75°: tan. 60o :: rad. 1: OG⇒ ,4641; hence GC = ,5359, GS = SP = 4 GC = ,26795, OS =,73205. And by trigonometry PS: S. 2 POS 21° :: OS : sine of 78° 16′ or 101° 44′ = OPS; ergo 4 OSP 80° 44′ or 57° 16', and ▲ OCD = 1 4 OSP 40° 22′ or 28° 38', the compli ment of which is 49° 38′ or 61° 22′ = ▲ DQC; then

as tan. 75°: tan. 4 DOC :: rad. : sine of 18° 22' 24" or 29° 23′ 29′′ the required latitude. Vide Prob. 151, Emmerson's Algebra.

1

Remark. The angle B OC will be a maximum when OP is a tangent to the semicircle at P; in which case the sine of the latitude will be equal to); that 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.

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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 LF meeting EB produced in F; draw FH parallel to A B cutting the circle in C, join A, C ; B, C; and ACB is the A required.

Demonstration. Join AE and AD; then by the similar triangles ALD, HEF, HF: HE:: LD:ĽA= LB, and alternately, HF: LD: HE: LB:: 4 HE : 4 LB; therefore HF x 4 LB = 4 LD × HE = ACCB; (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

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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