TRANSLATION AND IMITATION OF A LATIN POEM. Supplement, Vol. IV. P. 320. Poetical Magazine. Where shepherds gaily chant melodious strains, The warbling songsters make the groves resound, But ah! too soon will autumn quit the field, And show'rs of hail and snow come rushing forth; sorrow, Nor wish, too deep, to pry into the morrow. Surfleet, May 11, 1811. J. BROWN, VOL. N ADDRESS TO MY FIRST GRAY HAIR. BY MR. THELWALL. "AND thou hast chang'd thy hue, companion staid, Mute monitor! Of many a change approaching; that shall soon, Dappling at first with many a wintry spot, Well! my firm mind, That many a less expected change hath borne, Can bear that also. Hovering Winter, hail! Hail to thy wrinkled front, and hoary brow! Thou not refuse to them, my infant buds, Oh! give me yet, The sheltering fence of competence, to guard My waning strength, that each successive day Mathematical Department. MATHEMATICAL QUESTIONS IN NO. IV. ANSWERED. 1. Qu. (31) answered by Mr. T. Ford, Owstwick. x-10 PUT x number of gallons at first, then 3 what the officer seized; and £9.18s. £8.2s. = £ 1. 16s. is their value; therefore, x: £9. 18s. :: x-10 3 : £1.16s. Now multiply means and extremes, by reduction 22, and 9s. is the price per gallon. True answers were also sent by Messrs. Anglicus, J. Baines, B. Brooke, W. Bruster, W. Dunn, J. Gawthrop, A. Hirst, W. Harrison, J. Hine, Lucinda, R. Maffett A. Nesbit, Philpot, Rylando, J. Tomlinson, and J. Winward. 2. Qu. (32) answered by Mr. W. Bruster, Donningtei. By the nature of compound interest as 100: 16) + the amount of 1. for one year. 2x:: 1: 100+2x 100 pds. at 2x per cent. for 3x years, and x × = 5x by the, question; hence ( 100 3x 100 + 2x 3x = 5, 100 and extracting the cube root (100+2)=1.709976, from whence, by trial and error, x cmes out = 5.312728 the principal sought. The same by Mr. J. Baines, jun. Hort, ry Bridge. By the question, (1+.02x) x xc, or (1+.02.x) 5, whence 3x x log. (1 + .02x) = .69897, and by approximation x = 5.3127. Other solutions were sent by Mers. Anglicus, Dunn, Ford, Gawthrop, Hine, Hirst, Mett, Nesbit, Putsey, Rylando, Tomlinson, and Winward. 3. Qu. (33) answered by Anglicus, and Mr. Bruster. Put the versed sine A E50= , and DE = x; then DA = (AE+DE')=√(a1+r1). Also AE a: AD:: AD: AB = a x2 + the diameter. Now, by a A B C reduction 60x2 – 1920x = 70400. from whence = 53.806525, and AB = a + a 107.90284265 is the diameter required. Otherwise Mr. Ford, ar Mr. J. Gawthrop. Put the chord su... g the given arc = 2x, v = 50, and a = 160; then by known rules a, whence x = ✓ 53.806; whence 81/(x2—v3)—2x 3 9a2. - 6402 a2 a + 100 10 60 108 nearly. In the same manner the question was answered by Mr. Baines, Mr. Dunn, Mr. Hine, Mr. Hirst, Mr. Maffett, Mr. Nesbit, Rylando, Mr. Tomlinson, and Mr. J. Winward. 4. Qu. (34) answered by Mr. Putsey, and Rylando: Let DEF be a section of the given sphere, and ACB a section of the circumscribing cone. Put O Dr, and CD=x; then CE(CD × CF) = √(x2 — 2rx), and by similar triangles CE EO:. CD: DBA rx ; therefore, 2rx); 4r2x X by the ques √(x2 tion is a minimum, which in fluxions, and reduced, we get x4r; hence A B = 4/2, and the solidity of the cone 32 x 8 x .2618 67.0208, as required. Cor. The sphere is to its least circumscribing cone as 1 to 2. The same by Messrs. Eruster, Dunn, Hirst, and Nesbit. Let DEF be a section of the sphere, and ACB that of the cone, which will be the least when Cn - 2 n D, or Dn2n F; therefore, 2 F D = 8 = CD, and CÉ =√√√(CO2 — E O2) — ✓✓ (36 — 4) = 5.6568542, which is also A B, the base of the cone; hence its solidity is easily found 67.0208; whence it appears that the solidity of the cone is twice that of the sphere. Answers to this question were sent also by Anglicus, Mr. Baines, Mr. Ford, Mr. Gawthrop, Mr. Hine, Mr. Maffett, Mr. Tomlinson, and Mr. Winward. |