As ccntinel, laid at his feet. Poor Tray watch'd the flock on the plain; ,c And, pour'd from the thicket's retreat, Was heard the mellifluous strain! Suspended, his crook, on the tree, Hung ready his hand to receive; The ballad was plac'd on his knee, Which taught his fond bosom to heave. But broken is Corydon's reed, Ah! ne'er shall we hear it again! No longer his lambkins to feed, The shepherd shall traverse the plain. But though he to death is consign'd, And no more the lov'd bard shall we see, His song in a wreath is entwin'd, And that wreath forms a Garland for me! TO GRAY. Next see ethereal Gray, And oft when caution penn'd the guarded, fold, And listen'd pensive as his "curfew tolPd With ling'riug step, at midnight's awful noon, Explor'd with him the spot with grass o'ergrown, Oft shall his numbers soothe me to repose, His moral lay the hallow'd truth disclose, TO COLDSMITH. |(|ir. .t.t Next hapless Auburn's friend my bosom cheers, Poet belov'd! my vanquish'd heart is thine, TO BURNS. And whae is he that syngs sae weel, His gabby tales I love to hear, I wad advise, when rankled care And should auld mokie sorrow freeten, . And, while I breathe, whene'er Ise scant* I'll tak his beuk, and read awhile, mathematical Department. MATHEMATICAL QUESTIONS IN NO. VI. ANSWERED. 1. Qu. (75) Answered by Mr. Brewer, Private 2nd R. L. Militia. The area of the garden is readily found = 3130320 inches; and putting x for the depth of the pond, its diam. will be 8 s, and by mensuration (48x + x2) X ,5236* = 25,6564x3 is its concavity; also 64 x' x ,7854 — 50,2656a2 is the'surface of the pond. Hence, 3136320 — 50,2656 x* ± 25,6564 a;3, from whence x is found = 48,98515 int hes = 4,0820954 feet, and 81, or the diameter, = 32,65676 feet. Again, by Mr. B. Brooke. Let 8 * and x denote the diam. and depth of the pond; then its capacity will be 25,6564x', and its surface ~ 50,2656i'; but by thequestion 25,6564a;5 + 50,2656i' — 3136320 inches — the area of the garden; whence z is 48,9848, therefore the diam. is 10,8855, and the depth 1,3607 yards. In the same manner it was answered by Messrs. J. Baiues, Reading; J. Bamford, Holthead; IV. Bruster, Donington; J. Butterworth, Haggate; J. Cattrall, Ply. mouth; J. Cummins, Holbeck; W. Dunn, Broughton; J. Ely, Doveridge; J. Gawthorp, Leeds; W. Harrison, 2. Qu. (76) Answered by Messrs. Bamford, Cummins, and Ely. By adding the given surfaces together, we hare 314,16 for the whole surface; from whence we find the diam. = 10, and the circumference = 31,416; then, by mensuration, G| is the depth, and 4,713 the radius of the segment immersed; therefore its solidity, •or the quantity of water displaced, is 387,851; but, by the principles of hydrostatics, 1728 : 387,851 :: 1000: 14,028 lbs. = the weight of the globe. Again, by Messrs. Brooke, Bruster, and Webster. The diameter of the globe is = V -—— '— ° 3,1410 209 44 = 10, hence •■,.'■ =64 is the height of the im 3,1416 x 10 3 .90—40 . 400 „„_ mersed segt. and x,52S6 x ~r = 387,851 is its solidity. Therefore, by hydrostatics 1728 : 387,851 :: 1000 : 224.45 ounces the weight required. The same, by Messrs. Macann and J. Smith. Since the curve surface of the globe is given = 314.16 inches, its diam. is easily found == 10 in. and as the superficies of spherical segments are as their altitudes, we 10 x 2 have the height of the immersed segment = «^—5—, 10472 and its solid content — ■ cubic inches, is the con tent of the water displaced by the globe; therefore, by . 10472 hydrostatics, 1728 : 1000 :: -—-—: 224,4513 ounces, is the weight, as before. It was answered also by Messrs. Armitage, Barnes, Brewer, Cattrall, Gawthorp, Hine, Jones, Maffett, flesbit, Rylando, Tomlinson, and Winward., 3. Qu. (77) Answered by Mr. J. Smith, Alton Park. -<'A Let ABCD and ABK represent vertical sections of the conic frustum and the cone completed, EF the dividing section ; and drawLC parallel to GK. Then by sim. triangles LB = 2 : LC = 12 :: GB = 5 : GK = 30, therefore, IK = GK—GI = 18. Put ,7854 sz n, then, by mensuration, AB2 x GKx; = 1000«, is the content of the cone ABK, and DC X IK X ^= 216w, is the solidity of DCK. Half the difference of these = 392n is the solidity of the frustum EFCD, which, added to 216n, gives 608ra for the solidity of the cone EFK. Now, the cones ABK, EFK, eing similar solids. we have 1000b, : ^60&n fins :: GK = 30: HK = 3() J-fi— = 3 ^608 = 6 ^76 = 25.41494; therefore, HI = HK — IK = 7,41494 is the distance required. Again, by Mr. Armitage, Rochdale Academy. First, 10—6 : 12 :: 6.: 18 = what the cone wants in length to be complete, and 18 + 12 = 30 = length of the whole cone; then, by sim. solids, as 4/7S5,4 (= content of the whole cone) : 30 :: ^477,5232 (= the solidity of the top cone and half frustum) : £5,41494, and the distance from the bottom = 25,41494 — 18 = '7,41494, as before. Otherwise, by Mr. J. Cattrall, Fifer 2nd R. L. Militia. By similar triangles LB : LC :: BG : GK = 30; consequently, 785,4 is the solidity of the whole cone, anc| $15,753.6 is the solidity of the frustum, whence "785,4 — 610,7536 = 169,6464 = solidity of DCK, therefore, 169,6464 + - 477,5232 is the so |