LINES FO MY WIFE ON THE ANNIVERSARY OF OUR WEDDING DAY. ELIZA, 'tis just eighteen years Since thou and I first came together; Whatever cares disturbed my breast, Thine was the wish those cares to smother; When free from cares, my heart elate When sick, 'twas then thy constant care Then how can I reward thy love, Except by equal love returning? Some who have lost a tender wife, Sing doleful strains with much palaver, Boston, Feb. 1812. W. M. Mathematical Department. MATHEMATICAL QUESTIONS IN NO. IV. ANSWERED. 1 Qu. (43) answered by Mr. James Cattrall, Fifer, 2nd R. L. Militia. PUT r his age, then by the question 12; by cubing each side, we have x2-18x3-36x2+432r-1728, or x3—37x2+432x=1710; from whence r is found 19 years. The same by Mr. C. De Weight, Parsondrove. Let x+12 his age, then its square is '+24x+144, and, by the question, x2+24x+144-18=x3, or x3—x2 -24x126; whence x=7, and his age is 19 years. It was answered also by Messrs. W. Allen, Burstwick; J. Baines, Horbury Bridge; W. Bean, Ridgemont; B. Brooke, Headingley;, J. Butterworth, Haggate; J. Cummins, Holbeck; W. Dunn, Broughton; E. S. Fyres, Liverpool; J. Gawthrop, Leeds; P. Grey, Hornsey; J. Hine, Plymouth; A. Hirst, Marsden; R. Maffett, Plymouth; 4. Neshit, Farnley; W. Putsey, Pickering Rylando; Silvanus Shaw, Wortley; T. Thornton, Ridgemont; J. Tomlinson, Liverpool; and J. Wynward, Ply mouth. B 2. Qu. (44) Answered by Mr. J. Gawthrop, Leeds. Let AIIC be the radius of the segment's base, BH= AH-HF=DH, that of the given hemisphere, and OD D E F H G Р CAF; let fall the perp. EP OG that of the inscribed sphere; through O draw A E which will bisect the and join EF. Now, D being the point where the circles touch, DO and DH are in the same right line, both of them being perp. to a tangent at D. HG Put DH=10=a, and DO=r; then, H 0—a—1, (a2-2ax), AG=a+√(a2-2ax) and AO= √(AG2+OG3)=√ {2a2+2a√√(a2—2ax)—2ax+x2}, also by sim. triangles AO: OG :: AF : EF= but AE(AF'+EF')= AO OG :: AE: EP= 2ax ᎪᎤ 2a2+2a/(a2-2ax), and АО But EHP ZIAH (for EHP=22EAP Eu. 20.3), and AIH=2HP E, also A H≈H E, therefore HI-EP, and BH-HI=BH-EP=BI the alt. of the segment a 2a3x+2ax/(a2—2ax) AO2 Again, put .5236 p, then the solidity of the segment is BI'× (3A F-2BI) x p, which by the question is equal to the solidity of the sphere 8px'; now by substituting the value of A O, and dividing by p, we 2a2x+2ax/(a2—2ax) shall have (a (4a+ = 2a+2a√(a2ax)—2ar+x2) 4a2x+4ax/(a2-2ax) 2a2+2a1/(a2—2ax)—2ax+x3) = 8x3. From this equation r is found=4.542, then BI= 3.7825, and the solidity of the segment=3921. When the ends of the frustum are parallel; put x the height of the segment, then a-x is the diam. of the greatest globe that can be inscribed in the remaining frustum; also the solidity of the segment is (6ax2-2x3) xp (a-x)'xp, or 300x+30x-x3-1000, from whence x-2.6796; which is much less than BI 3.7825, found above. Answers to this question were transmitted by Messrs. -W. Allen, J. Baines, W. Bean, B. Brooke, J. Butterworth, J. Cattrall, J. Cummins, W. Dunn, T. Ford, J. Hine, A. Hirst, R. Maffett, A. Nesbit, W. Putsey, Rylando, S. Shaw, T. Thornton, J. Tomlinson, and J. Winward. 3. Qu. (45) Answered by Mr. J. Cummins and Let the base AB=26, CP-a, and put CO=r; then by the property of the parabola a : b2 : : x : b=DO;; b'x a therefore, 40 D2 x OP x .7854 = ·4ab2x—4b2x2、 a -4b2x2) × .7854 is the content A E B of the cylinder, to which adding the part DCE= şab2x2· 2a we have their sum = (4ab3x—2b3x1) × 7854— ab2.7854 = half the paraboloid by the question; hence x-a-2a/2, and the radius of the cylinder's base is b(1). Note. If the part D C E were to be left also, the alt. of the cylinder taken away is in that case = =2, and the radius of its base √✅‡ ba. a The same by Mr. J. Baines and Mr. B. Brooke. Put AB 2a, CP=b, .7854=p and CO=r; then, by the parabola, b: a2 :: x: DO=2; therefore, the a2x content of the cylinder 4pa2x-pa2, and the b content of the top part CDE=2par; but the b sum of these contents is, by the question=the content of half the paraboloid, that is, 4pa3x— 2pa2x2 b 4 pa2x2 =pa2b, vle nce 4bx-2x2-b2, and rb b √2 1b2, therefore, OP/2 and DE=a/(4—2/2). 2 Ingenious answers to this question were sent by Messrs. Butterworth, Dunn, Gawthrop, Hine, Hirst, Maffett, Nesbit, Putsey, Rylando, Tomlinson, and Win ward. 4. Qu. (46) Answered by Messrs. S. Shaw and Put the diameter of the sphered, the alt. of the segment and .5236=p; then, the solidity of the segment is 3pdx-2pr3, and its curve surface is=6pdr, 3pdx-2px3 3dx-2x therefore the ratio is 6pdx or 6 d which will be a max. when 3dx-2x is so, that is 3dx—4xx =0 and x=3d. Exactly in the same manner the question was answered by Messrs. Baines, Brooke, Butterworth, Dunn, Gawthrop, Hine, Hirst, Maffett, Nesbit, Putsey, Rylando, Tomlinson, and Winward. G 5. Qu. (47) Answered by Mr. Gawthrop, Mr. J. Whitley, and Rylando. Construction. Let ABCD D be the given rectangle. Draw the diagonal BD, and bisect the angles DBC, BDC, by the straight lines BO, DO; and from their intersection O draw OE perp. to BC, and BC will be the breadth of the walk. A H F Demonstration. Draw the lines as in the figure; then, by the construction O is evidently the centre of the circle inscribed in the triangle DBC; therefore, the figure OHBE=OEBF. In like manner we have OHDI OGDI: consequently, OGDI+OF BE+ OECI OHDI+OHBE+OECI; that is, the area of the walk is equal to the area of the triangle DCB, or equal the area of half the rectangle ABCD. Again, by Mr, J. Butterworth, Haggate; and Mr. Let ABCD represent the B garden, and, upon A D continued, take DG AB, and bisect DG in I, on AG, AI, describe semicircles, and from A the intersection F of DC with OD I |