LINES FO MY WIFE ON THE ANNIVERSARY OF OUK WEDDING DAY. Eliza, 'tis just eighteen years Since thou and I first came together ; In sunshine and in stormy weather. Thine was the wish those cares to smother; The kindest, fondest, tenderest mother. With joy, and mirth, and glee was thrilling, By every wish to please, fulfilling. To keep my feeble heart from drooping; And lay my pillow softly sloping. Except by equal love returning? And gently wipe thy tears when mourning. Sing doleful strains with much palaver, I'll praise Eliza while I have her! W. M. mathematical Department. MATHEMATICAL QUESTIONS IN NO. IV. ANSWERED. is 19 years. I Qu. (43) answered by Mr. James Cattrall, Fifer, 2nd R. L. Militia. Put r=his age, then by the question (7-18)=r12; by cubing each side, we have x2–18=r-36.r’+ 432x-1728, or 23--37x?+432x=1710; from whence r is found=i9 years. The same by Mr. C. De Weight, Parsondrove. Let x+12=his age, then its square is r?+24x+144, and, by the question, 2+24x+144-18=x, or x_x -- 24.x=126; whence x=7, and his age 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; A. Nesbrit, Farnley; W. Putsey, Pickering; Rylando; Silvanus Shaw, Wortley; T. Thornton, Ridgemont; J. Tomlinson, Liverpool; and J. Wynward, Plymouth. 2. Qu. (44) Answered by Mr. J. Gawthrop, Leeds. Let A INIC be the radius C В. of the segment's base, BH= A H=HF=DH, that of the I given hemisphere, and OD =OG=that of the inscribed AL F sphere; through O draw A E H G P which will bisect the LCAF; let fall the perp. EP and join EF. Now, D being the point where the circles touch, DO and D H are in the same right line, both of them being perp. to a tangent at D, Put DH=10=, and DO=r; then, HO=ano HG=v(a-2ar), AG=a+v(a’-2ax) and AO= ✓(A Gʻ+OG')=v{2a’+2avla-2ax)-2ax+x"}, 2ax also by sim. triangles AO: OG :: AF: EF= 2a+2a(a-2ar) but A E=»(AF: +EF)= and AO 2a*x + 2axıla' - 2ax) AO: OG :: AE : EP AO But LEHP= LIAH (for LEHP=2_EAP Eu. 20.3), and LAIHELHPE, also A H=HE, therefore HI-EP, and BH-HI=BH-EP=BI the alt. 2a'r+2ar/a-2ar) of the segment = a Again, A O2 put .5236=p, then the solidity of the segment is Blix (3AF-2BI) p, which by the question is equal to the solidity of the sphere=8px'; now by substituting the value of AO, and dividing by , we shall have (a 2a'r + 2ax/(a-2ar) 2a2+2a/(a-2ax)-2ar+ri 4a'x+4arva-2ar) ( 4a + 2a’+2a/(a−2ar)—2ar+x=) = 8®. From this equation x is found=4.542, then BI3.7825, and the solidity of the segment=3924. When the ends of the frustum are parallel ; put x =the height of the segment, then a-s is the diam. of the greatest globe that can be inscribed in the remaining frustum; also the solidity of the segment is (bar-2x) x p=!a-x)'xp, or 300.x+30x*-*=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. Cunmins, W. Dunn, T. Ford, J. Hine, A. Hirst, R. Maffett, A. Nesbit, W. Putsey, Rylando, S. Shaw, T Thornton, J, Tomlinson, and J. Winward. X E a 3. Qu. (45) Answered by Mr. J. Cummins and Mr. S. Shaw, Let the base AB=2b, CP-a, and put CO=r; then by the property of box DA the parabola a : 6 :: :: =D Oʻ; therefore, 40 D X OP X.7854 = -4abox-4b?r? X.7854 is the content A B of the cylinder, to which adding the part DCE= tab2x2 4abx—262ras we have their sum = X.7854 2a ab? x.7854 = half the paraboloid by the question; hence r=a-2a/2, and the radius of the cylinder's base is boll-V). Note. If the part DCE were to be left also, the alt. of the cylinder taken away is in that case = 2 { b P a b The same by Mr. J. Baines and Mr. B. Brooke, Put A B= 2a, CP=b, .7854=p and Co=r; then, by the parabola, b: e? :: 1 : DO="*; therefore, the content of the cylinder = 4pa’s – 4pa's, and the b content of the top part CDe=2pa's'; but the sum of these contents is, by the question=the content of half the paraboloid, that is, 4pa’r – 4 pa'r? b =pa'b, V 23 nce 4bx—2r2 = , and x=b{bx/2, therefore, O P=b/2 and DE=ar (4—2/2). Ingenious answers to this question were sent by Messrs. Butterworth, Dunn, Gauthrop, Hine, Hirst, Maffett, Nesbit, Putsey, Rylando, Tomlinson, and Wine 2pa'r? + ? b 1256 ward. or 4. Qu. (46) Answered by Messrs. S. Shaw and C. De Weight. Put the diameter of the sphered, the alt. of the segment=r and .5236=p; then, the solidity of the segment is 3pdr2--2prs, and its curve surface is=opdr, 3pdr?m-2p23 therefore the ratio is 3dx2r3 , which 6pdt 6d will be a max. when 3d.r-2x2 is so, that is 3d3-4r: =0 and r=id. 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. 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 G the angles DBC, BDC, by the straight lines BO, DO; H and from their intersection 0 draw O E perp. 10 BC, and А F BC will be the breadth of the walk. Demonstration. Draw the lines as in the figure; then, by the construction O is evideutly the centre of the circle inscribed in the triangle DBC; therefore, the figure OHBE=O EBF. In like manner we have OHĎI=OGDI: consequently, OGDI+OFBE+ OECI=OHDI+OHBETOECI; 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 A.r. A. Hirst, 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 Al the intersection F of DC with H F |
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