LINES : FO MY WIFE ON THE ANNIVERSARY OF OUR WEDDING DAY. ELIZA, 'tis just eighteen years Since thou and I first came together; In sunshine and in stormy weather. Whatever cares disturbed my breast, Thine was the wish those cares to smother; The kindest, fondest, tenderest mother. With joy, and mirth, and glee was thrilling, 'Twas thine to heighten all my bliss, 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. 1V. ANSWERED. 2nd R. L. Militia. The same by Mr. C. De Weight, Parsondrove. Let x+12=his age, then its square is 2? +24x-+-144, and, by the question, x? +24x+144-18=x, or 23-** -24.r=126 ; whence x=7, and his age is 19 years. It was answered also by Messrs. W. Allen, Burstwick; J. Baines, Horbury Briuge; 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. Nesbit, Farnley; W. Putsey, Pickering; Rylando; Silvanus Shaw, Wortley; T. Thornton, Ridgemont; J. Tomlinson, Liverpool; and J. Wynward, Plymouth. 2. Qu. (44) Answered by Dir. J. Gawthrop, Leeds. Let AI=IC be the radius B/ of the segment's base, BH= A H=HF=DH, that of the given hemisphere, and OD =OG=that of the inscribed a sphere; through O draw A E which will bisect the CAF; 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=a, and DO=r; then, HO=2-18 HG=v(a’- 2ax), AG=atv (a’-2ax) and AO= ✓(AG’+OGʻ)=v {2a’+2av(a-2ar)-2artx}, also by sim. triangles AO: OG :: AF: EF=240; but A E=v(AF:+EF*)=2a2+27(0—2ar), and AO: OG :: AE : EP=2 - 2a'r + 2ax(a' - 2ar) AO But LEHP=LIAH (for ZEHP=2LEAP Eu. 20.3), and LAIH=LHPE, also A H=HE, therefore HI-EP, and BH-HI=BH-EP=BI the alt. of the segment = a - 2a*r +2arna-2ar) SAO? put .5236=p, then the solidity of the segment is BI' X (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 p, we 2aRx + 2axx (a'- 2ur) shall have (a - 20:42 att 2a’ +2a/la-2ax)—2arti? faéx+4arv(a —2ar) (4a + 2a T 2a +2av(a-2ax)-2ax+x) From this equation x is found=4.542, then BI3,7825, and the solidity of the segment=3921. When the ends of the frustum are parallel ; put r the height of the segment, then amd is the diam. of the greatest globe that can be inscribed in the remaining frustum; also the solidity of the segment is (bar-2x3) x p=!a—x)'Xp, or 300x+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. Cummins, W. Dunn, T. Ford, J. Hine, A. Hirst, R. Maffett, A. Nesbit, W. Putsey, Rylando, $. Shaw, 7. Thornton, J. Tomlinson, and J. Winward. 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 the parabola a : 62 :: :6=DO’; D ot therefore, 40 D: X OP x .7854 = 4abʻx-4b°r . a x .7854 is the content A P of the cylinder, to which adding the part DCEfab x2: -4ab'r—26?r?y... 20, we have their sum =( *.7854= ab? x.7854 = half the paraboloid by the question; hence x=a-2av2, and the radius of the cylinder's base is b/(1- 1). Note. If the part DCE were to be left also, the alt. of the cylinder taken away is in that case , and the radius of its base v1 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: d’:::DO’="*; therefore, the content of the cylinder = 4pa? , and the .content of the top part CDE ; but the sum of these contents is, by the question=the content of half the paraboloid, that is, 4pa’r 4 paʻx?, 2pa?r? b =pa’b, V 23 nce 46x-28° = 6, and x=b }2, therefore, OP= and DE=aN(4-21/2). Ingenious answers to this question were sent by Messrs. Butterworth, Dunn, Garthrop, Hine, Hirst, Maffett, Nesbit, Putsey, Rylando, Tomlinson, and Wine ward. 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 3pdr2-2px?, and its curve surface is=6pdr, 3pdr2p.x3 3dx-2x therefore the ratio is P 6pdx , or 6ds which will be a max. when 3d.r—2x2 is so, that is 3d3-4r =0 and r= d. 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 c the angles DBC, BDC, by the straight lines BO, DO: and from their intersection o draw O E perp. to BC, and A FB BC will be the breadth of the walk. 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=O EBF. In like manner we have OHDI=OGDI: consequently, O GDI+OFBE+ OECI=OHDITOHBETOECI; that is, the area of the walk is equal to the area of the triangle DCB, or equal the area of half the rectangle ABC D. Again, by Mr, J. Butterworth, Haggate; and A.r. A. Hirst. Let ABCD represent the B garden, and, upon A D conti- 1% nued, take DG=AB, and bisect DG in I, on AG, AI, describe semicircles, and from ALT the intersection F of DC with |