Mathematical Questions and Solutions in Continuation of the Mathematical Columns of "The Educational Times" Volume 18

Mathematical Questions and Solutions in Continuation of the Mathematical Columns of "The Educational Times" Volume 18

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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1873 edition. Excerpt: ...probable; hence the mean range is 3817. (Proposed by M. Collins, B.A.)--A body will describe a circle by a force = v2 rl directed towards its centre. Hence prove that if a circle be drawn round any other interior point C as centre of attraction, the force of attraction towards C must vary as PC-r (the cube of the distance of P from the polar of C); P being any position of the point moving in the circumference; and show that this last property is true for all conic sections. Solution by J. J. Walker, M.A. 1. Let CT be the perpendicular on the tangent at P, and O the centre of the circle. If f be the force along PC, its component along PO must be v'2 CT 1 CP equal to--; that is, fx--oc----, or fee cp op. But in the circle CT is to the perpendicular from P on the polar of C as CO to radius; therefore, &c. 2. Let K be the centre of a conic, and KL, KM the semi-diameters conjugate to the directions KC, KP. If O be the centre of curvature at CP CP P, from above, fx Qp cc CT3 gM3-But & is readily shown that the perpendicular from P on the polar of C, a fixed point, is proportional to CT. KM; thus is proved the last elegant extension of the theorem, which is due to the late Sir Wm. Rowan Hamilton. II. Solution by the Proposer. The force urging P towards C multiplied by cos CPO is equal to the force urging the moving body P towards the centre O = 1 =; B Kp2 hence the force towards C Bp2 cos CPO Now let OC. OC= E2 = OP2; then CT perpendicular to OCC and to PT is polar of C; then since OC: OP = OP: OC, the angle OPC = OCP; that is, / PCY = CPT; hence the triangle CPY is similar to PCT; therefore CY: PT = PC: PC, which is constant, since C and C are inverse points relative to the circle. Hence the force acting on P towards C, and causing P to...show more

Product details

  • Paperback | 30 pages
  • 189 x 246 x 2mm | 73g
  • Rarebooksclub.com
  • United States
  • English
  • black & white illustrations
  • 1236976568
  • 9781236976567