Now, according to the problem we have a = r' = R, P = 90°. .. equations (1) and (2) become a × {2R × ... .. (2). 226. We generalize this problem by stating it, Let P (Fig. 76,) be the pole of the great ○ ABM of a given sphere whose centre is C; then supposing the quadrant AP to revolve uniformly about the axis PC with a given velocity, whilst a point from P moves along the quadrant also with a given uniform velocity, it is required to find the surface APQM traced out by the point P in moving through the whole quadrant. ບ Let v and nv be the velocities of P and of A (which measures the angular velocity of the quadrant), and let PQP, Pqb be any two successive positions of the quadrant, and Q, q of the point P. Then S denoting the surface APB, dS=QqbB. Now QBbq Zone corresponding to 'x (CN) :: Bb : ABMA = 2′′r. But the zone = 2xrx, and Bb = nd. PQ = nrdx and SC + nr √ r2 = x2 = nr √ r2 — x2 nr2 = (2x — n)r3. In the problem n = 4, when M coincides with A, and the surface of the hemisphere is divided into the two parts 47a = (2r)1 and 2. (x-2). The curve of double curvature AQM is called the Spiral of Pappus. 227. Let a, b, c, be the sides of the spherical ▲, and a, ß, y, their chords, and suppose A, A', the included by b, c, and B, y. Then it is well known that Hence, by substitution and reduction, we get cos. A' cos. A + sin. sin. b 2 C 2 b whence cos. A' is always > cos. A (for sin. sin. can never be negative), and consequently A is always > A' whether the spherical A be isosceles or not. 228. Measuring the co-ordinates of the semicircle S, and the sector S', we have dS = ydz, and dS' = — √ dx3 + dy3 and when z is indefinitely diminished, cos. z = 1 sin. z=z z) sin. x z cos. x sin. x. a and altitude a x, is max. and putting dV = 0, we get a =, which determines the cone required. 231. Let C (Fig. 77,) be the centre of the earth, AB = 9 feet, the height of the person's eyes from the ground, and CA = radius of the earth 4000 miles. Then if from B the tangent BD be drawn, the arc AD will be the distance required. Now CB (which is evidently a straight line) = 4000 miles + But since C is very small compared with 1, we may neglect the may easily be converted into miles, furlongs, poles, yards, &c. 232. Let A (Fig. 78,) be the given, and with AD, the given line drawn from the vertex A bisecting the opposite side of the A, describe a circular arc MN. Then ABC being any A of which BC is bisected in D by AD, and Abc that whose side bc is bisected by Ad at right angles, and .. touches MN, we have ▲ Abc > ABC. For rDs being drawn parallel to bc meeting the from C and B in r, s, it is evident that A Cec> CDs > DrB and à fortiori > bBe, and the part AbeC is common to the two ▲ Abc, ABC; ... Abc is > ABC. But ABC is any A whatever, .. Abc is the greatest A required. 233. Let AB. (Fig. 79,) the given linea, and suppose the velocity of the point A moving along AC from A = 4 × vel. of B moving from B along BA; then B', A', being any contemporaneous positions of A, B, required the curve to which A'B' is perpetually a tangent. Let ba be the position of A' B' immediately successive to A'B', and intersecting it in P. Then P is evidently the point at which B'A' touches the curve. Now let the equation to the line B' A' referred to origin A by co-ordinates x, y, parallel to AC, AB be Now at the point P, y and x belong to the curve as well as to the tangent; and since for the next position ab, they may be considered constant, we have the equation to the curve required. See pp. 47, 48. 234. Let a be the given line, and Q' the given square, then a being one of the required parts, and .. a - x the other, we have (a− x) x = Q2 a ± √ a2 4Q, which determines the rectangle re 2 quired. Also since 4Qa cannot exceed a2, ( ( - ) 2 α is the greatest square the rectangle can equal. |