MISCELLANIES. where situated in the plane of the parabola, whose equation referred to the same co-ordinates originating in its vertex is y2 = px Then the distance required is easily shewn to be √ (y − ß)2 + (x—a)2 = max. (y — B)2 + (x − a)2 = max. and 2dy. (y B) + 2dx (x — α) = 0 which more than demonstrates the problem. See No. 215. ao α m =x+m+ (1 − m) x−1 + Ax¬2 + &c. 2m2 by the Binomial Theorem. Hence the straight line whose equation is m and the given curve continually approach as x increases, and touch when x = a, i. e., the line This is Sterling's method (See Linea Tertü, &c., p. 48,) of determining the asymptote, which ought to be adopted in all elementary treatises as the most simple. The reader may verify the above result by the common method. 243. Let AB, BC= y; then TB tan. BCA = 1, and tan. TCA = tan. (TCB which reduces to y2 = cr+1 the equation to the curve required. = When n = 1, y Cxx, the equation to a circle, .. in the ACB = 4 ACT, which may easily be demonstrated geometrically. 244. Supposing the cylindrical vessel open at top, let a be the thickness of its bottom and side; let also m the given capacity, and y + a the radius of the base, and x + a the alti tude of the vessel. Then the whole volume V is expressed by V = z . (y + a)2 . (x + a) . . . . and the hollow part m by . (1) (2) m = min. and putting its differential = 0, we get, after reductions, by the extracting of which, the value of y, and .. of x (by eq. 2) determining the cylinder required, may be found. 245. Cavalerius, in his method of Indivisibles, first shews that if a cylinder, cone, and hemisphere have equal bases, and altitudes, and sections in each be made by planes parallel to their bases, cutting the axis of the cone at a distance from its vertex those from the centres of the bases of the hemisphere and cylinder, the section of the cylinder = the sum of the sections of the hemisphere and cone. Hence, since the method supposes solids to consist of indivisible plane surfaces, the cylinder hemisphere + cone. 246. The equation to the ellipse referred to its centre b √1-e2 cos.20 where b conjugate axis, and e the eccentricity, we have But if be the between and the tangent, we have P and tan. (between p and normal) = tan. (p± 90°) — — cot. P ρ = max. sin. cos. 1 €2 cos.20 = max. which will determine the points at which are situated the maximum angles required. 247. Let A, B, C; a, b, c, be the angles and opposite sides of the spherical A, and suppose A and C constant. Then, and supposing da positive or a increasing, db is positive or negative, according as cos. c is, i. e., according as c is < or > 248. This problem is equivalent to the following one, Required to prove, that if two parabolas of the same parameter, revolve round their axes, which are in the same straight line, thereby generating two paraboloids one within the other, and a plane be drawn touching the interior paraboloid in any point whatever of its surface, this plane will cut off from the exterior paraboloid a constant volume." Let the cutting plane touch the interior paraboloid in p. (Fig. so.) and suppose the plane of the generating parabolas QAq to be brought into a situation cutting plane QP'q, and intersecting it in Qg; also let Q'q', P'M' be the intersections of a plane axis Aa, with those of QAq, QP'q. Then since the planes Q'P'q', QP'q are QAq, P'M' is Qq, Q'q' at their point of inter section M'. Now Q'P'q', QP'q being the intersection of the planes with the surface, we have Q'P'q' a circle from the nature of the revolution, and .•. P'M'2 = Q'M' × M'q' = QM' × M'q × (by property P of the parabola) p and P being the principal parameter, and that corresponding to ordinates parallel to Qq. Hence the section QP'q is an ellipse whose axes (a, b,) are But, drawing pP parallel to Aa, and putting Aa = m, we readily prove that Ppm, and Qq (parallel to tangent at P) is bisected in p... |