Put 2p = y — 4; then (3) becomes y — y3 — 64; whence = 27 ·(3). ЗУ 32. "There are two spheres of equal size and of exactly the same appearance, one solid silver and galvanized with gold and the other hollow and made of gold. Required some means by which one may be determined from the other." SOLUTION BY PROF. D. TROWBRIDGE, WATERBURGH, N. Y. Convert the spheres into pendulums, and let a be the distance from the point of suspension to the centre of the spheres, and 7 the distance from the same point to the centre of oscillation in the solid sphere and l'in the hollow one, k the principal radius of gyration for the solid sphere, and k' for the other. Then If r and r' be the radii of the external and the internal surface of the ... k'>k, and l'>1; or the solid sphere will vibrate in the less time. [H. T. Eddy, Prof. of Math. and Astronomy, Cincinnati, Ohio, has favored us with an elaborate solution of this question. Prof. Eddy supposes the spheres to be rolled down an inclined plane and shows that the solid ball will roll more rapidly than the other and reach the bottom more quickly. He obtains the eq., t' = √(k12 + p2) ÷ (k2 + y2jt, in which t is the time and k the radius of gyration of the silver sphere, and 'the time and k' the radius of gyration of the golden sphere; and, assuming the spheres to be of equal weight, he finds k' = r2 and k" 38, and hence t'= 1.044t, nearly. We intended to publish this solution in full, but find it too extended for our space.-ED.] 33. "A circle is referred to rectangular axes passing through the center. A tangent and ordinate are drawn from any point of the circumference. A distance n times the abscissa of this point is measured upon the axis of x and from this point a perpendicular falls upon the tangent. Required the equation of the locus of the intersection of the perpendicular and tangent." SOLUTION BY WERNER STILLE, MARINE, ILL. Also, when NS === MR, because x = r cos 0, CS = n.r.cos A. (2)......... ..p.sin(+0) = n.r.cos.sin. (2) can be written in the form, p.sin(+0) From this we easily deduce = n.r.sin 20. 4,02 ·sin3 21, The double sign shows that the curve consists of two equal branches in symmetrical position. The figure represents the curve, approximately, when " = 4. [This solution gives the equations between the radius-vector and the angle subtended by the axis of x and the perpendicular upon the tangent. From it, however, we may readily obtain the equation between the radius-vector and the perpendicular (P) upon the tangent. We have Pncos' - I = = n(1 -— sin2)) nsin; and from I = 12 I Or, because: P:: sin: sing we have psing = Psin@, or, by substitution, which corresponds with the eq. obtained by Prof. Evans by a different process of reasoning.-ED.] DR. JAS. MATTESON, of De Kalb Center, Ill., has sent us an elaborate solution of No. 5, which we are sorry our space will not permit us to publish. Putting x, y and z for the three sides of the triangle; and 5, 7 and 9 for the three bisecting lines, and, moreover, putting mx = y and nx = z, the Dr. gets, m = .5814480988564, and n = 7991785252715. 37. BY G. SHAW, Kemble, Ontario, Canada.-Divide unity into three parts such that if each part be increased by unity the sums shall be three rational cubes. 38. BY ASHER B. EVANS, A. M., LOCKPORT, N. Y.-Prove that when x is infinite where e is the base of Napierian logarithms. 39. BY PROF. C. M. WOODWARD, ST. LOUIS, Mo.-Two weights are connected by a fi..e string which passes over a pulley; if the weights be 50 and 72 lbs., determine what stationary weight the string must be able to support, that it may just escape breaking during the motion. 40. BY THE EDITOR.-P and Q are two points, distant apart a miles on the bank of a straight canal, through which there flows, from Q towards P, a uniform current with the velocity of m miles per hour. R is a point on the opposite bank of the canal, at right angles from P with the line PQ. Two men, A and B, start at the same moment from the two points P and R; A starts from P and walks directly toward Q with the velocity of n miles per hour, and B starts from R, with a boat, and rows, with a constant effort that in the absence of a current, would carry him miles per hour, and endeavors to join A by rowing continually directly towards him, and succeeds in joining A at the moment he arrives at Q. What is the width of the canal? or, distance between P and R? The deduction of the motion of the planets, in accordance with the laws of Kepler, from the principle of universal gravitation, is important, not only on account of the extensive role this theory plays in Astronomy, but also for its interest, in a historical point of view, as Newton's principal discovery. Hence it is desirable that the demonstration be made as elementary and as brief as possible, in order that it may be brought within the comprehension of the largest number of persons. The polar equation of the conic section, referred to a focus as pole, a(1 - e2) r = I + ecos(λ—w)' is well known; a denotes half the greater axis, e the eccentricity and the angle made by the axis with the line from which is measured. It will be advantageous to replace a(1 − e) by p, p being the parameter, also to put a = ecoso, ß = esinw. Thus the equation becomes r+arcos + ẞrsind = p. Hence it is plain that the equation, in terms of rectangular coordinates, the origin being at a focus, but the axes of coordinates having any direction we please, is √x2 + y2 + ax + By = p.. .(I). |