We must now solve the equation a3 + b3 = t. y* + z3 Suppose y2 — yz + z2 = m3, and y and z prime to each other; then 3 -- nz ( n + 1)z) 3 + (( n + 1) y -- n2) * = (w2 + n + 1)[(n + 2)y m When n= 0, Hence we may take a = Put y m У =v+w, z = v w; then y2 - yz + z2 = v2 + 3w2: = m3, which is satisfied by v = p(p + 39)(p -- 39), w = 39(p + q)(p − q), m = p2 + 392; ···y=p(p + 39)(Þ — 39) + 39(Þ + q)(Þ − q), z = p(p + 39)(Þ — 39) — 39( Þ + q)(p − q). Take 11, q= 2; then y = p = 9 a = 14.04, 1404, b = 1637. See Mathematical Miscellany, pp. 118-123, whence the substance of SOLUTION BY E. B. SEITZ, GREENTILLE, OHIO. By a well known theorem, 39.-"Two weights are connected by a fine 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." SOLUTION BY HENRY GUNDER, GREENVILLE, OHIO. = As illustrated by Attwood's Machine, we have a moving force = 22g", and a mass to be moved = 122; hence the velocity acquired in one second is 228 g. The 72 pound weight will have acquired this velocity in one second, but moving freely under the influence of gravity its velocity would have been g. ... there must have been an opposing force of g, or 5 of 72, on the string 40.—(See ANALYST No. 9.) = 59 lbs. SOLUTION BY PROFESSOR ASHER B. EVANS, LOCKPORT, N. Y. We will take for the axis of y that side of the canal along which A walks and for the axis of x a line perpendicular thereto at an arbitrary point, the canal being on A's right hand side and in the direction of x positive. Let the time t be computed from an arbitrary instant, and let x' and y' be the coordinates of B when A is on the axis of y at a distance from the origin expressed by b÷nt, b being an arbitrary constant depending on the situation of the origin and the instant at which the time t is supposed to commence. Then the right line joining B and A will have for follows that this line will make with the axes of x and y angles whose constant forces are acting upon B, the force r due to his rowing and acting in the direction BA, and the force m due to the current and acting in the direction opposite to y positive. The components of the force r in the direction of v and y are r(y' — b―nt) rx' and Considering that the former of these compo √ x12 + (y' − b — nt)2 nents constantly tends to diminish the coordinate x', and that the latter tends to diminish y' when y'>(b + nt) and to increase it when y' <(b + nt), for the motion of B we shall have, from well known principles of mechanics, on omitting the accents as henceforth useless, Differentiating (3) and dividing by dt, we obtain Integrating (8) d(tany) VI+tan' . (8). log(~)" =log { tan↓ + V1 + tan3μ }, or better (~~)" = tang + √1 + tan3ų (9), where c' is an arbitrary constant. To determine this constant, we observe that at a certain time the line joining B and A is perpendicular to the side of the canal, or to the axis of y. Taking the origin, which we have not hitherto completely fixed, at the point where this perpendicular meets the axis of y, denoting the width of the canal by c, and taking the instant when A is at the origin of coordinates for the origin of t, we have when t = 0, b = 0, y = 0 and x = c; whence from equation (3) = I and c' = c the tan¢ = 0, and then from equation (9) (~)” = (~)” = dyrsine + m dx rcoso cose = 2 m (10). = ("1")(N)-(*") (2).......(II). Integrating equation (11), remembering that x=c when y = o, we have NOTE. The foregoing general solution includes the problem proposed by Werner Stille on p. 145 of THE ANALYST. For putting m = 0, r = n2 and c = the line AB, we have from equation (12) for the equation of the curve pursued by the dog. [Messrs. A. L. Baker, of Lafayette College, Easton, Pa., and H. Heaton, of Des Moines Public Schools, solved No. 40 by determining the Eq. to the curve which represents the trace of the boat in the water, in which case the question may be solved exactly as Mr. Stille's question, by substituting m + n for the velocity of the man (= 1 in Mr. Stille's quest.) and r for n, the velocity of the dog. The foregoing elegant solution of No. 40 is all that could be desired as a solution to questions of this kind. We add, however, the following solution based on geometrical considerations, hoping that it may assist the student in comprehending the solution of these questions: B, and draw b'c parallel to CQ. Put Chx and AB = y. As the distance traversed by the boat from B towards A, if not influenced by the current, would be proportional to the time, we shall have Bcrdt; but during the time dt the current will carry the boat in the direction of the line co' a dittance cf= mdt, the resultant of the two forces being the curve Bf; ... b'f = dx. Draw Bi perpendicular to = dx mdt. b'c, then is b'imdt, mdt, ..ci = b'f= dx, and fi If while B describes the curve Bf, A advances to h, we shall have Ah =ndt. Join fh and draw Ag and ce each perpendicular to fh. Then by similar triangles we have, Bc: ci::cf: ef, or rdt: dx:: mdt: mdx ref. Also Bc: ci:: Ah: gh, or rdt: dx:: ndt: ndx÷r gh. Now the decrement of AB is obviously ef+gh - Bc, therefore we (m + n) ÷ r = q and integrating we have |