SOLUTION OF A PROBLEM IN "CURVES OF PURSUIT." By DR. JOEL E. HENDRICKS, Newville, Indiana. PROBLEM.-Three foxes are supposed to be at F, and a dog at S, forty rods south of F. They all start at the same instant, the foxes running with the same uniform velocity, and the dog just twice as fast. The first fox runs east, along the line FE; the second one runs north, along the line FA; and third runs west, along the line FW. The dog runs directly towards the first fox, and overtakes it at E. He immediately pursues the second one, and overtakes it at N; then pursues the third, and overtakes it at W. How far must the dog run to catch the three foxes?-Communicated by Prof. Roor (Vol. I., p. 249). Solution. Let D represent the position of the dog when the first fox is at G, and draw DK parallel to SF, and dk infinitely near and parallel to DK. Also draw DG tangent to the curve at D, and dg tangent at the point d, and draw GI perpendicular to dg, and d H perpendicular to DK. Put SF(=40)=a, FK=w, FG=x, DG=y, and curve SD=2; and let the corresponding lines for the second and third curves be denoted by w1, 21, y and W2, X2, Y2, 22. Then is Kkdw, Gg=dx, Ig-Dd=dy, and Dd=dz; and by the similar triangles HDd and IGg we have 21, dz: dw:: dx (dz): dw= Ig...dy=dw — dz. But dz 2 dx, ..dy dydw2dx, .. dx = dw-dy. Whence by integrating we have (1) x = {w − + y + C.. (4) X2 = = FN, ..C=}FN(2+y2). By substituting for C, C1, C2, in (1), (2), (3), we have (5) (6) Now, when the dog overtakes the fox in running on each of the curves, the points D and G will coincide; and consequently y will become zero, and w will become x. Hence, if x', x1, x2 denote the whole distance run by the three foxes respectively, we shall have, x = = x2 + 1a; x1 = {xí + } α (2 + √ž) ; By the question, z=2(x,'—x') = 2a († (2 + √2) — 3) = $a (1 +√2)= 68.061 z=2(x-x1')=2a (2 (2+√2)2 — § (2+√2)) = a(6+5√2)=154.916 Therefore the whole distance run by the dog = = 276.310 rods. SECOND SOLUTION, by Prof. O. ROOT, Hamilton College, Clinton, N. Y. Take the origin at F, and let ry be the co-ordinates of any point in the curve described by the dog in the pursuit; then will y dx dy (dx2+dy) be the space passed over by the dog, and xthat passed over by the first fox in the same time; but as these spaces are as 2 to 1, we have =a, If now we put SF 40 rods = a, then, when y = a, we have dx = 0, and hence Clog a; and therefore (1) becomes * We omit the remainder of the solution, as the student will find no difficulty in completing it. It will be observed that DR. HENDRICK'S solution involves a differential equation of only the first order, while PROF. ROOT's, which is the solution usually given, involves one of the second order. Is DR. HENDRICK'S ingenious solution new? We have never seen it before. EDITOR. NOTE ON DIACAUSTICS. By REV. THOMAS HILL, President of Antioch College, Yellow Springs, Ohio. WHILE acknowledging the justice of Mr. PATERSON's remarks (Math. Monthly, p. 40, Vol. I.) on the waste of time involved in rediscovering that which was known centuries ago, I beg permission also to amuse myself with mathematical trifles even when I have not time to look up the results of previous investigators. The following note on Caustics is offered to the "Monthly," because it is interesting to me; but I do not know whether any or all of it is to be found in other places, with the exception of the corollary concerning the boundary of the space run through by a line in a rolling circle, which may be found, having no connection with caustics, in CHASLES. ୧ 1. Let of be the equation of any plane curve, being the radius of curvature, and the angle it makes with a fixed axis. Let be the angle which parallel rays of light make with the axis of v. 2. If the distance from the curve to the caustic, measured on the reflected ray, be called d, we have, by trigonometry, dd + ę dv sin (0 — v) = § de̱ cos (0 — v) + sin (0 — v) ḍ v ; § § d ያ and this divided by 2 dv gives the radius of the caustic, (2) = † (De̱ cos (0 — v) + 3 g sin (0 — v)). ያ 4. From (1) it follows that a caustic is at its maximum distance from the generating curve, when the parallel rays strike the curve at right angles, the distance being half the radius of curvature of the curve. 5. From (2) it follows that the radius of a caustic, when parallel rays strike the generating curve at right angles, is one fourth the radius of the evolute of the curve. But where the rays strike the curve tangentially, the radius of the evolute is three fourths that of the curve itself. 6. Equation (2) may be reduced for any particular curve to the form g1 =fv1, by putting V= Į (1⁄21⁄2 + 0 — 1 π). 4 R sin v; and we have 7. For example, let o 4 R sin g 1=R cos cos 0+4 R cos v sin v sin 3 R sin2v cos That is to say, the caustic of a cycloid, for parallel rays, is concentric with a cycloid of half the radius, coinciding with the small cycloid when 0 = 1 π. 8. Let the circle generating the original cycloid be again rolled through it, and it is manifest that the reflected ray will make, with a diameter through the generating point, the constant angle π — 0. Hence a diameter of the generating circle is a sliding tangent on the small cycloid, and any chord in the circle is a sliding tangent on a curve concentric with a half-size cycloid. 9. The caustic for = A sin" is (see GOULD'S Astr. Journal), = A(n sin" cos (v) cos v + 3 A sin" v sin (0 — v)) ; which, by putting ✪ = π, reduces to n (3) P1= 3 + n Asin" v cosv= (-) * * ' (3 + ") 4(1—cosv,);}(1+cos v1)'. 10. The substitution in the last equation of n=1 gives for the cycloid, as in Art. 7, |