Solution.-Take the given point as the origin, let the diameter of the circular field he represented by (a); put 0 for any angle of elevation and & for the angle of azimuth so taken that when the projectile will fall on the circumference of the field we shall have 20. Now since any por tion of the surface of a hemisphere whose radius is (a) (the diameter of the given circle) and whose center is at the given point is expressed by cos θα θα ψ, therefore the favorable cases will be expressed by the integral for the chance that the projectile will fall on the circular field. TANGENCY OF HYPERBOLOIDS OF REVOLUTION. BY PROF. C. M. WOODWARD, ST. LOUIS, MO. In his Applied Mechanics p. 430, under the head of skew-bevel wheels, Prof. Rankine says: "If two hyperbo loids, equal or unequal be placed in the closest possible contact, they will touch each other along one of the generating straight lines of each which will form their line of contact." This matter of tangency is stated without proper limitation, but the graphical method given later for finding the obliquities and the gorge circles of the required hyperboloids involves the condition of possibility of such a tangency, which I propose to deduce directly from two tangent hyperboloids by the methods of descriptive geometry. Let r, and r, be the radii of the two gorge circles, and i, and i, the obliquities of the two surfaces (i. e. the angles which the elements make with their respective axes); then the required condition is Take, as in the Fig., the plane through the axis of the first surface, parallel to the axis of the second as the vertical, and a plane perpendicular to the axis of the first surface, as the horizontal plane. Since the surfaces are tangent, and the plane tangent to both at the point where their gorge circles touch each other contains the common element and is parallel to the vertical plane, the common element is parallel to the vertical plane and its vertical projection is a common asymptote to the vertical projections of the surfaces. 1 At any point of the element of contact, as at N, (n, n') draw a common normal to the two surfaces- It intersects both axes, one in O, and the other in P. Since M N is parallel to vertical plane, o' n' p' must be perpendicular to m'n'. Now the horizontal projecting plane of M N divids the perpendicular between the axes into r, and r,, and the normal into O N and NP; hence The obliquities of the surfaces are given in full size in the vertical jection; that is Hence, and i1 = o' m' n' and i, = n' m' p'. i, pro Q. E. D. SOLUTIONS OF PROBLEMS IN NOS. 3 AND 4. Solutions of problems in No. 4, have been received as follows: From Prof. W. E. Arnold, 16 and 18; R. J. Adcock, 17; Marcus Baker, 16, 17, 18 & 19; S. J. Child, 16 & 18; George L. Dake, 16; Prof. A. B. Evans, 16, 17, 18 & 19; Heury Gunder, 16, 17, & 18; Prof. E. W. Hyde, 16, 17, & 18; William Hoover, 16; Phillip Hoglan, 18; Prof. I. N. Jones, 16; J. B. Mott, 16; L. E. Newcomb, 16; A. W. Mason, 16; Prof. A. W. Phillips, 16, 17, 18 & 19; L. Regan, 16; John W. P. Reid, 16; Henry A. Roland, 17 & 19; Miss Kittie Robinson, 16; Prof. Selden Sturges, 16, 17 & 18; E. B. Seitz, 16, 17, 18 & 19; S. W. Salmon, 16, 17, 18 & 19; R. L. Selden, 16; Walter Siverly, 16, 17, & 19; Elias Schneider, 16; Jas. Stott, 16 & 18; Prof. D. M. Sensenig, 16, 17, 18 & 19; Prof. M. C. Stevens, 16, 17 & 18; and Prof. David Trowbridge, 16, 17, 18 & 19. 14. "Two equal particles attracting each other with forces varying inversely as the square of the distance, are constrained to move in two straight lines at right angles to each other; supposing their motions to commence from rest, to find the time in which each of them will arrive at the intersection of the two straight lines." SOLUTION BY PROF. J. M. GREENWOOD, KIRKSVILLE, MO. Take the lines as eoördinates, the origin being at their intersection; let a and b represont the distances of the particles from the origin, and x and their distances at the time t. У From the problem (1) a2 + b2 = c2, (2) x2 + y2 = r2. When x = 0, y = 0, r = 0, the particles arrive at the origin simultaneously. To find the time when the force varies inversely as the square of the distance we have (Smith's Mechanics, page 207). 15. "Two points are taken at random in the surface of a given circle and a chord drawn through each at random; show that the chance of their intersecting is + 1 5 SOLUTION BY E. B. 8EITZ, GREENVILLE, OHIO. Let CED be the given circle, O its center, A, B the random points, CD the random chord through A, CHD an arc through B, M its center. Put A x, NH = = y, CN=m, O G = 1, < OAD= 0, < CMN= , area seg. CH D= 8, area sector CMD= V, area CMD=t. Then ON x sin 0, NE=1 Ν + sin 0, NG= 1-x +x m2 + y2 2 y 9 area seg. CED=л sin-1 mm x sin 0, area seg. C G D sin -1 m 8 = V-t. Now when Bis in the segment CED the limits of y are 0 and 1 + x sin 0, of x, 0 and 1, and of 0, 0 and ment C G D the limits of y are 0 and 1 ; and when B is in the seg is csin 0, of x, 0 and 1, and of y= 1 -x sin y=0 y=1 -x sin 2лds + y=0 2πας = = [sin−1 m]2 + m2 } 2 x d x d 0 *** [π2 — 4 02 + 2 02 cosec2 0--4 0 cot 0+ 5 + cos 2 0] d 6 16. "What are the sides of a right angled triangle the area of which is 5 acres, and one of its acute angles 22° 30'?" Let SOLUTION BY E. B. SEITZ, GREENVILLE, OHIO. the side adjacent to the given angle; then a tan 22° 30′ = the side opposite, and the area of the triangle is a pr + (a + b) p − 1 + (a + 2 b) p2 −2+ -2 (a + n b) = S”. SOLUTION BY PROF. D. TROWBRIDGE, WATERBURGH, N. Y. Multiply the given series by r and subtract the given series from the product, then S (r − 1) = a r2 + 1 − n b + b r (p −1 + zor-2 + _1) p n pn If 60, we have the ordinary geometrical series. 18. "Find the differential coefficient of the function (1+ ")".(1+x")TM and express the answer in its simplest form." |