About her sister's queenly forehead, May twines a wreath of buds and leaves, And autumn gathers buds and sheaves. Oh! thus through life's glad spring and summer Let thought grow into noble deed, Let love ennoble aspiration The harvest time shall bring our meed.—Ladies' Home Magazine. NIGHT. Mysterious night! when our first parent knew Yet, 'neath a curtain of translucent dew, And lo! creation widened in man's view! Who could have thought such darkness lay concealed THE LONDON TIMES' "LOG." In the office of the London Times, there is a bureau that, one would think, must be a terror to not a few. Every man whose life is deemed worth taking has a place in a certain "pigeon-hole," wherein the record is kept constantly "written up" to the latest possible period; a sort of "log" to be displayed the minute life's voyage is ended. Deeds done and words said in the heat of passion are there; acts performed at long intervals are brought into startling proximity, and all, of necessity, divested of the glow of action, the touch of nature as it were, that made us regard them in the living actor with a lenient, if not a loving eye. Such a man dies to-night! the "pigeon-hole" gives up its dead, and to-morrow morning, even before the subject has begun to lie in state, the leading acts and incidents of his life are spread all abroad to the world. Quick work they make of fame; -Chicago Journal. Mathematical Department. Solution of Problem No. 6.—In the accompanying figure, let B C be a portion of the path of the pursued; and A D the corresponding portion of the path of the pursuer. Then the line D C will be tangent to the curve A D at D. Let B C=z; A D-s and D C=t. We must now find some relations between the quantities z, s, t, and the angle A B C. If C F be an infinitely small portion of the path of the pursued, which may be represented by d z, the corresponding portion of the path of the pursuer= d 8, may be measued upon the tangent D C, and is represented in the figure by D E. Draw the lines E F and E G, making the angle C EF= the angle F E G. Draw Ca perpendicular to E B E D A G H = F; Fb parallel to Ca and Fc perpendicular to E Ca, b Fc=CE a Ca÷ EC= d z cos. ☐ t But the angle b F G=□ (2). Also, if m = the ratio 2 has been increased by C F b..: do= of the velocity of the pursuer to that of the pursued, we shall have s=m z (3). These are the relations sought; and they apply to all questions of a similar nature. To integrate them, we have from equation (2), t d□ = dz cos.. Multiplying this by cos. □ we have t d□ cos. □d z cos. dz (1—sin.2 □)—d z-dz sin.2-dz-sin. □ dz sin. □ by (1), dz-sin. □ (ds+d t)=d 2-sin. □d s-sin. □d t― by (3) d z―d t sin.——m d z sin. by (1), d z—d t sin. ——m(d s—d t). Hence, tdcos. —d z— -m(d s+d t). Transposing, t dcos. d t sin. m(d s+d t). D= +d t sin. —d z— Integrating this last equation, we have t sin. —2—m(d 8+t)+C. o, let b be the value of t, and p that of. Hence b sin. Hence C = b(m+sin. p). Hence t sin. When s = C-b m. sin. p). p= z—m(s+t)+b (m+ When the pursuer overtakes the pursued, t=o, and we have o—2—M 8 +b (m+sin. p) — z—m2z+b (m+ sin. p). Hence z=b = employ the angle a F C instead of the angle a C F, we shall have sin. p= m+cos. F cos. F, and z=b ̄ ̄ ̄m2-1' From this result may be drawn for this species of problem, the following rule: Divide the velocity of the pursuer by that of the pursued; to this ratio add the cosin of the angle made by the intersection of their courses drawn through the starting points; divide the sum by the square of the ratio, less one; and multiply by the starting distance, which gives the distance the pursued will have run when overtaken by the pursuer. In our question the ratio is 153. Our distance is mile; and our DEAR SIR:-There are some very profound criticisms in the April Number of the Journal, upon my solution of problem 6. Mr. Jas. M. Ingalls, the gentleman who makes the criticisms, seems to have doubts about the equation y2=(x2—a2) tan. A tan. A' being the equation of an hyperbola. To convince him that it is, I respectfully refer him to Davies' Analytic Geometry, B. 6, prop. 3. But Mr. Ingalls carries the idea that I ought to prove that tan. A tan. A' is constant. Is it not proof enough to know that we have already two variables in the equation, viz.: x and y? Or is Mr. Ingalls ignorant of the fact, that every equation between three variables must refer to, and only to a surface? Again: Mr. Ingalls says I committed an error in confounding the coordinate of a particular point with the general co-ordinate of the curve. Here I must put in a plea of obtuseness; for I cannot understand what the gentleman means. If he means that it is unmathematical to substitute the co-ordinates of a particular point in a line, for the general co-ordinate of the line, then I have the consolation of knowing that I am not alone in this error. Davies does the same thing in his Analytic Geometry, Book Second, propositions 4, 5, 6, 8, 9, and many others. In respect to the objections which Mr. Ingalls makes to equation (4) in my solution, I have to say, that not one of them, except the last, is well founded; and some of them, it appears to me, were made without due reflection. The integration which produced (4) was performed under the supposition that the co-efficient of dx was constant; this, of course, was an error; but at the most it was but carelessness in performing an operation, and does not in the least effect the form of the solution, it only changes the form of the equation (4). The correct form of (4) is ≈ — a2x2-b2x2- 4 a2x2-a+b2 A. W. WHITCOM. dx A general equation in place of (4)-30lution of problem No. 6-may be obtained thus: From the equation of the curve we have a2y2—b2x2—a2b2. Hence a1y2— I usually integrate such by developing the radical by the binomial theorem, then multiplying each term by dx, then integrate each term separately. There are, however, other methods of integrating them. A. W. WHITCOM. Yours truly, Problem No. 29.-Let A and B be two given points, MQ a straight B line given in position. Let AV and B V be drawn such that by drawing V P making given angles, with M Q the ratio of O P to PD shall be given. The locus of the vertex V, is required independently of fluxions. JAS. KENNEALY, Jr. [Mr. K. has sent a solution of problem 18, somewhat different from that given by Mr. Campbell in the April Number, but producing the same result.-ED.] We have several solutions from Mr. L. Campbell and others, which we are obliged to postpone this month. Will our correspondents please not send us problems which are published in the text-books, they are not as useful as original ones.-ED. NORMAL SCHOOLS. PROCEEDINGS OF THE BOARD OF NORMAL SCHOOL REGENTS, AT THEIR ANNUAL MEETING, APRIL, 13, 1859. BOARD met at 9 o'clock, A.M. Present, Hon. C. C. Sholes, Pres., Hon. W. E. Smith, Messrs. Bean, Cook, Chapman, Clark, Maxon, Robbins, and Draper. The Board proceeded to elect officers for the ensuing year. Hon. C. C. Sholes was elected President, Hon. W. E. Smith, Vice-President, and S. Chapman, Secretary. The Report of the Visitors appointed to examine the institutions asking aid from the Normal School Fund was then heard. On motion of Mr. Smith it was 66 Resolved, That no institution be allowed to draw from the income fund of this Board for any student who has not pursued at least three of the studies mentioned either in the course for the first or second year, as heretofore prescribed, for the requisite number of days, exclusive of elementary sounds, and vocal music." On motion by Dr. Cook it was "Resolved, That the ancient and modern languages be not regarded as belonging to the three studies in the foregoing resolution." The Board then took into consideration the reports from the institutions claiming from the Income Fund, and allowed an apportionment of eighteen dollars to each pupil, reported and allowed to the following institutions: |