Required 43. Since 23 X 43 = 344; and 73 = 343 we have 344 + 147 2 43387) +2=3.503401; the true root being 3503398. This method is equally applicable to fractional numbers. 5.456789 1 7 + X = 2.335977; true to the last figure inclusive, 3 FIRST PRINCIPLES OF THE DIFFERENTIAL CALCULUS. BY S. W. SALMON, MOUNT OLIVE, NEW JERSEY. LET a point A start from O and move uniformly along the line OX, and let de represent its rate of motion. At the same instant that A starts from 0, let an- any (1) in which a and b are constants. When n = 1, B will move uniformly; for other value of n B moves at a rate which either increases as x increases or decreases as x increases. Our problem is to find a general expression for the rate of motion of B, which we will denote by du. Let Q and S, at distances u' and ' from O respectively, be any two contemporaneous positions of A and B. Hence at these points we have (2) Let du' be the rate at which b passes through the point Q; now when du' is known we can find a point in the line OU, R suppose, from which another Point C must start simultaneously with A and B so that, moving uniformly along OU at the rate du', it shall coincide with B at the point Q. Let OR=c. The time in which B passes over the space u'-b, being the same as that in which A passes over the space a', -space ÷ rate = x' ÷ dx. which C passes over in this time time x rate The space In the immediate vicinity of Q, Cwill coincide with B only at the point Q, for since C moves uniformly at the same rate at which B passes through Q, if B's rate is increasing, B will from this point move faster than C, and if decreasing slower. Let C pass over the space y moves uniformly the relation between x and y must be expressed by an c while A passes over the space x, since C Substituting this value of e, and the value of c from (3) in (4), we get Subtracting (2) from (1), member from member, we have Dividing (6) by (5), member by member, we have This equation is true for every corresponding value of u, y and x; but when See Davies' Bourdon p. 257. Dropping the accents, since the solution depends on no condition belonging to the particular values of x' and u' more than to any other corresponding values of x and u, we have du anx"-1 dx. Suppose the relation between u and x to be expressed by the equation u = 4(x), where y(x) is a symbol representing any algebraic expression involving x. We have 4(x) = 4[x' + (x − x′)]. Suppose the second member of this equation to be developed, and arranged according to the ascending powers of (x x'), then 4(2) = x[x' + (x − x')] = y(x') + P(x — x') + Q(x — x')3 + &c.,. where P, Q, &c., are algebraic expressions containing r'. The development must commence with the zero power of (x - x') since it must reduce to (x') when x=x'; whence u―u'=y(x)—y(x') = P(x—x') + Q(x—x')3 +&c. In the same manner as before we get du' y — u′ ( 4(x) — 4(x' ) ) = 3 — W' (P + Q(x − x') + &c.) ; = dx U- ·u' xx whence, making xx', we get น - NOTE ON THE CORRECTION OF AN ERROR IN THE THEORY OF POLYCONIC PROJECTIONS, BY PROF. W. W. JOHNSON. IN the publications of the Coast Survey* it is erroneously stated that in the Polyconic Projection of the Sphere the projected meridians and parallels cut each other at right angles throughout the entire area. The Polyconic Projection is that in which a central meridian and the equator are represented by straight lines, perpendicular to one another, and correctly subdivided on the same scale; while the parallels are represented by arcs of circles described with radii proportional to the cotangents of the latitudes, such being the radii of the parallels considered as developed each from the tangent cone of which it forms the line of contact with the sphere. Letting p denote an arc of longitude measured on the equator, pcos L will denote an arc of the same longitude in latitude L, and as in the projection this arc is measured on a circle whose radius is cot L, its circular measure (which is denoted by 0) is 0 = p sin L. Adopting the central meridian in the projection as axis of x, and the equator as axis of y, the co-ordinates of the projection of a point in latitude Land longitude p are readily shown to be Making p constant these become the co-ordinates of the projected meridian. Now the inclination to the axis of x of a tangent to this curve being denoted by, we shall have dy ÷ dx; tan = and if the curve cuts the parallel at right angles we shall have the radius of the parallel coinciding with the tangent, and Y would be the supplement of tan 0 = dy ÷ dx. 0, or *Coast Survey Report for 1853 page 99, “a projection results in which all intersections of parallels and meridians take place at right-angles," and again, "Over the entire area of this projection all parallels and meridians intersect at right-angles." The same error occurs in Church's Descriptive Geometry, Art. 233, on the Polyconic Projection. "This has the advantage that the representatives of the parallels and meridians are perpendicular to each other, as in To test the truth of this, we find do=p cos LdL, hence dy= dx = pcot L cos 0 cos L + cosec L sin Putting this expression equal to tan 0, we deduce or that is p cos L sin20 cot L sin @= p cos sin @psin L. But this is not true, since 0=p sin L; therefore the meridians and parallels do not generally intersect at right angles. The result indicates however that when is small, that is, when either p or L is small, there is a close approximation to perpendicularity. There is also a close approximation when L is near to 90°. NOTE ON THE SOLUTION OF PROB. 78, BY DR. H. EGGERS.-THE published solution of problem 78, short as it is, seems to be rather artificial. The necessity of introducing into the solution the parallelogram equivalent to one-half the triangle does not appear. I beg to present here an other solution, founded on principles of modern geometry. Let ANM be the given triangle and MO the bisecting line from M, if P is situate within the angle AMN. From this it is evident that the required bisecting line through P will pass between 0 and N, and between A and M. The required line therefore has to cut from the given angle NAM a triangle of equal area with A triangle AOM. Any other bisecting line, not passing through P, will cut on AN and AM respectively two distances a and y, reckoning from A, so that the rectangle xy is constant and will mark on AN and AM two series of homographic points. Drawing from P lines to every pair of corresponding points we shall have two homographic pencils of rays with the common center P. The two coincident rays of the two pencils will be the required bisecting lines. The homography of the two series of points is determined by three pairs of corresponding points, that is, to the points A, 0, ∞ on line AN correspond the points o, M, A on line AM. Therefore the two pencils are P(A, 0, ∞) and P(∞, M, A). Construct now the two coincident rays according to Steiner's method by means of an arbitrary circle through P; these will solve our problem. For the characteristic property of the double ray is, that the two corresponding points which it marks on AM and AN, and point P are on one line. The same Solution applies when P is within the triangle, or, at an infinite distance. DETERMINANTS. BY PROF. DEL. KEMPER, M. A. HAMPDEN SIDNEY COLLEGE, VIRGINIA. § 1. If we have the two simultaneous linear equations we can, performing the elimination by any of the common methods, obtain values for the variables in terms of the coefficients: thus we find where the symbol means the Determinant of the four quantities enclo sed within the vertical lines, that is to say, the algebraic sum of their combinations in sets of two and two, with the condition that each combination contains one, and only one, of the quantities in each row or column: and that the signs are determined by the following rule. The Diagonal of the Determinant, viz., the combination got by reading from the left-hand upper corner to the right-hand lower corner, is affected with the sign plus: and then the sign plus or minus is affixed to each of the remaining combinations, according as it may be derived from the Diagonal by an even or by an odd number of interchanges among the suffixes attached to the quantities a1, by &c. The truth of this rule is rendered evident, in the case before us, by an inspection of the Determinant in its expanded form. § 2. Now consider the three equations α1x + b1y + c1z = d1 From these we obtain, by the common methods of elimination, d, b2c3 d1bзc2 d2bc1 — d2b1cз + dзb12 — dzb201 cz . ძეხვი, 3 1 and similar expressions for the values of y and z. The denominator of this value of a (and also of y and z) is a determinant, that is, may be symbolically exhibited as |