and 100 = 10 × 10, it follows that if we have an equation composed of terms which are the logarithms of these three numbers, it may be resolved into another, the terms of which shall be the logarithms of the number 11 and other smaller numbers. Now by the preceding formula, if we put 99 for n, we have 2 log. 99 - log. 98 log. 100 = N. that is, substituting log. 9+ log. 11 for log. 99, log. 2 + 2 log. 7 for log. 98, and 2 log. 10 for log. 100, 2 log. 9 + 2 log. 11 - log. 22 log. 7-2 log. 10 = N, and hence by transposition, &c. log. 11 = N + log. 2+ log. 7 - log. 9+ log. 10; and in this equation, The first term alone of this series is sufficient to give the logarithm of 11 true to 14 places. 111. When it is required to find the logarithm of a high number, as for example 1231, we may proceed as follows: log. 1231 = log. (1230+ 1) = log.{ 1230(1+1230) log.(1+ ;) 1 1230 Again, log. 1230 = log. 2 + log. 5 + log. 123, and log. 123 = log. {120(1+0) log. 120 = log. (23 × 3 × 5) = 3 log. 2+ log. 3 + log. 5. Therefore log. 1231 = 4 log. 2+log. 3+2 log. + log. (1+10). 5+log. (1+40) Thus the logarithm of the proposed number is expressed by the logarithms of 2, 3, 5, and the logarithms of 1 1 40' 1230' 1+ 1+ all of which may be easily found by the formulæ already delivered. 112. When it is required to interpose one logarithm between a series of equidistant terms in a table, it may be effected upon the principle of interpolation by means of the well-known theorem; viz. Thus, suppose there were given the logs. of 101, 102, 104, and 105, and that of 103 were required.. Here the number of equal intervals is 4, and of terms 5; so that the general form becomes a-4b+6c-4d+e=0; and c = [4 (b+d) -(a + e)] a=2-0043214 b=2-0086002 d=20170333 e=2-0211893 4(b+d) = 16-1025340 6)12-0770233 Log. of 103 =2-0128372 EXAM. 1. Given the logs. of 999 and 1000. Required the log. of 1001. EXAM. 2. Given the logs. of 51, 53, 57, and 59; to find the log. of 55. On this interesting and important subject, consult the Introduction to Dr. Hutton's Mathematical Tables, and Hellins's Mathematical Essays. TO FIND THE POINTS OF INFLEXION, OR OF CONTRARY FLEXURE IN CURVES. curve changes from concave to convex, or from convex to concave, on the same side of the curve. Such as the point e in the annexed figures, where the former of the two is concave towards the axis AD, from A to E, but convex from E to F; and on the contrary, the latter figure is convex from A to E, and concave from E to F. 114. From the nature of curvature, as has been remarked before at art. 85, it is evident, that when a curve is concave towards an axis, then the fluxion of the ordinate decreases, or is in a decreasing ratio, with regard to the fluxion of the absciss; but, on the contrary, that it increases, or is in an increasing ratio to the fluxion of the absciss, when the curve is convex towards the axis; and consequently those two fluxions are in a constant ratio at the point of inflexion, where the curve is neither convex nor concave; that is, is to y in a constant ratio, or - or is a constant quantity. But constant quantities have no fluxion, or their fluxion is equal to nothing: so that, in this case, the fluxion of i y is equal to nothing. And hence we have this or of general rule : 115. Put the given equation of the curve into fluxions; from which find either or y Then take the fluxion of i this ratio, or fraction, and put it equal to 0 or nothing; and from this last equation find also the value of the same or y Then put this latter value equal to the former, which i will form an equation; from which, and the first given equation of the curve, x and y will be determined, being the absciss and ordinate answering to the point of inflexion in the curve, as required. EXAMPLES. EXAM. 1. To find the point of inflexion in the curve whose equation is ax2 = a2y + x2y. This equation in fluxions is 2axi = a2y + 2xy + x2ỷ, which gives i a2 + x2 ý Zar-2xy Then the fluxion of this quantity made = 0, gives 2xi (ax-ry) = (a2+x2) × (aż-żу-хỳ); Lastly, this value of being put equal to the former, gives : a2+x3 a2+r2 a3-xa a-y 1 ; and hence 2x2 = a2-x, 2x or 3x2 = a2, and x = a, the absciss. Hence also, from the original equation, =a, the ordinate of the point of in EXAM. 2. To find the point of inflexion in a curve defined by the equation ay = a √ ax + x2. EXAM. 3. To find the point of inflexion in a curve defined by the equation ay2 = a2x + x3. ABCE D ΕΧΛΜ. 4. Το find the point of inflexion in the Conchoid of Nicomedes, which is generated or constructed in this manner: From a fixed point r, which is called the FGHI pole of the conchoid, draw any number of right lines, PA, PB, PC, PE, &c. cutting p the given line FD in the points F, G, H, I, &c.; then make the distances FA, GB, HC, IE, &c. equal to each other, and equal to a given line; then the curve line ABCE, &c. will be the conchoid; a curve so called by its inventor Nicomedes. TO FIND THE RADIUS OF CURVATURE OF 116. THE Curvature of a Circle is constant, or the same in every point of it, and its radius is the radius of curvature. But the case is different in other curves, every one of which has its curvature continually varying, either increasing or decreasing, and every point having a degree of curvature peculiar to itself; and the radius of a circle which has the same curva. ture with the curve at any given point, is the radius of curvature at that point; which radius it is the business of this chapter to find. 117. Let AEC be any curve, concave towards its axis AD; draw an ordinate DE to the point E, where the curvature is to be found; and E e d A D B G C suppose EC perpendicular to the curve, and equal to the radius of curvature sought, or equal to the radius of a circle having the same curvature there, and with that radius describe the said equally curved circle BEe; lastly, draw Ed parallel to AD, and de parallel and indefinitely near to DE: thereby making Ed the fluxion or increment of the absciss ad, also de the fluxion of the ordinate DE, and Fe that of the curve AE. Then put x = AD, y = DE, z = AE, and r = CE the radius of curvature; then Ed = i, de = ý, and Ee = ż. Now, by sim. triangles, the three lines Ed, de, Ee, which vary as therefore .. ż, ý, ż, GE, GC, CE ; GC i = GE. j; and the flux. of this eq. is GC.+GC.=GE. Y + GE. ÿ, or because GC = -BG, it is GC.2-BG.GE.Y+GE.. But since the two curves AE and BE have the same curvature at the point E, their abscisses and ordinates have the same fluxions at that point, that is, Ed or i is the fluxion both of AD and BG, and de or y ∝ the fluxion both of DE and GE. In the equation above therefore substitute i for BG, and ý for GE, and it becomes GCX- ii = GFY + ÿÿ, or Gcr - Gry = 2 + j = ż2. Now multiply the three terms of this equation respectively radius of curvature, for all curves whatever, in terms of the fluxions of the absciss and ordinate. 118. Further, as in any case either z or y may be supposed to flow equably, that is, either i or y constant quantities, or # or y equal to nothing, it follows that, by this supposition, either of the terms of the denominator, of the value of r, may be made to vanish. Thus, when i is supposed constant, ä being then = 0, the value of r is barely . |