Flux (sin x cos x)=x cos x- sin x = & cos 2x. (1+cos x) = cos x, we have flux/(1+cos x)= flux cos Since ✓ 2 x= sin. 2 Similarly we shall find that flux (cos log x)= flux log x sin log == sin log; and that flux ( sin x)=x sin x+zx cos x. APPLICATION OF FLUXIONS TO THE THEORY OF CURVES. 25. Of all the problems that can be proposed respecting a curve, the most simple is that which requires us to draw a tangent to any point of the curve. Suppose the curve to be AM, its axis, AP; its coordinates AP and PM; it is evident that to draw a tangent to the point M, we have only to determine the subtangent PT. Let us imagine the arc Mm to be infinitely small, draw the ordinate mp, infinitely near to AP, and sup pose Mr parallel to Pp. Let, as usual, AP=x, PM=y, and we shall have Pp, or Mr, mr-y; and by similar triangles Mrm, TPM we have mr: rM::MP: PT or y:::y: PT= Consequently we y have only to throw the equation of the curve into fluxions, in order to obtain the value of . and then to substitute this value in the y formula for the subtangent just found, and PT 26. The expression for the tangent MT is ✔ (~+y^); that of the subnormal PN is y3 PT MN =√(2*+91); and if through the point A we draw the line AQ parallel to MP, we shall have PT: AT:: MP: AQ, xy that is :: y: AQ=y- These values of AQ and yx. yx x x y y AT enable us to find the asymptotes of the curve AM if it have any; for if, after having substituted in these two values that of obtained from the equation of the curve, we suppose a infinite, there will be as many asymptotes as there are different values of the lines AQ and AT. The position of the asymptotes will always be determined by the points T and Q. We shall now apply these for mula to a few examples. yx y The equation of the circle is ya-; therefore yy=-xx, and or the subtangent= y1__ (aa—x3). The sign indicates that the subtangent must be taken in the same direction as the abscissa, because in the construction of the formula it was taken in a contrary direction. If we had taken the vertex of the curve for the origin of the abscissæ, the equation would have been y2 = 2ax-XX, and this would have given a positive result as in the formula. The equation y1—a— gives or the subnormal = x; and the normal = √ ( y2 + y2 ya) = √(x2+3)=a=the radius, as it evidently should be. y In the parabola y=pr; therefore 9p, the subnormal; and a + x an expression which is reduced the quantity a, if we suppose a infinite. same supposition we find that AQ= y − z ? = y - b2x On the (a+x)= a2y -), becomes simply b. These two values of AT and AQ give the position of the asymptotes the same as we found in article (49) conic sections. In the logarithmic curve, we have = A log y, and z = Ay. y Therefore A. Consequently its subtangent is always equal to y the modulus (art 69, page 429) a á 27. Let there be any curve BOC with another curve BMA, such, that if we prolong the ordinates OP of the first till they meet with the second curve, the line MO may be some function of arc BO; it is proposed to draw through the given point M the tangent MT. Conceive the ordinate mp indefinitely near to MP, and Mr parallel to the tangent at the point O; if we make the arc BO = 2, MO =u, we shall have mr — u, rM-Oo-z, and uz::u: OT= = by taking the fluxions of the equation which expresses the given relation; and consequently TO, or the point T will be determined, whence it will be easy to draw the tangent MT. b a bz a Suppose, for example, that u 2, we shall have u = and OT=z=arc BO. If BOC is a circular arc, then AMB, is a cycloid, and this construction is the same as we have already given. In the quadratrix, if we compute the abscisse from the centre, we have (art 76 page 432) y=cot; therefore y= cot a cx a cxx ved by the two similar triangles MOT, and the ffuxional triangle formed by drawing an ordinate indefinitely near to MP. (We employ-y, because y diminishes, as x augments.) CO = PM=y== cot, we shall have CT = a a When CM=CT, or at the point D, we have, as before explained, the base CD=2, and consequently CT = αα c CM CD We must there fore take CT a third proportional to the base CD and the radius CM; this will give the point T, through which and through the point M if we draw the line MT, it will be the tangent required. 28. To draw tangents to spirals, we must resolve the following problem. Let there be described a circle with any radius CA, and let there be a curve CKM such, that drawing the radius CMN, the line CM may be any function of the arc ABN, it is required to draw through the given point M the tangent MT. Conceive the two radii CMN, Cmn indefinitely near to each other, and the little arc Mr described from the centre C, and with the ra dius CM; and then draw CT perpendicular to CM. This done, let CM=y, ABN=x, CA a; we shall then have MQO'. and CT= y_axy_xy α y In the hyperbolic spiral whose equation is xy-ab, we shall have 29. In the logarithmic spiral in which the angle CMT is con stant, conceive the radii CM, T Let t-tan Mmr, we shall have t= flux (log z); or + a constant quantity C; because the at x fluxion of the equation log = is the same as that of log == +C. Now the equation log z=+C, shews I°, that this spiral makes at an infinite number of revolutions about its centre, as well in approaching towards it, as in receding from it; for in place of a we may successively substitute +, x + 2x, x + 3%, &c. -2x+x, &c. « being the circumference ANB. |