Page:Encyclopædia Britannica, Ninth Edition, v. 13.djvu/69

 INFINITESIMAL CALCULUS equation of the locus is (a-r)- = 1ulta (see fig. 13). Hence the =dr / H -- JJTS ; so that, if s 1 be the arc contained lie- arc ds Fig. 13. tvveen the values a and ^a + e ( = AG) of r, and s 2 the arc between the values a- c ( = AN) and %a, we have r-(a-r) 2 (&quot;-i I ?-(- r) J a*b- &quot; 2 y _ . f a&quot;l&amp;gt; 2 Now, putting in the former r=ka + z, and in the latter we find whence we conclude, as Bernoulli did, that even in curves whose rectification has not yet been effected, parts may be assigned which are equal though dissimilar ; such as BG = AN, and GI = NI, where AI&amp;lt;=4. John Bernoulli, following up this discovery of his brother, pro posed to find for a given curve another, such that the sum or the difference of two arcs, one on each curve, may be expressed by arcs of circles. In the case of the cubical parabola he noticed that the two curves reduce to one curve, in which, without effecting the recti fication, pairs of arcs could be found whose difference is rectifiable. 189. The Count Fagnani next proposed in 1714 the problem, &quot;given a portion of the parabola whose equation is x* = ?/, to find another portion of it such that the difference of these two parts may be rectifiable. &quot; In the following year, not having received a solu tion, Fagnani published his own, with greater generality, as follows. 771+2 Taking a constant, and m any real number, let y = m + 2 the equation of a parabolic .curve (compare Ex. 2, 166). the portion of the tangent be tween any point (xy) of the curve and the axis of abscissee. Then by the equation of the curve, Let t be t ~1 and the arc of the curve 0! V a) dx . 1 I Integration by parts gives (see fig. 14) m /-. I dx =-arcPP t - if P, P! have the abscissa; x a, x ; also TO /- ! dz m + 2 1+ if Q&amp;gt; Qi have z , z-j, as abscissae. Nov/ if the former integral can be transformed by introducing a variable ^ so that the function under the sign of integration may remain unaltered, and thus the former integral may pass into the latter, we may equate arc PP X - ( P 1 E 1 - PR) = arc QQj - (Q^ - QS). Thus the question is to determine an integral of the differential equation dx - = 0. or of Fagnani gave the following solutions. For m = 4 the curve is the cubical parabola, and the relation between a; and z is xz = a-. For m = 3 the equation is satisfied by the relation For m = 6 the curve is as in the problem proposed, and the relation is 190. Passing over Fagnani s investigations relative to the 1cm- niscate, such as his discoveries of the method of doubling or halv ing any arc, of dividing the quadrant into equal parts in number 2.2 1 &quot;, 3.2 &quot;, or 5.2 m, discoveries which prompted to their author the wish, since executed, that on his tomb a lemniscate should be in scribed in mcmoriam, we must mention his well-known geometrical theorem, that on the circumference of an ellipse, in innumerable ways, pairs of arcs can be determined having their difference ex pressible by a right line (first published 1716). His method is as follows. If we have h, I, f, g constants, and ( fhx-^) + (fix*) = 0, then, first, the sum fdx^hx* + 1 fdz VAe* + 1 _ hxs. . - I A ix-^ = &quot;*&quot;/ .//=-., ./ 7; Ui T 1 - U^&amp;gt; J V/ar + y J Nfz- + cj V - ft and, secondly, the same sum = a + -- j= when s= - 1. In the former case the relation gives _f -flx -gl Introducing this into the first integral, and the corresponding valuo of x into the second, the sum of the integrals becomes /dxf - ~V7 But differentiating the relation, and dividing by Ifxz, we find whence, substituting, the sum is found to be as stated. A liko treatment yields the formula when s=-l. This theorem is applied in its former part to elliptic arcs. Let us call the axis major la, the parameter^, and the abscissa x ; then, if h=p2a, the element of the arc AB (fig. 15) corresponding to the abscissa CD = a; can be shown to be If now Z=2a 3, /= - la, and &amp;lt;/ = 2 3 , this becomes the former differ ential in (1) ; and it appears that, taking another abscissa C E = ~ = - wo have Jixz, T arc AB + arc AF = - 2 + K. To determine the value of the constant K, leta; = 0; then AF becomes the entire arc AG, hence arcAB-arcGF=-|~. The second part of the theorem is applied to the hyperbola (fig. 16). Calling IIA = 2i7, the parameter p, and x the variable abscissa CD, and putting h=p + 2a, the element of the arc AIJ is easily found to be expressed by dx/hx 2 - 2a 3