The solution of a memorable problem by a special artifice of calculation

arXiv:0710.3956v1 [math.HO] 21 Oct 2007 The solution of a memorable problem by a special artifice of calculation∗ Leonhard Euler 1. The problem which I take up for solving here can be thus enunciated: To find a curved line AM (Fig. 1), with coordinates CP = x, PM = y, the arc AM = s and chord CM = √xx + yy = z, so that the integral formula R vds obtains a maximum or a minimum value, with v any fixed function of z. 2. In general, if a relation is sought between two variables x and y, and it is put dy = pdx, and V is any function of these x, y and p, so that its differential has this form: dV = Mdx + Ndy + Pdp, then the integral formula R V dx will have a maximum or a minimum value when it happens that Ndx = dP, so that this equation would express the sought relation between x and y. 3. While the entire problem is reduced to this equation, it will however be useful to further consider another equation which is equivalent to the first. For when Ndx = dP, which multiplied by p becomes Ndy = pdP, substituting in this value will produce dV = Mdx + Pdp + pdP = Mdx + d · Pp. It therefore follows that Mdx = d · (V −Pp). This equation is the most accom- modating for use in our analysis. 4. Let us now transfer these general precepts to the proposed problem. And first putting dy = pdx we will have ds = dx√1 + pp. Next, since it is z = √xx + yy, it will be dz = xdx+ydy z . Then indeed, as v is a function of z, let us put dv = qdz and it will then be dv = q(xdx+ydy) z . Now therefore for the formula of the maximum or the minimum we will have V = v√1 + pp. By differentiating this we will get dV = q(xdx + ydy)√1 + pp z + vpdp √1 + pp, ∗Presented to the St. Petersburg Academy on March 22, 1779. Originally published as Solutio problematis ob singularia calculi artificia memorabilis, M´emoires de l’acad´emie des sciences de St-P´etersbourg 2 (1810), 3–9. E731 in the Enestr¨om index. Translated from the Latin by Jordan Bell, Department of Mathematics, University of Toronto, Toronto, Canada. Email: jordan.bell@utoronto.ca 1 A M P B C Fig. 1 and with the differential compared to the general form it will be M = qx√1+pp z , N = qy√1+pp z and P = vp √1+pp. 5. Therefore, the equation comprising the total solution of our problem, which was Ndx = dP, now assumes this form: qydx√1+pp z = d · vp √1+pp. As well, the other equation which will be used, since V −Pp = v √1+pp, may be expressed thus: qxdx√1 + pp z = d · v √1 + pp. Hence, since the first equation is d · vp √1 + pp = vdp √1 + pp + p · d · v √1 + pp = vdp √1 + pp + qxdy√1 + pp z , we will be led to this: q(ydx −xdy)√1 + pp z = vdp √1 + pp and hence it will be ydx −xdy = vzdp q(1 + pp), and this is the equation which we employ for getting the solution of our problem. 6. Let us divide this equation by zz = xx + yy, so that we will obtain this form: ydx −xdy xx + yy = vdp qz(1 + pp), where it is clear that the integral of the first side is Atang x y . In truth it seems here that little gain be obtained, since the latter part of the equation is com- pletely intractable. In the meanwhile however let us put ϕ to be the angle whose tangent is x y , so that our equation would be dϕ = vdp qz(1+pp). 7. With this angle ϕ introduced, we will be able to render the coordinates x and y susceptible to calculation. For since it is x y = Atang ϕ, it will be 2 x = z sin ϕ and y = z cos ϕ, by means of whose values the letter p can also be extracted. Indeed because p = dy dx, it will be p = dz cos ϕ −zdϕ sin ϕ dz sin ϕ + zdϕ cosϕ. Now let us put dz = tzdϕ, so that it becomes p = t cos ϕ −sin ϕ t sin ϕ + cos ϕ = t −tang ϕ 1 + t tang ϕ. This expression clearly expresses the tangent of the difference of two angles, the tangent of the first of which is = t, while the latter angle is = ϕ. 8. Thus for p equal to the tangent of any angle whatsoever, let us put p = tang ω, and ω will be the difference of these angles, namely ω = Atang t−ϕ, whence dω = dt 1+tt −dϕ. Also indeed, because p = tang ω and so ω = Atang p, it will further be dω = dp 1+pp. Hence our equation to be resolved will be dϕ = vdω qz . Moreover from the preceding, since the form was dω = dt 1+tt −dϕ, hence qzdϕ v = dt 1 + tt −dϕ or dϕ(1 + qz v ) = dt 1 + tt. 9. Also, because we have put dz = tzdϕ, it will be dϕ = dz tz , and having substituted in this value our equation takes the form: dz z (1 + qz v ) = tdt 1 + tt. Then, since qdz = dv, the integration can be done most conveniently by loga- rithms; for it will be lz + lv = l √ 1 + tt −ln, and consequently we will obtain this integrated formula: vz = √1+tt n . 10. Now let us investigate the value of t from this equation, in which it will be t = √nnvvzz −1. For since t = dz zdϕ, we gather from this equation that dϕ = dz z√nnvvzz −1, which is a differential equation of the first degree between the angle ϕ and the distance CM = z, if indeed v is a function of z itself. Indeed for the angle it is noted that tang ϕ = x y and hence x = z sin ϕ and y = z cos ϕ, so that now the two coordinates x and y can by exp

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