Proposition 40

PROPOSITION 40. Problem

361. To find the general rule, following which a curve ought to be disposed, in order that a body descending on that curve arrives at any point on the curve in the shortest possible time.

Solution

Let AMC (Fig. 45) be a curve of this kind, upon which a body arrives at C from A in a shorter time than on any other curve crossing from A to C.

Therefore for any 2 point M and μ between these, the curve intercepted must be compared, in order that its motion along the arc AMC is completed in a shorter time than any other, if it should be intercepted. Now the nearby points M and μ are joined by the two elements Mm, mμ , and the time to pass through Mmμ is a minimum, or by the rules of the method of maxima and minima, equal to the time for the nearby elements Mn, nμ . [p. 175] The applied lines MP, mp, and μπ are drawn to the axis AP and with the elements Pp, p π equal to each, or also MG = mH and pm, if there is a need, is produced to n and mn in made infinitely small with respect to the elements Mm and nμ . Therefore it must be : t .Mm + t .mμ = t .Mn + t .nμ .

[The time to traverse the elements Mm and mμ is equal to the time to traverse the infinitesimally close elements Mn and nμ , the condition for the time variable to be a stationary point w.r.t. the independent variable x, for which Pp = pπ = dx. Meanwhile the curve is described at these points M , m , and μ by y , y + dy , and y + dy + ddy ] Let the speed that the body has at M correspond to the height v, with which therefore both the elements Mm and Mn are traversed. Moreover the speed that the body has at m, corresponds to the height v + du and the speed that it has at n, corresponds to the height v + du + ddw [note that du includes contributions from the vertical force, as well as from a general horizontal force and some tangential force acting along the curve; the final speed includes second order contributions from the horizontal and tangential forces, as we are shown eventually in the corollaries. Recall that in Euler’s analysis, an increment is always added at the start of an element, and holds a constant value within the increment : thus, the speed at M corresponds to v throughout Mm, the speed at m corresponds to v + du, and is held constant throughout mμ , etc]; moreover with that [former] speed the element mμ is traversed, as now with this [latter] speed so the element nμ is traversed. Hence therefore this equation is obtained : now hence with the centres M and μ and with the arclets mg and nh drawn, there arises :

Again, …

From which, with the negligible parts ignored, on substituting there arises Now on account of the [four] similar triangles nmg, mMG and nmh, μmH , it follow that, and …

On account of which the above equation is given by : Which equation is homogeneous and determines the nature of the curve AMC, the so- called brachistochrones, [p. 176] upon which the body in the shortest time arrives at C from A. Q.E.I

Corollary 1

362. Therefore, if the lengths are called : …then … With these substituted, there is obtained :

In which, if from the forces acting v, du, and ddw can be determined, and the equation for the brachystochrone curve can be obtained. But always ddw thus involves mn, as from the calculation it exceeds mn.

Corollary 2

363. Let the radius of osculation of the curve Mmμ = r ; and this lies in the region directed away from the axis AP, given by : but dy as the differential of ds ;] [as dx2 + dy2 = ds2, ddx = 0, and dsdds = dyddy, giving dx r hence this equation is produced : Where it is to be noted that 2rv is the centrifugal force, which is sent along the normal to the curve at M .

Corollary 3

364. If, from the forces acting, it follows that [we now interpret this equation as the sum of the works done per unit mass for increments dx with the force P, dy with force Q, and force R along ds ] : then … since with the point m translated to n the small distance dy is increased by mn and ds by dy .mn the small amount ng. [p. 177] Moreover since ng = ds , then with which substituted this equation is obtained :

Scholium 1

365. From the solution, it is understood that the formula found extends the widest and can be extended to any forces acting, and also including resistance. For whatever forces should be acting, so du as well as ddw can be determined, and which values put in place gives the equation for the brachystochrone sought. But yet these are only in place, if the directions of the forces are in the same plane; for the curve found is situated in the same plane.

Yet no less, if the forces should not be acting in the same plane, a brachystochrone curve can be found in a given plane with the help of this formula. Indeed in any given particular plane it pleases, a brachystochrone curve is given, whatever the forces acting should be. Now the other question is, if a brachystochrone curve is sought from among all the curves joining two given points, also not situated in one given plane. Now as often as the directions of the forces acting in the same plane are put in place, there is no doubt that the brachystochrone curve is in the same plane. For if the curves are not in the same plane, oblique forces are acting and therefore they do not give the body as great an acceleration as is possible. [p. 178] Therefore from this solution, so the brachystochrone curve is found completely, if the directions of the forces acting are in the same plane, and also a brachystochrone is present in a given plane, whatever should be the forces acting. [However, we understand that the time will be larger for such curves where the forces are not coplanar.]

Scholium 2

366. This question about brachystochrone curves or lines of quickest descent was first set out by the most celebrated Joh. Bernoulli [actually Leibniz and Jas. Bernoulli] and many solutions of this are extant, as in the Act. Lips., so also in the English Transactions, and in the Comm. Acad. Paris, and in others, where the solution for this problem, under the hypothesis of forces acting directed downwards and for centripetal forces, have been given. But no one has solved the fundamental problem that we have given here, set out so widely, so that it can also be extended to include forces of any kind, as well as resistance. For all have supposed that ddw = 0 , because that is always wrong, unless the direction of the force is MG or mH.