On the curve of fastest descent in whatever resistent medium
Euler determines various properties of the brachistochrone curve subject to friction, taking the form of (v/c)n, where v is the squared velocity and c is the squared velocity at which the forces of friction and gravity cancel. He derives the brachistochrone property 2v/r = N, where r is the radius of curvature and N is the component of force that is perpendicular to the curve. Euler recalls that the frictionless solution reads s2 = 2a(s–x), for some constant a, which is the already-established equation for a cycloid. He further considers the frictionless case with centripetal force taking the form ym, where y is the distance to the attracting point. If a is the initial distance from this point, the solution reads Az2 = am+3 – ym+3 for some constant A. Now, looking at the brachistochrone with friction, Euler calls the horizontal coordinate y and the vertical coordinate x. Defining p = ds/dx, Euler finds that the equation 1/(p3nqn) = ngn-1/(2n-1cn) ∫ (p2–1)n-1dp/p2n = P-n must always hold, which reduces conveniently for n=1. He then finds the solution s = c∙ln((s–ax–ac+c)/(c–ac)), after which he proceeds to look for some expressions for the involved arc length. He ends by looking at how other methods would have influenced the solution and how this solution relates to the tautochrone.
Original Source Citation
Commentarii academiae scientiarum Petropolitanae, Volume 7, pp. 135-149.
Opera Omnia Citation
Series 1, Volume 25, pp.41-53.