18th Putnam 1958

Problem B6

A particle of unit mass moves in a vertical plane under the influence of constant gravitational force g and a resistive force which is in the opposite direction to its velocity and with magnitude a function of its speed. The particle starts at time t = 0 and has coordinates (x, y) at time t. Given that x = x(t) and is not constant, show that y(t) = - g x(t) ∫0t ds/x'(s) + g ∫0t x(s) / x'(s) ds + a x(t) + b, where a and b are constants.



We can write the equations of motion as x'' = - f(x',y') x', y'' = - f(x', y') y' - g. If x'(t) = 0 at some t, then a possible solution to the equations would be x = constant (and y found by integrating y'' = - f(0, y') y' - g). But solutions are unique, so this would be the solution. But we are told that x is not constant. So x'(t) is never zero.

We do not know anything about f, so we resolve perpendicular to it to get x'' sin θ - y'' cos θ = g cos θ, or x'' tan θ - y'' = g. But tan θ = dy/dx = y'/x', so x'' y'/x' - y'' = g (*).

Dividing by x' gives x'' y'/(x')2 - y''/x' = g/x'. Integrating, y'/x' = A - g ∫ dt/x'. So y' = Ax' - g x' ∫ dt/x'. Integrating again, y(T) = A x + B - g ∫0T x' ∫0t ds/x' dt. Now we can integrate the last integral by parts: ∫0T x' ∫0t ds/x' dt = ∫t=0T0t ds/x' dx = x(T) ∫0T ds/x' - ∫0T x/x' dt. So we get finally the expression in the question.



18th Putnam 1958

© John Scholes
25 Feb 2002