3rd Putnam 1940

Problem B1

A stone is thrown from the ground with speed v at an angle θ to the horizontal. There is no friction and the ground is flat. Find the total distance it travels before hitting the ground. Show that the distance is greatest when sin θ ln (sec θ + tan θ) = 1.



Straightforward, except for a tricky integral.

Take coordinates (time and distance) zeroed on the peak of the trajectory. After time t, the stone travels a distance x = v t cos θ horizontally and a distance 1/2 g t2 vertically. So the trajectory is a parabola 2y = k x2, where k = g/(v cos θ)2. The stone is on the ground at t = +/- v/g sin θ, at a horizontal distance +/- a from the peak, where a = v2/g sin θ cos θ.

The length of the parabola y = k/2 x2 between x = -a and x = a is 2 ∫0a (1 + k2x2)1/2 dx.

To do the integral it helps to remember that 1 + sinh2z = cosh2z. So substituting kx = sinh z, will essentially give us the integral of cosh2z. That is doable, using the analog of the double angle formulae. So setting I = ∫ (1 + k2x2)1/2 dx, and substituting x = sinh z, we have I = 1/k ∫ cosh2 z dz = 1/(2k) ∫ (cosh 2z + 1) dz = 1/(4k) sinh 2z + 1/(2k) z = 1/(2k) sinh z cosh z + 1/2 z = x/2 (1 + k2x2)1/2 + 1/(2k) sinh-1(kx). We have k = g/(v cos θ)2, and a = v2/g sin θ cos θ, so ka = tan θ. So the required path length is p(θ) = a(1 + k2a2)1/2 + 1/k sinh-1a = v2/g (sin θ + cos2θ sinh-1 tan θ).

To find the maximum, we differentiate, getting p'(θ) = cos q - 2 cos θ sin θ sinh-1 tan θ + cos2θ (1 + tan2θ)-1/2 sec2θ = 2 cos θ(1 - sin θ sinh-1tan θ). In the range [0, p/2], tan θ is monotone increasing. Sinh-1z is strictly monotone increasing for positive z, so sinh-1tan θ is strictly monotone increasing on (0, π/2). Indeed it evidently tends to ∞ as θ tends to π/2. Hence (1 - sin θ sinh-1tan θ) is strictly monotone decreasing on (0, π/2) and crosses zero once. Cos θ is monotone decreasing and positive, so p'(θ) is strictly monotone decreasing and crosses zero once. Hence p(θ) has a single maximum on [0, π/2], which is achieved for the value φ in (0, π/2) for which (1 - sin φ sinh-1tan φ) = 0. Rearranging, sinh(1/sin φ) = tan φ. Squaring, adding 1, and taking the square root: cosh(1/sin φ) = sec φ. Adding the last two equations: e1/sin φ = tan φ + sec φ, or 1/sin φ = ln(tan φ + sec φ), or sin φ ln(tan φ + sec φ) = 1.



3rd Putnam 1940

© John Scholes
15 Sep 1999