17th Putnam 1957/A2 solution

17th Putnam 1957

Problem A2

k is a real number greater than 1. A uniform wire consists of the curve y = e^x between x = 0 and x = k, and the horizontal line y = e^k between x = k - 1 and x = k. The wire is suspended from (k - 1, e^k) and a horizontal force applied at the other end, (0, 1) to keep it in equilibrium. Show that the force is directed towards increasing x.

Solution

If d is the density per unit length, then the total mass of wire is d(1 + ∫₀^k √(1 + e^2x) dx ). The moment about the y-axis is d(k - 1/2) + d ∫₀^k x √(1 + e^2x) dx. We require that the x-coordinate of the centre of mass exceeds k - 1. In other words, k - 1/2 + ∫₀^k x √(1 + e^2x) dx > (k - 1) ( 1 + ∫₀^k √(1 + e^2x) dx ).

Set f(k) = ∫₀^k x √(1 + e^2x) dx - (k - 1) ∫₀^k √(1 + e^2x) dx. Then we require f(k) > -1/2 for all k > 1. Clearly f(0) = 0. We try differentiating, and get f '(k) = √(1 + e^2k) - ∫₀^k √(1 + e^2x)dx.

One possibility is to evaluate the integral explicitly (substitute y = e^2x, then z² = 1 + y, we end up with a messy but doable integral). But it is maybe easier to approximate. We have 1 + e^2x < 1 + e^2x + e^-2x/4 = (e^x + e^-x/2)². So ∫₀^k √(1 + e^2x)dx < ∫₀^k (e^x + e^-x/2) dx = (e^x - e^-x/2) |₀^k = e^k - e^-k/2 - 1/2 < e^k < √(1 + e^2k). So f '(k) > 0. Hence f(k) > f(0) > 0 > -1/2.

17th Putnam 1957