k is a real number greater than 1. A uniform wire consists of the curve y = ex between x = 0 and x = k, and the horizontal line y = ek between x = k - 1 and x = k. The wire is suspended from (k - 1, ek) 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.
If d is the density per unit length, then the total mass of wire is d(1 + ∫0k √(1 + e2x) dx ). The moment about the y-axis is d(k - 1/2) + d ∫0k x √(1 + e2x) dx. We require that the x-coordinate of the centre of mass exceeds k - 1. In other words, k - 1/2 + ∫0k x √(1 + e2x) dx > (k - 1) ( 1 + ∫0k √(1 + e2x) dx ).
Set f(k) = ∫0k x √(1 + e2x) dx - (k - 1) ∫0k √(1 + e2x) dx. Then we require f(k) > -1/2 for all k > 1. Clearly f(0) = 0. We try differentiating, and get f '(k) = √(1 + e2k) - ∫0k √(1 + e2x)dx.
One possibility is to evaluate the integral explicitly (substitute y = e2x, then z2 = 1 + y, we end up with a messy but doable integral). But it is maybe easier to approximate. We have 1 + e2x < 1 + e2x + e-2x/4 = (ex + e-x/2)2. So ∫0k √(1 + e2x)dx < ∫0k (ex + e-x/2) dx = (ex - e-x/2) |0k = ek - e-k/2 - 1/2 < ek < √(1 + e2k). So f '(k) > 0. Hence f(k) > f(0) > 0 > -1/2.
17th Putnam 1957
© John Scholes
25 Feb 2002