a and b are positive reals and a > b. Let C be the plane curve r = a - b cos θ. For what values of b/a is C convex?
Solution
The curvature is (x'y'' - x''y')/(x'2 + y'2)3/2. So for convexity we require x'y'' ≥ x''y'. Putting k = b/a, c = cos θ, s = sin θ, we have x/a = c - k c2, x'/a = - s + 2kcs, x''/a = - c - 2ks2 + 2kc2 and y/a = s - ksc, y'/a = c - kc2 + k s2, y''/a = -s + 4ksc. Thus the condition becomes after a little cancellation, 1 + 2k2 - 3kc ≥ 0. c takes values in the range -1 to 1, so for the condition to be true for all points of the curve we require 1 + 2k2 - 3k ≥ 0. But 1 + 2k2 - 3k = (2k - 1)(k - 1). We are given that k < 1, so we must have k < 1/2.
For small k, the curve is approximately a circle centred on the origin. As k increases it develops a flattening near x = a, y = 0. For k > 1/2, this becomes a dimple in the surface, so that convexity is broken. For k = 1, the depression in the surface extends as far as the centre (the origin). For k > 0, the curve intersects itself at the origin so that it comprises two ovals one inside the other.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002