A weather station measures the temperature T continuously. It is found that on any given day T = p(t), where p is a polynomial of degree ≤ 3, and t is the time. Show that we can find times t_{1} < t_{2}, which are independent of p, such that the average temperature over the period 9am to 3pm is ( p(t_{1}) + p(t_{2}) / 2. Show that t_{1} = 10:16am, t_{2} = 1:44pm.

**Solution**

Let t be time after 9am and T be 6 hrs after 9am, so that t = 0 represents 9am and t = T represents 3pm. We may write the polynomial p(t) as at^{3} + bt^{2} + ct + d for some a, b, c, d. The average temperature is 1/T ∫_{0}^{T}(at^{3} + bt^{2} + ct + d) dt = T^{3} a/4 + T^{2} b/3 + T c/2 + d. We wish to find t_{1}, t_{2} so that this equals (t_{1}^{3} + t_{2}^{3}) a/2 + (t_{1}^{2} + t_{2}^{2}) b/2 + (t_{1} + t_{2}) c/2 + d for all a, b, c, d.

The terms in d match. The term in c matches provided we take t_{1} + t_{2} = T. The term in b matches provided we take t_{1}^{2} + t_{2}^{2} = 2T^{2}/3. These two equations already determine t_{1} and t_{2}. In fact, solving the quadratic, we get t_{1} = T(1 - 1/√3)/2 and t_{2} = T(1 + 1/√3)/2.

For the terms in a to match we need t_{1}^{3} + t_{2}^{3} = T_{3}/2. But we can easily check that the values above satisfy this relation also.

Finally note t_{1} = 6(1/2 - (√3)/6 ) hrs = 3 - √3 = 3 - 1.732 = 1.268 hrs = 1 hr 16 min (approx), so t_{1} represents about 10:16am and t_{2} about 1 hr 16 min before 3pm or 1:44pm.

© John Scholes

jscholes@kalva.demon.co.uk

25 Aug 2001