The monthly average temperature T (in °C) in a certain city, may be approximated by
T(t) = a cos[ b(t - d) ] + c
where t is the time in months, t = 0 corresponds to January 1 st, and we assume that function T(t) has a period of 12 months.
a) Find a, b (b > 0), c and d if T has a maximum of 22.4 °C in the middle of July (t = 6.5) and a minimum of - 10 °C.
b) In which month is T minimum?
c) Over how many months is T greater than 18 °C?
Answers
Solution
a) The minimum and maximum values of T , Tmax and Tmin respectively, allow us to find a and c as follows
We are given: Tmax = 22.4 and Tmin = -10
c = (Tmax + Tmin) / 2 = 6.2
|a| = (Tmax - Tmin) / 2 = 16.2 , we take a > 0 and equal to 16.2
The graph of T is that of a a cos(bt) shifted 6.5 to the right. Hence
T(t) = 16.2 cos[ b(t - 6.5) ] + 6.2
We now use the period to find b.
The period is equal to 12 = 2π / b , hence b = π / 6
T(t) is given by
T(t) = 16.2 cos[ (π / 6)(t - 6.5) ] + 6.2
b) Find t for which T is minimum. The minimum value of T is equal to -10. Hence t that makes T minimum is found by solving
16.2 cos[ (π / 6)(t - 6.5) ] + 6.2 = -10.2
cos[ (π / 6)(t - 6.5) ] = - 1
(π / 6)(t - 6.5) = π
t = 12.5
NOTE We could have answered part b) using the fact that the distance between a maximum and the following minimum in a sinusoidal function is half a period and therefore T is minimum at t = 6.5 + (1/2)12 = 12.5 days
t = 12.5 is larger than one period which is 12. Hence 12.5 corresponds to a period of 12 months (one year) and 0.5 month of January. So the temperature is minimum in mid January.
c) (See graph below) We first find the times t1 and t2 at which T = 18 by solving the equation
16.2 cos[ (π / 6)(t - 6.5) ] + 6.2 = 18
cos[ (π / 6)(t - 6.5) ] = (18 - 6.2) / 16.2
(π / 6)(t - 6.5) = arccos((18 - 6.2) / 16.2)
t = (6/π) arccos((18 - 6.2) / 16.2) + 6.5 = 7.94 months
Examine the graph of T below and the horizontal line y = 18 (graphical interpretation of the equation above), we note that the solution t = 7.94 is larger that 6.5 which corresponds to t2 = 7.94. By symmetry t1 may be calculated as follows
Graph of y = T(t) and y = 18
t1 = 6.5 - (t2 - 6.5) = 6.5 - (7.94 - 6.5) = 5.06 months
The number of months for which T is greater than 18 is equal to
t2 - t1 = 7.94 - 5.06 = 2.88 ; approximately 3 months.
