The number of hours of daylight H in a certain area is approximately given by the function
H(t) = 2.5 cos[ b(t - d) ] + 11.5
where H is in hours and t in days, and the function has a period of one year (365 days).
a) Find b (b > 0) and d if H is maximum on June 21st (month of February has 28 days).
b) Which day is the shortest (has the smallest number of hours of daylight)?
Answers
Solution
a) Since the period is known and equal to 365 days, then .
365 = 2π / b , hence b = 2π / 365
If we set d = 0 in the function H(t) = 2.5 cos[ b(t - d) ] + 11.5, it becomes H(t) = 2.5 cos[ b t ] + 11.5 which has a maximum at t = 0.
In our problem, the maximum happens on the 21 st of June corresponding to
t = 31 + 28 + 31 + 30 + 31 + 21 = 172 (number of days from January 1st to 21 st of June)
Therefore the horizontal shift (translation) of 172 is to the right and d = 172. Hence
H(t) = 2.5 cos[ 42.2(t - 172) ] + 11.5
b) The shortest day corresponds to t that gives H minimum which is equal to 11.5 - 2.5 = 9. Hence we need to solve the equation
2.5 cos[ (2π / 365)(t - 172) ] + 11.5 = 9
cos[ (2π / 365)(t - 172) ] = (9 - 11.5) / 2.5 = - 1
(2π / 365)(t - 172) = π
t = 365π/(2π) + 172 = 354.5 days
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 h is minimum at t = 172 + (1/2)365 = 354.5 days
For the first 11 months (from January to November) of the year, t is given by
t = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 = 334
Number of days in December
354.5 - 334 = 20.5 number of days in December
which approximately corresponds to the 21 st of December.
Answer:
b) December 21st
Step-by-step explanation: