   SEARCH HOME Math Central Quandaries & Queries  Question from Marilyn, a student: Hi, could you please help me with this question? In a city (in the Northern Hemisphere) the minimum number of hours of daylight is 9.6 and the maximum number is 14.4. If the 80th day of the year (March 21) has 12 hours of daylight, determine a sine function which gives the number of hours of daylight for any given day of the year. (Jan 1 = 1, Jan 2 = 2, etc). Thank you! Hi Marilyn,

The average of 9.6 and 14.4 is 12. 14.4 = 12 + 2.4 and 9.6 = 12 - 2.4 so the number of hours of daylight cycles between 12 - 2.4 and 12 + 2.4. Since you are assuming that the periodic cycle is a sine function the equation can de written

h(d) = 12 + 2.4 sin(f(d))

where d is the day (d = 1 on Jan 1, d = 2 on Jan 2 etc.), h is the number of hours of daylight and f(d) is some unknown expression involving d. The maximum value of the sine function is 1 and the minimum value is -1 so h(d) will cycle between 12 + 2.4 and 12 - 2.4.

You know that h(80) = 12 so when d = 80 the value of the sine function is zero. Hence the expression f(d) might be d - 80 so that when d = 80, sin(d - 80) = sin(80 - 80) = sin(0) = 0. But this can't be correct! The period of the sine function is 360 degrees and hence sin(d - 80) would cycle after 360 days and you want it to cycle once a year which is 365 days. Thus you need a scale factor k to shift the cycle from 360 units to 365 units. That is you need f(d) = sin[k(d - 80)] for some number k. Can you see a scale factor k that will work?

Harley     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.