Someone please help me!!! I think I've found the correct formula for this but I don't understand exactly why this is a sine function and how to find the days of the year when Portland has 11 and 15 hours of daylight. PLEASE HELP!!
The equation I got was: D = 4 sin(2π t / 365) + 13. Is this right??
It is possible to use a sinusoidal function to model the amount of perceived daylight in a certain location over the course of a year. For Portland, Oregon, there is a minimum of 9 hours of “daylight” on the 1st day of winter and a maximum of 17 hours of “daylight” on the 1st day of summer. Let D represent the number of hours of “daylight” in Portland, Oregon, T days after the 1st day of spring (assume that T = 0 corresponds to March 20th). You may assume that 1 year has 365 days.
Find a formula for such a function, being sure to explain the practical meanings of any important pieces of the formula (amplitude,midline, and period). Use your formula to determine on what days of the year (month and day, not just T’s value) Portland has about 11 hours of “daylight” and about 15 hours of “daylight”. Please round to the nearest day, if not exact.
How do I solve the function I got for t?? I'm assuming I should plug in 11 and 15 for D but I don't know how to solve it from there and how to get an exact date.
Also, I'm not sure what the practical meanings of the period, amplitude, and midline are... Can someone explain them to me please?? Thank you!!
5 answers
D = 4 sin(2π t / 365) + 13
where D represent the number of hours of “daylight” in Portland
You want to find when D=11. So, solve
4 sin(2π t / 365) + 13 = 11
4 sin(2π t / 365) = -2
sin(2π t / 365) = -1/2
Now, sin(7π/6) = -1/2, so
2πt/365 = 7π/6
t = 7/6 * 365/2 = 212.9
So, you want the date 213 days after March 20.
Pulling out your handy Julian date calendar, you find that March 20 is day 79 of the year, so you want day 292, or Oct 19.
Do similar work for D=15. Or, considering the symmetric nature of sin(x), you can get the date another way.
The amplitude is the maximum deviation above or below that average.
Recall that the period of sin(x) is 2pi
for any value of x, sin(x+2pi) = sin(x)