can anyone help

Mr. red · #1 $Old$ June 25th, 2008, 06:11 PM

Many calendars number the days of the year. January 1st gets number one, january 2nd gets number 2, january 31st gets 31, february 1st gets 32, december 31st gets 365 in a non-leap year, etc. mary ann noticed that on the northern hemisphere island where she lived there was 14 hours, 42 minutes of sunlight on the longest day of the year and 9 hours, 18 minutes on the shortest. created a sin function or cosine function that would allow someone to input the day number and get the hours of sunlight. explain your reasoning for choosing the parameters that you have.

thelostchild · #2 $Old$ June 26th, 2008, 02:31 AM

NB this is my way of approaching the problem, it may not necessarily be correct so dont take my answer to be perfect

ok we need a sine function the most general form of this possible is
$y=A \sin(\omega x + \alpha) + B$

first of all it would need a period of 365 days (assuming we havn't got a leap year) therefore as
$\omega = \frac{2 \pi}{T}=\frac{2 \pi}{365}$

looking it up on wikipedia the longest day of the year is on 21st June, using your numbering system this would be day 172

also the shortest day is on 21st December, so that would be day 355.

we know the function must be a maximum on day 172, so therefore

$\sin(\frac{2 \pi}{365} \times 172 + \alpha)=1$

solving this for alpha gives us $\alpha = -\frac{323}{730} \pi$

ok were half way there!

the longest day of the year is 14 hours 42 mins = 14.7 hours

therefore when the function is a maximum y=14.7

sine has is max value of 1 so
$14.7=A+B$

the shortest day of the year is 9 hours 18 mins=9.3 hours

sine has a min value of -1 so
$9.3=-A+B$

solving the two equations gives us A=2.7 and B=12

so the final equation is

$y=2.7 \sin (\frac{2 \pi}{365}x-\frac{323 \pi}{730})+12$