just set it equal to 10
10 = 12+2.83sin((2?)/365(t-800)
-2 = 2.83sin((2?)/365(t-800)
-2/2.83 = sin((2?)/365(t-800)
-.706713... = sin((2?)/365(t-800)
using my calculator set to radians, taking the inverse sine of +.706713, I get a reference angle of .78484 radians
but for the sine to be negative, it must be in III or IV
first answer = ? + .78484 = 3.926435...
so ((2?)/365(t-800) = 3.926435...
t-800 = 3.926435...(365)/(2?)
= 228.09
t = appr 1028
but the period of the curve is 365, so adding or subtracting multiples of 365 will produce more answers.
possible answer in a year:
1028-2(365) = 298.1 day
second answer:
2? - .78484 = 5.498345..
((2?)/365(t-800) = 5.498345..
t-800 = 319.4
t = 1119.4
again, subtracting periods of 365
t = 24 th day
So it reaches 10 hrs of sunlight on day 24 until day 298
that is,
there is daylight for more than 10 hours for
298-24 or 274 days
I know your answer given is 34
but both of my answers work and are confirmed by Wolfram:
http://www.wolframalpha.com/input/?i=plot+y+%3D+12%2B2.83sin((2%CF%80)%2F365(x-800)
hover your mouse at y = 10 and you will see the x value at close to 20 and 300
In Philadelphia the number of hours of daylight on day t (t is the number of days after january 1) is modeled by the function
L(t)= 12+2.83sin((2*Pi)/365(t-800)
a. Which day have about 10 hours of daylight?
b. How many days of the year have more than 10 hours of day light?
the back of the book gave the answer
a. 34th day
b. 308th day
I want to know how they got this answer?
1 answer
2 answers