In a certain city number of hours of sunlight on summer sols
Solution
solution:
according to calender
June 21 is the 172nd day of the year, and December 21 (the winter solstice) is the 355th day of the year
if we assume 1 jan isfirst day of the year
now the time period should be 365 days
if i assume function
y=sin(B(t-C))+D
Since the range of t is from 1 to 365, and the range of the input should be 0 to 2*pi
B(365 - C) - B(1 - C) = 2*pi - 0
B * (365 - C - 1 + C) = 2*pi
B * 364 = 2*pi
B = 2*pi / 364
now
summersolistic means maximum daylight means 15.848 maxima at x=pi/2
(2*pi / 364) * (172 - C) = pi / 2
172 - C = 364 * pi / (2 * 2*pi)
172 - C = 364 * pi / (4*pi)
172 - C = 364 / 4
172 - C = 91
C = 172 - 91
C = 81
Now B and C are calculated now calculating A and D by putting maximum daylight at x=pi/2
and minimum daylight at x=-pi/2
15.848= A sin [2*pi / 364 * (172 - 81)] + D
15.848 = A sin [2*pi / 364 * 91] + D
15.848 = A sin [pi / 2] + D
15.848 = A * 1 + D
15.848 = A + D ..................(i)
and at minima
8.454 = A sin [2*pi / 364 * (355 - 81)] + D
8.454 = A sin [1.505 * pi] + D
now
sin [1.505 * pi] is approximately -1
8.454 = D-A ............(ii)
adding (i) and (ii)
D=12.151
A=3.697
now our sine function is
y=3.697sin[2*pi / 364 * (t - 81)]+12.151
b) now at t=91
y=3.697sin[2*pi / 364 * (91 - 81)]+12.151
y=3.697sin [0.1726]+12.151
y=3.697*0.1718+12.151
y=12.7861hours

