1 Using techniques of calculus determine which day of the ye

1. Using techniques of calculus, determine which day of the year your model predicts to have the least daylight and which day of the year is predicted to have the most daylight. Show your working, including a verification for your answers being the days of least/most sunlight using an appropriate calculus-based test.

2. Using techniques of calculus, determine the total number of hours of sunlight in a year and the average number of hours sunlight per day, as predicted by your model.

day10=8.9672h     day60= 11.12h       day120=14.2342h    day 200= 15.2164h day250= 12.9858h day 300=10.3965h day 350=8.7364h

The longest day is 15:40:27 (20th June) 15.67417 h.

The shortest day is 8:42:51 (21st December) 10457/1200 h.

Mathematical model:

h(t) = a sin (bt + c) + d

15.67417 = 3.48sin(2/366*172+C) + 12.19417     (C(172) = -1.38196)

Solution

Given mathematical model is
h(t) = a sin (bt + c) + d

15.67417 = 3.48sin(2/366*172+C) + 12.19417     (C(172) = -1.38196)
that means required model is
h(t) = a sin (bt + c) + d

f(t) = 3.48sin(2/366*t-1.38196) + 12.19417
where t is number of day in any year.
f(t) is the amount of light on a particular day.
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Now we have to use technique of calculus to determine say having least/most amount of daylight.
that is find derivative and set that =0 to find value of t.
f\'(t) = 3.48cos(2/366*t-1.38196)*(2/366)
0 = 3.48cos(2/366*t-1.38196)*(2/366)
0 = cos(2/366*t-1.38196)
we know that cos(x) is 0 at x=pi/2 or 3pi/2
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for pi/2
cos(pi/2) = cos(2/366*t-1.38196)
pi/2 =2/366*t-1.38196
t=172.000150047
t=172 means approx 20th June.
which is correct according to given description that \"The longest day is 15:40:27 (20th June) 15.67417 h.\"
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for 3pi/2
cos(3pi/2) = cos(2/366*t-1.38196)
3pi/2 =2/366*t-1.38196
t=355.000150047
t=355 means approx 21st december.
which is correct according to given description that \"The shortest day is 8:42:51 (21st December) 10457/1200 h.\"
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