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A model for the length of daylight (in hours) in Philadelphia on the t th day of the year is L(t)= 12+2.8 sin[2pi/365(t-80)] Us...Asked by ken
A model for the length of daylight (in hours) in Philadelphia on the tth day of the year is
L(t) = 12 + 2.8 sin 2π/365(t − 80)
.
Use this model to compare how the number of hours of daylight is increasing in Philadelphia on April 21 and June 5. (Assume there are 365 days in a year. Round your answers to four decimal places.)
April 21 L'(t) =
June 5 L'(t) =
L(t) = 12 + 2.8 sin 2π/365(t − 80)
.
Use this model to compare how the number of hours of daylight is increasing in Philadelphia on April 21 and June 5. (Assume there are 365 days in a year. Round your answers to four decimal places.)
April 21 L'(t) =
June 5 L'(t) =
Answers
Answered by
ken
I took the first derivative and plug in April 21 = 111, and June 5 = 156 in t. got 0.0482 but its not correct
Answered by
Reiny
Assuming April 21 ----> 111
L(111) = 12 + 2.8 sin 2π/365(111 − 80)
= 12 + 2.8 sin 2π/365(31)
= 13.424...
did you set your calculator to radians?
But whatever setting, you should have known that 0.0482 was wrong
since sin(anything) lies between -1 and +1
then 2.8 sin 2π/365(t − 80)
must lie between -2.8 and 2.8
and when 12 is added to that, the only answers possible are those between 9.2 and 14.8
L(111) = 12 + 2.8 sin 2π/365(111 − 80)
= 12 + 2.8 sin 2π/365(31)
= 13.424...
did you set your calculator to radians?
But whatever setting, you should have known that 0.0482 was wrong
since sin(anything) lies between -1 and +1
then 2.8 sin 2π/365(t − 80)
must lie between -2.8 and 2.8
and when 12 is added to that, the only answers possible are those between 9.2 and 14.8