In the mysterious lost city of Mim, the length of daylight (in hours) on the tth day of the year is modeled by the function      L(t)=12+2sin[((2pi)/(365))*(t-80)] Use this model to compare how...

In the mysterious lost city of Mim, the length of daylight (in hours) on the tth day of the year is modeled by the function

 

 

 

L(t)=12+2sin[((2pi)/(365))*(t-80)]

Use this model to compare how the number of hours of daylight is increasing on March 13 and June 3 (assume that this is a standard year, not a leap year).

a) Rate of increase on March 13=

b) Rate of increase on June 3 =

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to notice that the given function models the length of daylight, hence you need to differentiate the given function  with respect to t to find the change in length of daylight at different days of year.

`L'(t)=(12+2sin((2pi)/(365))*(t-80))'`

`L'(t)=2cos[((2pi)/(365))*(t-80)]*[((2pi)/(365))*(t-80)]'`

`L'(t)=2cos[((2pi)/(365))*(t-80)]*[((2pi)/(365))]`

You need to decide how many days have gone till March 13, hence 13 March is the 31+28+13 = 72nd day of the year.

You need to evaluate L'(72) such that:

`L'(72)=2cos[((2pi)/(365))*(72-80)]*[((2pi)/(365))]`

`L'(72)=[((4pi)/(365))]cos[((-16pi)/(365))]`

Since the cosine function is even, then `2cos[((-16pi)/(365))]= 2cos[((16pi)/(365))]`

You need to decide how many days have gone till June 3 , hence 13 March is the 31+28+31+30+31+3 = 154th day of the year.

You need to evaluate L'(154) such that:

`L'(154)=[((4pi)/(365))]2cos[((148pi)/(365))]`

Hence, evaluating the rate of increase of length of the day till March 13 and June 3 yields `L'(72)=[((4pi)/(365))]cos[((-16pi)/(365))] ` and  `L'(154)=[((4pi)/(365))]2cos[((148pi)/(365))].`

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