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In the mysterious lost city of Mim, the length of daylight (in hours) on the tth day of...

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soccerfan55 | Student | Honors

Posted March 21, 2012 at 1:35 PM via web

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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

Posted March 22, 2012 at 1:46 AM (Answer #1)

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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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