Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.3 - New Functions from Old Functions - 1.3 Exercises - Page 43: 25

Answer

$L(t) = 12+2~sin[\frac{2\pi}{365}(t-80)]$

Work Step by Step

We can see that the average number of hours is 12 hours and the amplitude of the graph is 2 hours. We can write a function for the number of hours of sunlight: $L(t) = 12+2~sin[\frac{2\pi}{365}(t-80)]$ We can check the model using March 31st, which is day 90 of the year: $L(t) = 12+2~sin[\frac{2\pi}{365}(t-80)]$ $L = 12+2~sin[\frac{2\pi}{365}(90-80)]$ $L = 12.34~hours$ The data shows that the amount of sunlight on this day is $6:18~pm - 5:51~am$ which is $12~h~27~min$. Note that this is $12.45~hours$
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