Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises - Page 236: 24

Answer

(a) $D'(t) = 2.39~m/h$ (b) $D'(t) = 0.93~m/h$ (c) $D'(t) = -2.28~m/h$ (d) $D'(t) = -1.21~m/h$

Work Step by Step

$D(t) = 7+5~cos[0.503(t-6.75)]$ $D'(t) = (-5)(0.503)~sin[0.503(t-6.75)]$ $D'(t) = -2.515~sin[0.503(t-6.75)]$ (a) We can find the rate of change of the tide at $3:00~am$: $D'(t) = -2.515~sin[0.503(t-6.75)]$ $D'(t) = -2.515~sin[0.503(3-6.75)]$ $D'(t) = 2.39~m/h$ (b) We can find the rate of change of the tide at $6:00~am$: $D'(t) = -2.515~sin[0.503(t-6.75)]$ $D'(t) = -2.515~sin[0.503(6-6.75)]$ $D'(t) = 0.93~m/h$ (c) We can find the rate of change of the tide at $9:00~am$: $D'(t) = -2.515~sin[0.503(t-6.75)]$ $D'(t) = -2.515~sin[0.503(9-6.75)]$ $D'(t) = -2.28~m/h$ (d) We can find the rate of change of the tide at $noon$: $D'(t) = -2.515~sin[0.503(t-6.75)]$ $D'(t) = -2.515~sin[0.503(12-6.75)]$ $D'(t) = -1.21~m/h$
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