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 $noon$:
$D'(t) = -2.515~sin[0.503(t-6.75)]$
$D'(t) = -2.515~sin[0.503(12-6.75)]$
$D'(t) = -1.21~m/h$
