Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 70

Answer

$$h\left( t \right) = - 2\cos 3t + 6$$

Work Step by Step

$$\eqalign{ & h'\left( t \right) = 6\sin 3t \cr & h\left( t \right) = \int {h'\left( t \right)} dt \cr & then \cr & h\left( t \right) = \int {6\sin 3t} dt \cr & find{\text{ the general solution}} \cr & h\left( t \right) = - 2\cos 3t + C \cr & {\text{using the initial condition }}h\left( {\pi /6} \right) = 6 \cr & 6 = - 2\cos 3\left( {\frac{\pi }{6}} \right) + C \cr & 6 = - 2\cos \left( {\frac{\pi }{2}} \right) + C \cr & C = 6 \cr & {\text{the solution to the initial value problem is}} \cr & h\left( t \right) = - 2\cos 3t + 6 \cr} $$
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