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 - Page 238: 86

Answer

$$s\left( t \right) = 2\sin t$$

Work Step by Step

$$\eqalign{ & v\left( t \right) = 2\cos t;{\text{ }}s\left( 0 \right) = 0 \cr & {\text{Calculating the position function}} \cr & s\left( t \right) = \int {v\left( t \right)} dt \cr & s\left( t \right) = \int {\left( {2\cos t} \right)} dt \cr & s\left( t \right) = 2\sin t + C \cr & {\text{Calculating the function for the initial position }}s\left( 0 \right) = 0 \cr & 0 = 2\sin \left( 0 \right) + C \cr & C = 0 \cr & {\text{Thenerefore,}} \cr & s\left( t \right) = 2\sin t \cr & {\text{Graphing the velocity and position functions}} \cr} $$
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