University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 273: 101


$v=\dfrac{1}{2 \cos t} +\dfrac{1}{2}$

Work Step by Step

We need to find the anti-derivative for $\dfrac{dv}{dt}= \dfrac{1}{2} \sec t \tan t $ Thus, we have $v=\dfrac{1}{2 \cos t} +C$ Apply the initial condition $v(0)=1$ in the above equation to solve for $C$. we get $\dfrac{1}{2}+C=1 \implies C=\dfrac{1}{2}$ Hence, $v=\dfrac{1}{2 \cos t} +\dfrac{1}{2}$
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