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: 100


$r=\dfrac{1}{\pi} \sin \pi \theta +1$

Work Step by Step

We need to find the anti-derivative for $\dfrac{dr}{d \theta}= \cos \pi \theta $ Thus, we have $r=\dfrac{1}{\pi} \sin \pi \theta +C$ Apply the initial condition $r(0)=1$ in the above equation to solve for $C$. we get $0+C=1 \implies C=1$ Hence, $r=\dfrac{1}{\pi} \sin \pi \theta +1$
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