University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Section 10.5 - Areas and Lengths in Polar Coordinates - Exercises - Page 585: 29



Work Step by Step

Here, we have $r=f(\theta)$ Then, $x=r \cos \theta \implies x=f(\theta) \cos \theta$ and $y=r \sin \theta \implies y=f(\theta) \sin \theta$ As we know that $L=\int_{a}^{b}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt$ or, $L=\int_{a}^{b} \sqrt{(f'(\theta) \cos \theta-f(\theta) \sin \theta)^2+(f'(\theta) \sin \theta+f(\theta) \cos \theta)^2}dt$ or, $L=\int_{a}^{b}\sqrt{(f'(\theta))^2 +f(\theta)^2}d\theta$ Thus, the length of the curve is given as: $L=\int_{a}^{b}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$
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