Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 658: 41


(a) $\pi$ (b) $\pi$

Work Step by Step

(a) Here, we have $L=\int_{0}^{\pi/2}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt=\int_{0}^{\pi/2} \sqrt{(-2 \sin 2t)^2+(2 \cos 2t)^2}dt$ $\implies L=\int_{0}^{\pi/2} \sqrt{4}dt=[2t]_{0}^{(\pi/2)}=\pi$ (b) Here, we have $L=\int_{-1/2}^{1/2}\sqrt{(\pi \cos \pi t)^2+(-\pi \sin \pi t)^2}dt$ This implies that $L=(\pi) \int_{-1/2}^{1/2} dt$ Therefore, $L=[\pi t]_{-1/2}^{1/2}=\pi$
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