University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 573: 41


a) $\pi$ and b) $\pi$

Work Step by Step

a) Since, $L=\int_{0}^{\pi/2}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $L=\int_{0}^{\pi/2} \sqrt{(-2 \sin 2t)^2+(2 \cos 2t)^2}dt=\int_{0}^{\pi/2} \sqrt{4}dt$ or, $L=[2t]_{0}^{\pi/2}=\pi$ b) a) Since, $L=\int_{-1/2}^{1/2}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $L=\int_{-1/2}^{1/2}\sqrt{(\pi \cos \pi t)^2+(-\pi \sin \pi t)^2}dt=\int_{-1/2}^{1/2} \pi dt$ or, $L=[\pi t]_{-1/2}^{1/2}=\pi$ Hence, a) $\pi$ and b) $\pi$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.