Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.4 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 674: 51

Answer

One loop of the curve $$ r=\cos 2\theta $$ is traced with $$ -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} $$ the length of one loop of the given curve is equal to $$ \begin{aligned} L &=\int_{-\pi / 4}^{\pi / 4} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\ & \approx 2.4221 \end{aligned} $$
1567087645

Work Step by Step

One loop of the curve $$ r=\cos 2\theta $$ is traced with $$ -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} $$ $$ \begin{aligned} r^{2}+\left(\frac{d r}{d \theta}\right)^{2}& =\cos ^{2} 2 \theta+(-2 \sin 2 \theta)^{2} \\ &=\cos ^{2} 2 \theta+4 \sin ^{2} 2 \theta \\ & =1+3 \sin ^{2} 2 \theta \end{aligned} $$ then the length of one loop of the given curve is equal to $$ \begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\ &=\int_{-\pi / 4}^{\pi / 4} \sqrt{1+3 \sin ^{2} 2 \theta} d \theta \\ & \approx 2.4221 \end{aligned} $$
Small 1567087645
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.