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: 52

Answer

the curve: $$ r=\tan \theta $$ is completely traced with $$ \theta =\frac{\pi}{6} \quad \text {to} \quad \theta= \frac{\pi}{3} $$ the length of the given curve is equal to $$ \begin{aligned} L &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3} } \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\ \\ & \approx 1.2789 \end{aligned} $$
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Work Step by Step

the curve: $$ r=\tan \theta $$ is completely traced with $$ \theta =\frac{\pi}{6} \quad \text {to} \quad \theta= \frac{\pi}{3} $$ the length of the given curve is equal to $$ \begin{aligned} L &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3} } \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\ \\ & =\int_{\pi / 6}^{\pi / 3} \sqrt{\tan ^{2} \theta+\sec ^{4} \theta} d \theta \\ & \approx 1.2789 \end{aligned} $$
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