Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.4 Area and Arc Length in Polar - Exercises - Page 625: 24

Answer

Length using integral in polar coordinates: $s = \tan A$ Length using trigonometry: $\overline {PQ} = \tan A$

Work Step by Step

Write $r = \sec \theta = \frac{1}{{\cos \theta }}$. It follows that $r\cos \theta = 1$. Since $x = r\cos \theta $. It implies that the polar equation $r = \sec \theta $ is a line $x=1$ in rectangular coordinates. From the figure attached, we see that the distance from the origin to a point in the line makes an angle $\theta$ with the $x$-axis. We have $r = f\left( \theta \right) = \sec \theta $, ${\ \ }$ $f'\left( \theta \right) = \sec \theta \tan \theta $ Using Eq.(7), we get the arc length for $0 \le \theta \le A$ $s = \mathop \smallint \limits_0^A \sqrt {{{\sec }^2}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } {\rm{d}}\theta $ $s = \mathop \smallint \limits_0^A \sqrt {{{\sec }^2}\theta \left( {1 + {{\tan }^2}\theta } \right)} {\rm{d}}\theta $ Since ${\sec ^2}x - {\tan ^2}x = 1$, the integral becomes $s = \mathop \smallint \limits_0^A \sqrt {{{\sec }^4}\theta } {\rm{d}}\theta = \mathop \smallint \limits_0^A {\sec ^2}\theta {\rm{d}}\theta $ From Eq. 14 of Section 8.2 (page 402) we have $\smallint {\sec ^m}x{\rm{d}}x = \frac{{\tan x{{\sec }^{m - 2}}x}}{{m - 1}} + \frac{{m - 2}}{{m - 1}}\smallint {\sec ^{m - 2}}{\rm{d}}x$ So, $s = \tan \theta |_0^A = \tan A$ Using trigonometry, we have the length segment from the points $P$ to $Q$ corresponding to $0 \le \theta \le A$ as $\tan A = \frac{{\overline {PQ} }}{{\overline {OP} }}$ $\overline {PQ} = \tan A$ Thus, the two results agree.
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