## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.4 Area and Arc Length in Polar - Exercises - Page 625: 32

#### Answer

Length $\simeq 3.235$

#### Work Step by Step

We have $r = f\left( \theta \right) = 1 + \theta$, ${\ \ }$ $f'\left( \theta \right) = 1$ Using Eq. (7) the length of the curve for $0 \le \theta \le \frac{\pi }{2}$ is $s = \mathop \smallint \limits_0^{\pi /2} \sqrt {{{\left( {1 + \theta } \right)}^2} + 1} {\rm{d}}\theta$ Let $u = 1 + \theta$. So, $du = d\theta$. The integral becomes $s = \mathop \smallint \limits_1^{1 + \pi /2} \sqrt {1 + {u^2}} {\rm{d}}u$ Let $u = \tan v$. So, $du = {\sec ^2}vdv$. The integral becomes $s = \mathop \smallint \limits_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} {\sec ^2}v\sqrt {1 + {{\tan }^2}v} {\rm{d}}v$ Since ${\sec ^2}v - {\tan ^2}v = 1$, we get $s = \mathop \smallint \limits_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} {\sec ^3}v{\rm{d}}v$ 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, $s = \frac{{\tan v\sec v}}{2}|_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} + \frac{1}{2}\mathop \smallint \limits_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} \sec v{\rm{d}}v$ From Eq. 13 of Section 8.2 (page 402) we have $\smallint \sec x{\rm{d}}x = \ln \left| {\sec x + \tan x} \right| + C$ So, $s = \frac{{\tan v\sec v}}{2}|_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} + \frac{1}{2}\ln \left| {\sec v + \tan v} \right||_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)}$ $s \simeq 3.235$

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