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: 37


total length $\simeq 6.68$

Work Step by Step

The three-petal rose in Figure 20 has polar equation $r = \cos 3\theta $. So, we have $r = f\left( \theta \right) = \cos 3\theta $, ${\ \ }$ $f'\left( \theta \right) = - 3\sin 3\theta $. When $\theta=0$, we have $r=1$. So, the curve starts at $\left( {1,0} \right)$. When $\theta = \frac{\pi }{6}$, we have $r=0$. So, the curve reaches the origin when $\theta = \frac{\pi }{6}$. From Figure 20, we see that the curve completed the upper-half of the petal for the interval $0 \le \theta \le \frac{\pi }{6}$. By symmetry, the entire curve is in the interval $0 \le \theta \le \pi $. Using Eq. (7) the total length of the curve is $s = \mathop \smallint \limits_0^\pi \sqrt {{{\cos }^2}3\theta + {{\left( { - 3\sin 3\theta } \right)}^2}} {\rm{d}}\theta $ $s = \mathop \smallint \limits_0^\pi \sqrt {{{\cos }^2}3\theta + 9{{\sin }^2}3\theta } {\rm{d}}\theta $ Evaluating it using a computer algebra system we obtain the result: $s \simeq 6.68$
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