Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 656: 47

Answer

$$ x=\sin t + \sin1.5 t, \quad y=\cos t, \quad 0 \leq t \leq 4 \pi $$the length of the curve is $$ \begin{aligned} L &=\int_{0}^{4\pi} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\ & \approx 16.7102 \end{aligned} $$

Work Step by Step

$$ x=\sin t + \sin1.5 t, \quad y=\cos t, \quad 0 \leq t \leq 4 \pi $$ then $$ d x / d t=\cos t + 1.5 \cos 1.5 t , \quad d y / d t=- \sin t $$ so $$ (d x / d t)^{2}+(d y / d t)^{2}=\cos ^{2} t+3 \cos t \cos 1.5 t+ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad + 2.25 \cos ^{2} 1.5 t+\sin ^{2} t $$ Thus the length of the curve is $$ \begin{aligned} L &=\int_{0}^{4\pi} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\ &=\int_{0}^{4 \pi} \sqrt{1+3 \cos t \cos 1.5 t+2.25 \cos ^{2} 1.5 t} d t \\ & \approx 16.7102 \end{aligned} $$
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