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: 45

Answer

$$ x=e^{t} \cos t \quad y=e^{t} \sin t, \quad 0 \leq t \leq \pi $$ The exact length of the curve is $$ \begin{aligned} L &=\int_{0}^{\pi} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\ &=\sqrt 2 (e^{ \pi}-1) \\ \end{aligned} $$
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Work Step by Step

$$ x=e^{t} \cos t \quad y=e^{t} \sin t, \quad 0 \leq t \leq \pi $$ then $$ d x / d t=e^{t}(\cos t-\sin t) , \quad d y / d t=e^{t}(\sin t+\cos t) $$ so $$ \begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} & =\left[e^{t}(\cos t-\sin t)\right]^{2}+\left[e^{t}(\sin t+\cos t)\right]^{2} \\ &=\left(e^{t}\right)^{2}\left(\cos ^{2} t-2 \cos t \sin t+\sin ^{2} t\right) \\ & \quad\quad +\left(e^{t}\right)^{2}\left(\sin ^{2} t+2 \sin t \cos t+\cos ^{2} t\right.\\ &=e^{2 t}\left(2 \cos ^{2} t+2 \sin ^{2} t\right) \\ &=2 e^{2 t} \end{aligned} $$ Thus the exact length of the curve is $$ \begin{aligned} L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\ &=\int_{0}^{\pi} \sqrt{2 e^{2 t} } d t \\ & =\sqrt 2 \int_{0}^{\pi} e^{ t} d t \\ &=\sqrt 2 [ e^{ t}]_{0}^{\pi} \\ &=\sqrt 2 (e^{ \pi}-1) \\ \end{aligned} $$
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