Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.3 Arc Length and Curvature - 13.3 Exercises - Page 909: 45


$\dfrac{1}{(\sqrt 2e^t)}$

Work Step by Step

As we are given that $x=e^t \cos t, y=e^t \sin t $ This yields, $x'=e^t \cos t-e^t \sin t, y'=e^t \sin t+e^t \cos t$ Also, $x''=-2e^t \sin t, y''=2e^t \cos t$ $\kappa=\dfrac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{[\dot{x}^2+\dot{y}^2]^{3/2}}$ Thus, $\kappa=\dfrac{|(e^t \cos t-e^t \sin t)(2e^t \cos t)-(e^t \sin t+e^t \cos t)(-2e^t \sin t)|}{[(e^t \cos t-e^t \sin t)^2+(e^t \sin t+e^t \cos t)^2]^{3/2}}$ or, $\kappa=\dfrac{|2e^t (\cos^2 t+sin^2t)|}{[(2e^t (\cos^2 t+sin^2t)]^{3/2}}$ or, $\kappa=\dfrac{|2e^t (1)|}{[2e^t (1)]^{3/2}}$ Hence, the result. $\kappa=\dfrac{1}{(\sqrt 2e^t)}$
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