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: 44


$\dfrac{ab\omega}{[a^2 \sin^2 \omega t+b^2 cos^2 \omega t]^{3/2}}$

Work Step by Step

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