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: 43



Work Step by Step

As we are given that $x=t^2, y=t^3 $ This yields, $x'=2t, y'=3t^2 $ Also, $x''=2, y''=6t $ From previous solution, we have $\kappa=\dfrac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{[\dot{x}^2+\dot{y}^2]^{3/2}}$ Thus, $\kappa=\dfrac{|(2t)(6t)-(3t^2 \cdot 2)}{[(2t)^2+9t^2]^{3/2}}$ or, $\kappa=\dfrac{|12t^2-6t^2|}{[4t^2+9t^4]^{3/2}}$ Hence, it has been proved. $\kappa=\dfrac{6t^2}{[4t^2+9t^4]^{3/2}}$
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