Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.4 - The Chain Rule - Exercises 14.4 - Page 816: 2


a. 0 b. 0

Work Step by Step

a) Using chain rule: $\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}=2x (-\sin t+\cos t)+2y(-\sin t -\cos t)$ and $2 (\cos t+\sin t) (-\sin t+\cos t)+2 (\cos t- \sin t)(-\sin t -\cos t)=0$ b. Using direct differentiation: We have, $w^2=x^2+y^2$ or, $(\cos t+\sin t)^2+(\cos t -\sin t)^2=2$ also, $\dfrac{dw}{dt}=0$ Then $\dfrac{dw}{dt}(0)=0$
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