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: 3


a. 1 b. 1

Work Step by Step

a. Using chain rule: $\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dt}=(\dfrac{-2}{z})\sin t cos t +(\dfrac{2}{z})\sin t cos t+(\dfrac{x+y}{z^2 t^2})$ and, $\dfrac{(\cos^2 t+\sin^2 t)}{(\dfrac{1}{t^2})(t^2)}=1$ Now, by using direct differentiation. we have $w=(\dfrac{x}{z})+(\dfrac{y}{z})=\dfrac{\cos^2 t}{(1/t)}+\dfrac{\sin^2 t}{(1/t)}t$ From above part, we get $\dfrac{dw}{dt}=1$ b. Thus, $\dfrac{dw}{dt}(3)=1$
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