Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.5 - The Chain Rule. - 14.5 Exercise - Page 945: 52

Answer

a) $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) (\cos \theta) +( \dfrac{\partial z}{\partial y}) (\sin \theta)$ b) $\dfrac{\partial z }{\partial \theta}=(r \cos \theta) (\dfrac{\partial z}{\partial y})-(r \sin \theta) (\dfrac{\partial z}{\partial x})$

Work Step by Step

a) Here, we have $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) (\dfrac{dx}{dr}) +( \dfrac{\partial z}{\partial y}) (\dfrac{dy}{dr})$ Plug in the given data, then we have $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) (\cos \theta) +( \dfrac{\partial z}{\partial y}) (\sin \theta)$ b) Here, we have $\dfrac{\partial z }{\partial \theta}=( \dfrac{\partial z}{\partial x}) (\dfrac{dx}{d\theta}) +( \dfrac{\partial z}{\partial y}) (\dfrac{dy}{d\theta})$ Plug in the given data, then we have $\dfrac{\partial z }{\partial \theta}=(r \cos \theta) (\dfrac{\partial z}{\partial y})-(r \sin \theta) (\dfrac{\partial z}{\partial x})$ Hence, we have: a) $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) (\cos \theta) +( \dfrac{\partial z}{\partial y}) (\sin \theta)$ b) $\dfrac{\partial z }{\partial \theta}=(r \cos \theta) (\dfrac{\partial z}{\partial y})-(r \sin \theta) (\dfrac{\partial z}{\partial x})$
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