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 817: 31


$-1$ and $-1$

Work Step by Step

Consider $F(x,y,z)=\sin (x+y)+\sin (y+z)+\sin (x+z)$ and $\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ $\implies \dfrac{\partial z}{ \partial x}=-[\dfrac{\cos (x+y)+\cos (x+z)}{\cos (y+z)+\cos (x+z)}]$ For the point $(\pi,\pi,\pi)$, we have $\dfrac{\partial z}{ \partial x}=-[\dfrac{\cos (\pi+\pi)+\cos (\pi+\pi)}{\cos (\pi+\pi)+\cos (\pi+\pi)}]$ or, $=-1$ Now, consider $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ . $\implies \dfrac{\partial z}{ \partial y}=-[\dfrac{\cos (x+y)+\cos (y+z)}{\cos (y+z)+\cos (x+z)}]$ For the point $(\pi,\pi,\pi)$, we have: $\dfrac{\partial z}{ \partial y}(\pi,\pi,\pi)=-[\dfrac{\cos (\pi+\pi)+\cos (\pi+\pi)}{\cos (\pi+\pi)+\cos (\pi+\pi)}]=-1$
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