University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.4 - The Chain Rule - Exercises - Page 711: 31

Answer

$-1,-1$

Work Step by Step

Since, we have $F(x,y,z)=\sin (x+y)+\sin (y+z)+\sin (x+z)$ $\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ ...(1) Now, need to plug the derivatives in equation (1). we have $\dfrac{\partial z}{ \partial x}=-\dfrac{\cos (x+y)+\cos (x+z)}{\cos (y+z)+\cos (x+z)}$ Now, plug the point $(\pi,\pi,\pi)$ Thus, we get $\dfrac{\partial z}{ \partial x}=-\dfrac{\cos (\pi+\pi)+\cos (\pi+\pi)}{\cos (\pi+\pi)+\cos (\pi+\pi)}=-1$ Now, $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ ...(2) Plug the derivatives in equation (2). we have $\dfrac{\partial z}{ \partial y}=-\dfrac{\cos (x+y)+\cos (y+z)}{\cos (y+z)+\cos (x+z)}$ and $\dfrac{\partial z}{ \partial y}(\pi,\pi,\pi)=-\dfrac{\cos (\pi+\pi)+\cos (\pi+\pi)}{\cos (\pi+\pi)+\cos (\pi+\pi)}=-1$
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