## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 58

#### Answer

(a) $L(x,y,z) =1+y+z-\dfrac{\pi}{4}$ (b) $L(\dfrac{\pi}{4},\dfrac{\pi}{4},0) =\dfrac{\sqrt 2}{2}-\dfrac{\sqrt 2}{2}x +\dfrac{\sqrt 2}{2} y +\dfrac{\sqrt 2}{2} z$

#### Work Step by Step

(a) $f_x (0,0, \dfrac{\pi}{4})=-\sqrt 2 \sin x \sin (y+z) =0 \\ f_y (0,0, \dfrac{\pi}{4})=-\sqrt 2 \cos x \cos (y+z)$ and $f_x (0,0, \dfrac{\pi}{4})=1 \\ f_z(0,0, \dfrac{\pi}{4})=\sqrt 2 \cos x \cos (y+z)$ or, $f_x (0,0, \dfrac{\pi}{4})=1$ and $L(x,y,z) =1+1(y-0)+(1)(y-0)+1(z-\pi/4)\\ =1+y+z-\dfrac{\pi}{4}$ (b) $f_x(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=- \dfrac{\sqrt 2}{2}$ and $f_y(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=\dfrac{\sqrt 2}{2}$ and $f_z(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=\dfrac{\sqrt 2}{2}$ So, $L(\dfrac{\pi}{4},\dfrac{\pi}{4},0) =\dfrac{\sqrt 2}{2}-\dfrac{\sqrt 2}{2}x +\dfrac{\sqrt 2}{2} y +\dfrac{\sqrt 2}{2} z$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.