## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 835: 39

#### Answer

(a) $$L=2x+2y+2z-3$$ b) $$L(x,y,z)=y+z$$ c) $$L(x,y,z)=0$$

#### Work Step by Step

(a) $f_x=y+z$ and $f_x(1,1,1)=1+1=2$ $$f_y(x,y) =x+z \\ f_{y}(1,1,1) =1+1=2 \\ L(x,y,z)=3+2(x-1)+2(y-1)+2(z-1) =2x+2y+2z-3$$ b) $$f(1,0,0)=0 \\ f_{x}(1,0,0)=0 \\ f_{xy}(x,y) =0$$ So, $L(x,y,z)=0+0(x-1)+(y-0)+(z-0) =y+z$ c) $$f(0,0,0)=0; f_{x}(1,0,0)=0 \\ f_{y}(0,0,0) =0 \\ f_{z}(0,0,0) =0 \\ L(x,y,z)=0$$

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