Thomas' Calculus 13th Edition

Published by Pearson

Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 26

Answer

a) $4x+4y+4$ b) $10x+10y-5$

Work Step by Step

a. As we know that $L(x,y)=f(x,y)+f_x(x-m)+f_y(y-n)$ Here, we have $f_x=2x+2y+2z$ and $f_x(0,0)=4$; Also, $f_y=2x+2y+2z$ and $f_y(0,0)=4; f(0,0)=4$ Thus, $L(0,0)=4+4(x-0)+4(y-0) \\\implies L(0,0)=4x+4y+4$ b. As we know that $L(x,y)=f(x,y)+f_x(x-m)+f_y(y-n)$ Here, we have $f_x=2x+2y+2z$ and $f_x(1,2)=10, f_y=2x+2y+2z ; f_y(1,2)=10$ and $f(1,2)=25$ Thus, $L(1,2)=25+10(x-1)+10(y-1)\\\implies L(1,2)=10x+10y-5$

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