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 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 25


a) $1$ b) $2x+2y-1$

Work Step by Step

a. As we know that $L(x,y)=f(x,y)+f_x(x-a)+f_y(y-b)$ Then $f_x=2x$ and $f_x(0,0)=0$; and $f_y=2y; f_y(0,0)=0$ Also, $f(0,0)=1$ Thus, $L(0,0)=1+0(x-0)+0(y-0)$ or, $L(0,0)=1$ b. As we know that $L(x,y)=f(x,y)+f_x(x-a)+f_y(y-b)$ Here, we have $f_x=2x$ and $f_x(1,1)=2;f_y=2y ;f_y(1,1)=2;f(1,1)=3$ Thus, $L(1,1)=3+2(x-1)+2(y-1)$ or, $L(1,1)=2x+2y-1$
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