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 835: 34


$$ 0.02$$

Work Step by Step

$f_x= x+y+3 $ and $f_x(2,2)=2+2+3=7 \\f_y(x,y) =x+\dfrac{y}{2}-3 \\f_{y}(2,2) =2+\dfrac{2}{2}-3=0 \\f_{xx}(x,y)=1\\ f_{yy}(x,y)=\dfrac{1}{2} \\f_{xy}(x,y) =1$ Now, $L(x,y)=3+1(x-2) -6(y-1) =7+x-6y$ $|E(x,y)| \leq \dfrac{1}{2} [ |x-2| +|y-2|)^2$ or, $ E \leq \dfrac{1}{2} (0.1+0.1)^2 = 0.02$
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