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: 33

Answer

$$0.06$$

Work Step by Step

$f_x= 2x-3y $ and $f_x(2,1)=1; \\ f_y(x,y) =-3x $ and $f_{y}(2,1) =-3 (2)=-6$ Now $f_{xx}(2,1)=2;\\ f_{yy}(x,y)=0 \\ f_{xy}(x,y) =-3$ So, $$L(x,y)=3+1(x-2) -6(y-1) =7+x-6y$$ Now, $|E(x,y)| \leq \dfrac{1}{2} \times 3 [ |x-2| +|y-1|)^2 $ or, $E \leq \dfrac{3}{2} \times (0.1+0.1)^2 =0.06$
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