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: 45

Answer

$0.0024$

Work Step by Step

$$f_x=xz-3yz+2 \\ f_x(1,1,2)=(1)(2)-3(1)(2)+2=-2 \\ f_y(x,y) =-3z \\ f_{y}(1,1,2) =-6 \\ f_z=x-3y \\ f_z(1,1,2)=1-3(1)=-2 \\ f_{xx}(x,y)=0\\ f_{yy}(x,y)=0 \\ f_{xy}(x,y) =0 \\ f_{yz}=-3$$ Error: $|E(x,y,z)| \leq \dfrac{1}{2} \times (3) [ |x-1| +|y-1|+|z-2|)^2$ or, $$E \leq \dfrac{3}{2} \times (0.01+0.01+0.02)^2 =0.0024$$
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