Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.4 - Tangent Planes and Linear Approximation - 14.4 Exercise - Page 935: 31


$\triangle z=0.9225$ and $dz=0.9$

Work Step by Step

We are given that $z=5x^2+y^2$ The differential form of the given function can be evaluated as follows: $dz=10x dx+2y dy$ Here, $\triangle x=1.05-1=0.05$ and $\triangle y=2.1-2=0.1$ Now, we have $dz=10 \cdot (1) (0.05)+2 \cdot (2) (0.1)$ or, $dz=0.9$ Thus, $\triangle z=f(1.05,2.1)-f(1, 2)$ or, $=5(1.05)^2+(2.1)^2-(5+4)$ Thus, $\triangle z=0.9225$
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