Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises - Page 975: 31


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

Work Step by Step

Write the differential form such as follows: $dz=(10x) dx+(2y) dy$ $\triangle x=1.05-1=0.05;\\ \triangle y=2.1-2=0.1$ Now, we have $dz=10 \times (1) (0.05)+2 \times (2) \times (0.1)=0.9$ Thus, we get $\triangle z=f(1.05,2.1)-f(1, 2)=5x^2+y^2=5(1.05)^2+(2.1)^2-(5+4)$ and $\triangle z \approx 0.9225$ Hence, $\triangle z=0.9225$ and $dz=0.9$
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