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: 32

Answer

$\triangle z=-0.7189$ and $dz=-0.73$

Work Step by Step

Write the differential form $dz=\dfrac{\partial z}{\partial x} dx + \dfrac{\partial z}{\partial y} dy + \dfrac{\partial z}{\partial z} dz$ The partial derivatives w.r.t $t$ and $x$ are: $z_x=2x-y$ and $z_y=6y-x$ At the given points, we have $z_x(3,-1)=7$ and $z_y(3,-1)=-9$ Here, $\triangle x=2.96-3=-0.04\\ \triangle y=-0.95-(-1)=0.05$ Further, $dz=\dfrac{\partial z}{\partial x} dx + \dfrac{\partial z}{\partial y} dy + \dfrac{\partial z}{\partial z} dz=7 \times (-0.04) +(-9) \times (0.05)=-0.73$ and $\triangle z=f(3,-1)-f(2.96,-0.95)=14.2811-15=-0.7189$
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