Calculus: Early Transcendentals 8th Edition

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

Chapter 14 - Section 14.6 - Directional Derivatives and the Gradient Vector - 14.6 Exercise - Page 957: 9

Answer

a)$\nabla f=<2xyz-yz^3, x^2z-xz^3,x^2y-3xyz^2>$ b)$\nabla f_{(2, -1, 1)}=<-3, 2, 2>$ c)$D_uf_{(2, -1, 1)}=\frac{2}{5}$

Work Step by Step

$\nabla f=<2xyz-yz^3, x^2z-xz^3,x^2y-3xyz^2>$ $\nabla f_{(2, -1, 1)}=<-3, 2, 2>$ $D_uf_{(2, -1, 1)}=\nabla f_{(2, -1, 1)}\cdot u=<-3, 2, 2>\cdot<0, \frac{4}{5}, -\frac{3}{5}>=\frac{2}{5}$
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