University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.5 - Directional Derivatives and Gradient Vectors - Exercises - Page 720: 15

Answer

$3$

Work Step by Step

Our aim is to take the first partial derivatives of the given function $f(x,y,z)$ with respect to $x$, by treating $y$ and $z$ as a constant, and vice versa: $f_x(1,-1,2)=y+z=1$ and $ f_y(1,-1,2)=x+z=3$ and $f_z(1,-1,2)=x+y=0$ Now, the gradient equation is: $\nabla f (1,-1,2)= \lt 1,3,0 \gt$ The directional derivative at that direction is $D_v f=\lt 1,3,0 \gt \times \dfrac{\lt 3,6, -2 \gt }{\sqrt {(3)^2+(6)^2+(-2)^2}} \gt=\lt 1,3,0 \gt \times=\lt \dfrac{3}{7}, \dfrac{6}{7}, \dfrac{-2}{7} \gt=3$
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