## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.5 - Directional Derivatives and Gradient Vectors - Exercises 14.5 - Page 826: 16

#### Answer

$$0$$

#### Work Step by Step

Now, $f_x=2x$ and $f_y=4y$ and $f_z=-6z$ and $\nabla f (1,0,1) =\lt 2,4, -6 \gt$ Now, $v=\dfrac{u}{|u|}=\dfrac{\lt 1,1,1 \gt}{\sqrt {3}}=\lt \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3} \gt$ Since, Directional derivative: $D_v \ h = \nabla h \ \cdot v$ or, $=\lt 2,4,-6 \gt \cdot \lt \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3} \gt$ or, $=0$

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