Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.5 - Directional Derivatives and Gradient Vectors - Exercises 14.5 - Page 826: 18



Work Step by Step

Directional derivative: $D_v h = \nabla h \cdot v$ Now, $$h_x= -y \sin xy +\dfrac{1}{x} \\ h_y= -x \sin xy +ze^{xy} \\ h_z=ye^{yz}+z^{-1}$$ So, $\nabla g (1,0,\dfrac{1}{2}) =\lt 1,\dfrac{1}{2},2 \gt$ Now, $v=\dfrac{u}{|u|}=\dfrac{\lt 1,2,2 \gt}{\sqrt {9}}=\dfrac{\lt 1,2,2 \gt}{3}=\lt \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} \gt $ Since, $D_v \ h = \nabla h \cdot v$ or, $=\lt 1,\dfrac{1}{2},2 \gt \cdot \lt \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} \gt$ or, $ =\dfrac{6}{3}$ or, $=2 $
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