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: 34



Work Step by Step

Since, $D_u f = \nabla f \cdot u$ $T_x= 2y \\ T_y=2x-z \\ T_z=-y$ $\nabla T(1,-1, 1) =-2i+j +k $ and $|\nabla T(1,-1, 1)|=\sqrt {(2)^2 +(1)^2+(1)^2}=\sqrt {6}$ or, $|\nabla T(1,-1, 1)|=\sqrt {6}$ This means that the minimum rate of change in temperature $T$ is $-\sqrt 6 \gt -3$ .
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