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: 33



Work Step by Step

Since, $D_u f = \nabla f \cdot u$ $ f_x= 2x-3y \\ f_y=-3x+8y $ $\nabla f =(2x-3y) \ i +(-3x+8y) \ j $ $\implies \nabla f (1,2)=-4i+13j$ and $|\nabla f (1,2)|=\sqrt {(-4)^2 +(13)^2}=\sqrt {16+169}=\sqrt {185}$ The maximum rate of change in function $f$ is less than $14$; this means that there will be no direction equal to $14$.
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