Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 13 - Partial Derivatives - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 969: 59

Answer

Vector along the gradient: $-\mathbf{i}+\mathbf{j}=\mathbf{V}$ Unit vector: $\frac{1}{\sqrt{2}}(-\mathbf{i}+\mathbf{j})=\mathbf{v}$ Rate of change: $\frac{1}{\sqrt{2}}=\|\nabla f(1,2,-2)\|$

Work Step by Step

$f$ increases more quickly in the positive direction of $\nabla f(x, y, z)$ \[ \begin{array}{l} \frac{1}{z} =f_{x}(x, y, z)\\ -\frac{1}{2}=f_{x}(1,2,-2) \\ -\frac{2 z}{y^{3}}=f_{y}(x, y, z) \\ \frac{1}{2} =f_{y}(1,2,-2) \\ -\frac{x}{z^{2}}+\frac{1}{y^{2}}=f_{z}(x, y, z) \\ 0=f_{z}(1,2,-2) \end{array} \] \[ \frac{1}{2}(-\mathbf{i}+\mathbf{j})=\nabla f(1,2,-2) \] Vector along the gradient: $-\mathbf{i}+\mathbf{j}=\mathbf{V}$ Unit vector: $\frac{1}{\sqrt{2}}(-\mathbf{i}+\mathbf{j})=\mathbf{v}$ Rate of change: $\frac{1}{\sqrt{2}}=\|\nabla f(1,2,-2)\|$
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