Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.6 - Directional Derivatives and the Gradient Vector - 14.6 Exercise - Page 957: 21

Answer

$\sqrt{65}, \lt 1, 8 \gt$

Work Step by Step

Formula to calculate the maximum rate of change of $f$: $D_uf=|\nabla f(x,y)|$ $\nabla f(x,y)=(\dfrac{2y}{\sqrt x},4\sqrt x)$ $\nabla f(4,1)=\lt \dfrac{2(1)}{\sqrt (4)},4\sqrt (4) \gt=\lt 1, 8 \gt$ $|\nabla f(4,1)|=\sqrt{1^2+ 8^2}=\sqrt{65}$ Hence, the required answers are: $\sqrt{65}, \lt 1, 8 \gt$
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