Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.6 Directional Derivatives and the Gradient Vector - 14.6 Exercises - Page 997: 21


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

Work Step by Step

Our aim is to determine the maximum rate of change of $f(x,y)$.In order to find this, we have : $D_uf=|\nabla f(x,y)|$ Given: $f(x,y)=4y \sqrt x$ $\nabla f(x,y)=(\dfrac{2y}{\sqrt x},4\sqrt x)$ From the given data, we have $f(x,y)=f(4,1)$ Thus, $\nabla f(4,1)=\lt \dfrac{2(1)}{\sqrt (4)},4\sqrt (4) \gt=\lt 1, 8 \gt$ or, $|\nabla f(4,1)|=\sqrt{1^2+ 8^2}=\sqrt{65}$ Therefore, the maximum rate of change of $f(x,y)$ and the direction is: $\sqrt{65}, \lt 1, 8 \gt$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.