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: 25

Answer

$\dfrac{3}{4}, \lt \dfrac{1}{4},-\dfrac{1}{2},-\dfrac{1}{2} \gt$

Work Step by Step

Formula to calculate the maximum rate of change of $f$: $D_uf=|\nabla f(x,y)|$ $\nabla f(x,y)=\lt 1/y+z , -x/(y+z)^2,-x/(y+z)^2 \gt $ $\nabla f(8,1,3)=\lt 1/1+3 , -8/(1+3)^2,-8/(1+3)^2 \gt=\lt \dfrac{1}{4},-\dfrac{1}{2},-\dfrac{1}{2} \gt$ $|\nabla f(8,1,3)|=\sqrt{(\dfrac{1}{4})^2+(-\dfrac{1}{2})^2+(-\dfrac{1}{2})^2}=\dfrac{3}{4}$ Hence, the required answers are: $\dfrac{3}{4}, \lt \dfrac{1}{4},-\dfrac{1}{2},-\dfrac{1}{2} \gt$
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.