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: 25

Answer

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

Work Step by Step

Our aim is to determine the maximum rate of change of $f(x,y,z)$.In order to find this, we have : $D_uf=|\nabla f(x,y,z)|$ Given: $f(x,y,z)=x/(y+z)$ $\nabla f(x,y,z)=\lt 1/y+z , -x/(y+z)^2,-x/(y+z)^2 \gt $ From the given data, we have $f(x,y,z)=f(8,1,3)$ $\nabla f(8,1,3)=\lt 1/1+3 , -8/(1+3)^2,-8/(1+3)^2 \gt=\lt 1,-2,2 \gt$ $|\nabla f(8,1,3)|=\sqrt{(\dfrac{1}{4})^2+(-\dfrac{1}{2})^2+(-\dfrac{1}{2})^2}=\dfrac{3}{4}$ Therefore, the maximum rate of change of $f(x,y)$ and the direction is: $\dfrac{3}{4}, \lt 1,-2,-2 \gt$
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