Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.5 - Directional Derivatives and Gradient Vectors - Exercises 14.5 - Page 826: 15



Work Step by Step

In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant to find the x-coordinate of the gradient vector, and vice versa: $f_x(1,-1,2)=y+z=1 \\ f_y(1,-1,2)=x+z=3 \\f_z(1,-1,2)=x+y=0$ Write the gradient equation. $\nabla f (1,-1,2)= \lt f_x,f_y, f_z \gt = \lt 1,3,0 \gt$ Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 3,6, -2 \gt }{\sqrt {(3)^2+(6)^2+(-2)^2}} =\lt \dfrac{3}{7}, \dfrac{6}{7}, \dfrac{-2}{7}\gt$ The directional derivative at that direction is given as: $D_v f=\lt 1,3,0 \gt \times \lt \dfrac{3}{7}, \dfrac{6}{7}, \dfrac{-2}{7}\gt=3$
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.