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 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 20


$\approx 0.57735$

Work Step by Step

As we know that, $u=\dfrac{v}{|v|}$ Here, $u=\dfrac{\lt 2,2,-2 \gt}{\sqrt{2^2+2^2+(-2)^2}}=\lt \dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{-1}{\sqrt 3} \gt$ Now $df=(\nabla f (3,4, 12) \cdot u) ds=[(\lt \dfrac{1}{\sqrt 3}\gt)(\lt \dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{-1}{\sqrt 3} \gt)](0.1)$ Thus, $\dfrac{1}{\sqrt 3} (0.1) \approx 0.57735$
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.