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: 19

Answer

$\approx 0.00076$ or, $\approx 0.0008$

Work Step by Step

As we know that, $u=\dfrac{v}{|v|}$ Thus, $u=\dfrac{\lt 3,6,-2 \gt}{\sqrt{3^2+6^2+(-2)^2}}=\lt \dfrac{3}{7},\dfrac{6}{7},\dfrac{-2}{7} \gt$ Now $df=(\nabla f (3,4, 12) \cdot u) ds=[(\lt \dfrac{3}{169},\dfrac{4}{169},\dfrac{12}{169} \gt)(\lt \dfrac{3}{7},\dfrac{6}{7},\dfrac{-2}{7} \gt)](0.1)$ Thus, $(\dfrac{9}{183})(\dfrac{1}{10}) \approx 0.00076$ or, $\approx 0.0008$
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