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: 12

Answer

$\dfrac{-11}{25\sqrt{34}}$

Work Step by Step

Formula to calculate the directional derivative: $D_uf=\nabla f(x,y) \cdot u$ or, $D_uf=\nabla f(x,y) \cdot \dfrac{v}{|v|}$ This implies At $(1,2)$ $D_uf (1,2)=\dfrac{3}{25} \times \dfrac{3}{\sqrt{34}}+\dfrac{-4}{25} \times \dfrac{5}{\sqrt{34}}$ Hence, $D_uf (1,2)=\dfrac{-11}{25\sqrt{34}}$
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