Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 802: 39

Answer

${D_{\bf{u}}}f\left( P \right) = \frac{1}{2}\sqrt 6 $

Work Step by Step

We have $f\left( {x,y,z} \right) = \sin \left( {xy + z} \right)$ and $P = \left( {0, - 1,\pi } \right)$. The gradient of $f$ is $\nabla f = \left( {y\cos \left( {xy + z} \right),x\cos \left( {xy + z} \right),\cos \left( {xy + z} \right)} \right)$ The gradient of $f$ at $P$ is $\nabla {f_P} = \left( {1,0, - 1} \right)$. So, $||\nabla {f_P}|| = \sqrt {{1^2} + {0^2} + {{\left( { - 1} \right)}^2}} = \sqrt 2 $ Using Theorem 4, the rate of change of $f$ in the direction of a unit vector making an angle $\theta = 30^\circ $ with $\nabla {f_P}$ is ${D_{\bf{u}}}f\left( P \right) = ||\nabla {f_P}||\cos \theta $ ${D_{\bf{u}}}f\left( P \right) = \sqrt 2 \cos 30^\circ = \frac{1}{2}\sqrt 6 $
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