Thomas' Calculus 13th Edition

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

Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 41


$$ \dfrac{\pi}{\sqrt 2}$$

Work Step by Step

Since, $D_u f = \nabla f \cdot u$ We have: $$r=\cos 3t i+\sin 3t j +3t k$$ $v(t)=\dfrac{dr}{dt}=-3 \sin 3t i+3 \cos 3t j +3t k$ or, $ v(\dfrac{\pi}{3})=-3 \sin 3t i+3 \cos 3t j +3t k=-3j+3k$ Now, $f(-1,0, \pi) =yz i+xz j +xy k$ or, $f(-1,0, \pi) =-\pi \ j$ $\nabla f \cdot u = (- \pi j) \cdot (\dfrac{-1}{\sqrt 2} j+\dfrac{1}{\sqrt 2} k) = \dfrac{1}{\sqrt 2} \pi$
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