University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Practice Exercises - Page 750: 41

Answer

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

Work Step by Step

$r=\cos 3t i+\sin 3t j +3t k$ and $v(t)=-3 \sin 3t i+3 \cos 3t j +3t k \implies v(\dfrac{\pi}{3})=-3j+3k$ $f(-1,0, \pi) =yz i+xz j +xy k \implies f(-1,0, \pi) =-\pi j$ We know that $D_u f = \nabla f \cdot u$ So, $\nabla f \cdot u = (- pi j) (\dfrac{-1}{\sqrt 2} j+\dfrac{1}{\sqrt 2} k) =\dfrac{\pi}{\sqrt 2}$
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