University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 13 - Section 13.5 - Directional Derivatives and Gradient Vectors - Exercises - Page 720: 11

Answer

$-4$

Work Step by Step

Our aim is to take the first partial derivative of the given function $f(x,y,z)$ with respect to $x$, by treating $y$ and $z$ as a constant, and vice versa: $f_x=2y $ and $f_y= 2x-6y $ The gradient equation is: $\nabla f = \lt f_x,f_y \gt =\lt 24 , 2x-6y \gt =\lt 2(5), 2(5)-6(5)\gt =\lt 10 , -20 \gt$ Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 4, 3 \gt }{\sqrt {4^2+3^2}} =\lt \dfrac{4}{5}, \dfrac{3}{5} \gt $ Directional derivative at that direction is: $D_v f=\nabla f \cdot v=\lt 10 , -20 \gt \times \lt \dfrac{4}{5}, \dfrac{3}{5} \gt =-4$
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