Thomas' Calculus 13th Edition

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

Chapter 14: Partial Derivatives - Section 14.5 - Directional Derivatives and Gradient Vectors - Exercises 14.5 - Page 826: 11



Work Step by Step

In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant to find the x-coordinate of the gradient vector, and vice versa: $f_x=2y \\ f_y= 2x-6y $ Write the gradient equation. $\nabla f = \lt f_x,f_y \gt = \lt 2(5), 2(5)-6(5)\gt =\lt 10 , -20 \gt$ Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 4, 3 \gt }{\sqrt {4^2+3^2}} =\lt \dfrac{4}{5}, \dfrac{3}{5}\gt$ The directional derivative at that direction is given as: $D_v f=\nabla f \cdot v=\lt 10 , -20 \gt \times \lt \dfrac{4}{5}, \dfrac{3}{5}\gt=-4$
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