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: 12



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=4x \\ f_y= 2y $ Write the gradient equation. $\nabla f (-1,1)= \lt f_x,f_y \gt = \lt 4(-1), 2(1) \gt =\lt -4,2 \gt$ Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 3, -4 \gt }{\sqrt {3^2+(-4)^2}} =\lt \dfrac{3}{5}, \dfrac{-4}{5} \gt$ The directional derivative at that direction is given as: $D_v f=\nabla f \cdot v=\lt -4,2 \gt \times \lt \dfrac{3}{5}, \dfrac{-4}{5} \gt =-4$
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