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: 14


$\dfrac{-3}{2\sqrt {13}}$

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: $h_x(1,1)=\dfrac{-y/x^2}{1+\dfrac{y^2}{x^2}}+\sqrt 3 \cdot \dfrac{(y/2)}{\sqrt {1-(xy/2)^2}}=\dfrac{1}{2} \\ h_y (1,1)=\dfrac{1/x}{1+\dfrac{y^2}{x^2}}+\sqrt 3 \cdot \dfrac{(x/2)}{\sqrt {1-(xy/2)^2}}=\dfrac{3}{2} $ Write the gradient equation. $\nabla f (1,-1)= \lt h_x,h_y \gt = \lt \dfrac{1}{2} , \dfrac{3}{2} \gt$ Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 3,-2 \gt }{\sqrt {(3)^2+(-2)^2}} =\lt \dfrac{3}{\sqrt {13}}, -\dfrac{2}{\sqrt {13}}\gt$ The directional derivative at that direction is given as: $D_u h=\lt \dfrac{1}{2} , \dfrac{3}{2} \gt \times \lt \dfrac{3}{\sqrt {13}}, -\dfrac{2}{\sqrt {13}}\gt =\dfrac{-3}{2\sqrt {13}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.