Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.6 Exercises - Page 968: 27

Answer

(a) $-\nabla f(\bf {x})$ (b) $\lt -12,92 \gt$

Work Step by Step

(a) Formula to calculate the directional derivative: $D_uf=\nabla f(x,y) \cdot u$ or, $D_uf=|\nabla f(x,y)||u| \cos \theta$ Minimum value of $\cos \theta=-1$ when $\theta =\pi or 180^\circ$ Thus, $D_uf=-|\nabla f(x)|$ or, $-\nabla f(\bf {x})$ (b) Formula to calculate the maximum rate of change of $f$: $D_uf=|\nabla f(x,y)|$ $\nabla f(x,y)=\lt 4x^3y-2xy^3,x^4-3x^2y^2 \gt$ $\nabla f(2,-3)=\lt 12,-92 \gt$ $|\nabla f(2,-3)|=\sqrt{(12)^2+(-92)^2}=\sqrt {\dfrac{9}{25}}=\dfrac{3}{5}$ From part (a) $D_uf=-|\nabla f(x)|=-\lt 12,-92 \gt =\lt -12,92 \gt$ Hence, the required answer is: $\lt -12,92 \gt$
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.