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: 10


$\lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt $

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=e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} \\ f_y= e^{x+y} \cos z +arcsin x \\f_z=-e^{x+y} \sin z $ Write the gradient vector equation. $\nabla f = \lt f_x,f_y,f_z \gt $ $\implies \nabla f = \lt e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} , e^{x+y} \cos z +arcsin x , -e^{x+y} \sin z \gt $ Thus, $\nabla f (0,0,\dfrac{\pi}{6}) = \lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt $
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