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: 9


$\lt -\dfrac{26}{27},\dfrac{23}{54}, \dfrac{-23}{54} \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: $f_x=-x(x^2+y^2+z^2)^{-3/2}+\dfrac{1}{x} \\ f_y= -y(x^2+y^2+z^2)^{-3/2}+\dfrac{1}{y} \\f_z=-z(x^2+y^2+z^2)^{-3/2}+\dfrac{1}{z}$ Write the gradient vector equation. $\nabla f = \lt f_x,f_y,f_z \gt $ $\nabla f = \lt -x(x^2+y^2+z^2)^{-3/2}+\dfrac{1}{x}, -y(x^2+y^2+z^2)^{-3/2}+\dfrac{1}{y} ,-z(x^2+y^2+z^2)^{-3/2}+\dfrac{1}{z}\gt $ Thus, $\nabla f (1,1,1) = \lt -\dfrac{26}{27},\dfrac{23}{54}, \dfrac{-23}{54} \gt $
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