Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 802: 50

Answer

$$ f(x,y,z)=x+3y+z+K $$

Work Step by Step

Given $$\nabla f=\langle 1, 3,1\rangle$$ Since $$ \nabla f=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle=\langle1, 3,1\rangle$$ Then \begin{align*} \frac{\partial f}{\partial x}&=1\ \ \ \Rightarrow\ \ f=x+C_1\\ \frac{\partial f}{\partial y}&=3\ \ \ \Rightarrow\ \ f=3y+C_2\\ \frac{\partial f}{\partial z}&=1\ \ \ \Rightarrow\ \ f=z+C_3\\ \end{align*} Hence $$ f(x,y,z)=x+3y+z+K $$
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