Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 14 - Vector Calculus - Review Exercises - Page 1147: 6

Answer

$F=\lt -x e^{-x^2-y^2-z^2},-y e^{-x^2-y^2-z^2}, -z e^{-x^2-y^2-z^2} \gt$

Work Step by Step

The vector field $F$ can be computed as: $F=\lt \dfrac{\partial \phi}{\partial x}, \dfrac{\partial \phi}{\partial y}, \dfrac{\partial \phi}{\partial z}\gt$ Here, we have $\phi (x, y,z)=\dfrac{1}{2} e^{-x^2-y^2-z^2}$ Thus,our required vector field $F$ is: $F=\lt \dfrac{\partial }{\partial x} (\dfrac{1}{2} e^{-x^2-y^2-z^2}), \dfrac{\partial}{\partial y} (\dfrac{1}{2} e^{-x^2-y^2-z^2}), \dfrac{\partial}{\partial z} (\dfrac{1}{2} e^{-x^2-y^2-z^2})\gt$ or, $F=\lt -x e^{-x^2-y^2-z^2},-y e^{-x^2-y^2-z^2}, -z e^{-x^2-y^2-z^2} \gt$
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