Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - Chapter Review Exercises - Page 971: 24

Answer

The vector field is conservative. The potential function is $f\left( {x,y,z} \right) = y {\tan ^{ - 1}}x + {z^2} + C$, where $C$ is a constant.

Work Step by Step

Write ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\frac{y}{{1 + {x^2}}},{{\tan }^{ - 1}}x,2z} \right)$. 1. We check if ${\bf{F}}$ satisfies the cross-partials condition: $\frac{{\partial {F_1}}}{{\partial y}} = \frac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ \ }$ $\frac{{\partial {F_2}}}{{\partial z}} = \frac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ \ }$ $\frac{{\partial {F_3}}}{{\partial x}} = \frac{{\partial {F_1}}}{{\partial z}}$ Since $\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + {x^2}}}$, we get $\frac{1}{{1 + {x^2}}} = \frac{1}{{1 + {x^2}}}$, ${\ \ \ \ }$ $0=0$, ${\ \ \ \ }$ $0=0$ From these results, we conclude that ${\bf{F}} = \left( {\frac{y}{{1 + {x^2}}},{{\tan }^{ - 1}}x,2z} \right)$ satisfies the cross-partials condition. Therefore, by Theorem 4 in Section 17.3, ${\bf{F}}$ is conservative. Thus, there is a potential function for ${\bf{F}}$. 2. Find a potential function for ${\bf{F}}$. Let the potential function for ${\bf{F}}$ be $f\left( {x,y,z} \right)$ such that ${\bf{F}} = \nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$. So, a. taking the integral of $\frac{{\partial f}}{{\partial x}}$ with respect to $x$ gives $f\left( {x,y,z} \right) = \smallint \frac{y}{{1 + {x^2}}}{\rm{d}}x = y{\tan ^{ - 1}}x + m\left( {y,z} \right)$ b. taking the integral of $\frac{{\partial f}}{{\partial y}}$ with respect to $y$ gives $f\left( {x,y,z} \right) = \smallint {\tan ^{ - 1}}x{\rm{d}}y = y{\tan ^{ - 1}}x + n\left( {x,z} \right)$ c. taking the integral of $\frac{{\partial f}}{{\partial z}}$ with respect to $z$ gives $f\left( {x,y,z} \right) = \smallint 2z{\rm{d}}z = {z^2} + o\left( {x,y} \right)$ Since the three ways of expressing $f\left( {x,y,z} \right)$ must be equal, we get $y{\tan ^{ - 1}}x + m\left( {y,z} \right) = y{\tan ^{ - 1}}x + n\left( {x,z} \right) = {z^2} + o\left( {x,y} \right)$ From here, we conclude that the potential function is $f\left( {x,y,z} \right) = y{\tan ^{ - 1}}x + {z^2} + C$, where $C$ is a constant.
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