Calculus: Early Transcendentals (2nd Edition)

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

Chapter 14 - Vector Calculus - 14.3 Conservative Vector Fields - 14.3 Exercises - Page 1085: 21

Answer

$xz+y$

Work Step by Step

For a vector field to be Conservative, $\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial x}$ and $\dfrac{\partial g}{\partial z}=\dfrac{\partial h}{\partial y}$ and $\dfrac{\partial f}{\partial z}=\dfrac{\partial h}{\partial x}$ We have: $f(x,y,z)=z, g(x,y,z) =1$ and $h(x,y,z)=x$ Thus, $\dfrac{\partial f}{\partial y}=0=\dfrac{\partial g}{\partial x}$ and $\dfrac{\partial g}{\partial z}=0=\dfrac{\partial h}{\partial y}$ and $\dfrac{\partial f}{\partial z}=1=\dfrac{\partial h}{\partial x}$ Therefore, a vector field $F$ is Conservative. Now, potential function $F=\nabla \phi$ So, $\dfrac{\partial \phi}{\partial x}=z \implies \phi(x,y,z)=xy+a(y,z)$ and $\dfrac{\partial \phi}{\partial y}=1=\dfrac{\partial a}{\partial y} \implies \dfrac{\partial a}{\partial y}=0 \implies a(y,z)=y+b(z)$ and $\dfrac{\partial \phi}{\partial z}=x =x+b'(z) \implies b(z)=c$ Thus, $\phi(x,y,z)=xz+y$
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