Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 15 - Vector Analysis - 15.1 Exercises - Page 1049: 32

Answer

$\mathbf{F}$ is conservative

Work Step by Step

Given $$\mathbf{F}(x, y)=\frac{1}{\sqrt{1+xy }}(y\mathbf{i}+x\mathbf{j}) =\frac{y}{\sqrt{ 1+xy}} \mathbf{i} +\frac{x}{\sqrt{1+xy}} \mathbf{j} $$ Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}\\ &=f_{x}(x, y)\mathbf{i}+f_{y}(x, y)\mathbf{j} \end{align} we get \begin{array}{l} {M=f_{x}(x, y)=\frac{\partial f(x,y)}{\partial x}=\frac{y}{\sqrt{1+xy}}=y(1+xy)^{-\frac{1}{2}}} \\ {N=f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=\frac{x}{\sqrt{1+xy}}=x(1+xy)^{-\frac{1}{2}}} \\ \end{array} This implies that $M$ and $N$ have continuous first partial derivatives. Since \begin{array}{l} {\begin{align} \frac{\partial M}{\partial y}&=(1+xy)^{-\frac{1}{2}}+y (-\frac{1}{2})x(1+xy)^{-\frac{3}{2}}\\ &=(1+xy)^{-\frac{1}{2}}+-\frac{1}{2}xy(1+xy)^{-\frac{3}{2}}\\ &=\frac{1}{2}(1+xy)^{-\frac{3}{2}}(2+2xy-xy)\\ &= (1+xy)^{-\frac{3}{2}}(2+ xy ) \end{align}} \\ {\begin{aligned} \frac{\partial N}{\partial x}&=(1+xy)^{-\frac{1}{2}}+x (-\frac{1}{2})y(1+xy)^{-\frac{3}{2}}\\ &=\frac{1}{2}(1+xy)^{-\frac{3}{2}}(2+2xy-xy)\\ &= (1+xy)^{-\frac{3}{2}}(2+ xy ) \end{aligned}} \\ \end{array} So, we get $$ \frac{\partial N}{\partial x}= \frac{\partial M}{\partial y}$$ And this implies that $\mathbf{F}$ is conservative
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