Textbook Problem

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Finding a Potential Function In Exercises 11-18, determine whether the vector field is conservative. If it is, find a potential function for the vector field.

$F\left(x,y\right)=-\frac{y}{x^2}i+\frac{1}{x}j$

To determine

Whether the provided vector field $F\left(x,y\right)=\frac{y}{x}+C$ is conservative or not. If it is conservative, then find a potential function for the vector field.

Explanation of Solution

Consider the function,

$F\left(x,y\right)=\frac{y}{x}+C$

Partially differentiate both sides of the expression $M=-\frac{y}{x^2}$ with respect to ‘y’.

$\begin{array}{c}\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left(-\frac{y}{x^2}\right)\\ =-\frac{1}{x^2}\frac{\partial }{\partial y}\left(y\right)\\ =-\frac{1}{x^2}\end{array}$ …… (1)

And, partially differentiate both sides of the expression $N=\frac{1}{x}$ with respect to ‘x’,

$\begin{array}{c}\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left(\frac{1}{x}\right)\\ =\frac{\partial }{\partial x}\left(x^{-1}\right)\\ =-x^{-2}\frac{\partial }{\partial x}\left(x\right)\\ =-\frac{1}{x^2}\end{array}$ …… (2)

Now, from the equation (1) and (2);

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

For the vector field $F\left(x,y\right)=Mi+Nj$, here both M and N are differentiable function, then $F\left(x,y\right)$ is said to be conservative, if and only if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$