Q. 42 - Calculus

Question

Determine whether or not each of the vector fields in Exercises 41–48 is conservative. If the vector field is conservative, find a potential function for the field.

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

Step-by-Step Solution

Answer

Since, $\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$ so the given vector is conservative.

A potential function for the field is $f(x,y)=x{\tan }^{-1}y$ 

Step 1. Given Information

We have to determine whether or not each of the vector fields in the given exercises is conservative. If the vector field is conservative, find a potential function for the field.

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

Step 2. Firstly finding the given vector field is conservative or not.

A vector field $\mathbf{F}(x,y)=\left(F_1\right(x,y),F_2(x,y\left)\right)$ is conservative if and only if $\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}.$

$$\begin{gathered} \frac{\partial F_1}{\partial y}=\frac{\partial }{\partial y}{\tan }^{-1}y \frac{\partial F_2}{\partial x}=\frac{\partial }{\partial x}\frac{x}{1+y^2} \\ \frac{\partial F_1}{\partial y}=\frac{1}{1+y^2} \frac{\partial F_2}{\partial x}=\frac{1}{1+y^2}\frac{\partial }{\partial x}x \\ \frac{\partial F_1}{\partial y}=\frac{1}{1+y^2} \frac{\partial F_2}{\partial x}=\frac{1}{1+y^2} \end{gathered}$$

Hence, $\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}.$

Step 3. Now finding the potential function for the field.

Since, $\mathbf{F}(x,y)=\left({\tan }^{-1}y, \frac{x}{1+y^2}\right)$

$$\begin{gathered} f(x,y)=\int {\tan }^{-1}ydx+B \\ f(x,y)={\tan }^{-1}y\int dx+B \\ f(x,y)={\tan }^{-1}y·x+\alpha +B \\ f(x,y)=x{\tan }^{-1}y+\alpha +B \end{gathered}$$

where $\alpha$ is an arbitrary constant and B is the integral with respect to y of the terms in $F_2(x,y)$ in which the factor x does not appear.

Step 4. In this case, that is all of F 2 ( x , y ) , so

$$\begin{gathered} B=\int \frac{x}{1+y^2}dy \\ B=x\int \frac{1}{1+y^2}dy \\ B=x{\tan }^{-1}y+\alpha \end{gathered}$$

where β is an arbitrary constant.

Step 5. Setting the constants equal to zero since they do not affect the gradient of f ( x , y )

We have, 

$f(x,y)=x{\tan }^{-1}y$