Question

Determine whether the vector field is conservative. If it is, find a potential function for the vector field.

(a). \( F(x,y)={e}^{x+y}i+{e}^{xy}j \)

(b). \( F(x)=\frac{xy^2}{(1+x^2)^2}i+\frac{x^2y}{1+x^2}j \)

Answer 1

Solution

(a). \( F(x,y)={e}^{x+y}i+{e}^{xy}j \)

\( M={e}^{x+y}=>\frac{\partial M}{\partial y}={e}^{x+y} \)

\( N={e}^{xy}=>\frac{\partial N}{\partial x}=y{e}^{xy} \)

\( => \frac{\partial N}{\partial x}\neq \frac{\partial M}{ \partial y} => F \)  in not conservative.

(b).  \( F(x)=\frac{xy^2}{(1+x^2)^2}i+\frac{x^2y}{1+x^2}j \)

\( M=\frac{xy^2}{(1+x^2)^2}=> \frac{\partial M}{\partial y}=\frac{2xy}{(x^2+1)^2} \)

\( N=\frac{x^2y}{1+x^2}=> \frac{\partial M}{\partial x}=\frac{2xy}{(x^2+1)^2} \)

\( => \frac{\partial M}{ \partial y}=\frac{\partial N}{\partial x}=> F \) is conservative. 

(*) Find the potential function f.

\( F=\bigtriangledown f=>(M,N)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})=> \)\( \begin{cases} M=> \frac{\partial f}{\partial x}=\frac{xy^2}{(1+x^2)^2} , (i)& \quad \\ N=> \frac{\partial f}{\partial y}=\frac{x^2y}{1+x^2} ,(ii)& \quad \ \end{cases} \)

From \( (ii) \) \( => \frac{\partial f}{\partial y}=\frac{x^2y}{1+x^2} =>f(x,y)=\frac{x^2y^2}{2(x^2+1)}+c(x) \)  \( ,(iii) \) , \( ,c\in R \)

From \( (iii) \) \( =>\frac{\partial f}{\partial x}= \frac{xy^2}{(x^2+1)^2}+c'(x)=N \)

\( =>\frac{xy^2}{(x^2+1)^2}+c'(x)=\frac{xy^2}{(1+x^2)^2}=>c'(x)=0=>c(x)=k,k\in R \)Thus, \( f(x,y)=\frac{x^2y^2}{2(x^2+1)}+k , k\in R \)

 

Answer

(a). F is not conservative.

(b). Therefore, the potential function is \( f(x,y)=\frac{x^2y^2}{2(x^2+1)}+k , k\in R \)