Question

Determine the map \( g\in C^\infty \) with \( g(0)=0 \) that makes the vector field.

\( F(x,y)=(ysinx+xycosx+e^y)i+(g(x)+xe^y)j \)

conservative. Find a potential for the resulting \( F \) .

 

Answer 1

Solution

\( F \) in conservative iff \( \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} \)

\( M=ysin(x)+xycos(x)+e^y=>\frac{\partial M}{\partial y}=sin(x)+xcos(x)+e^y \)\( N=g(x)+xe^y=>\frac{\partial N}{\partial x}=g_x(x)+e^y \)

\( \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=>sin(x)+xcos(x)+e^y=g_x(x)+e^y=sin(x)+xcos(x) \)\( => g(x)=\int sin(x)dx+\int xcos(x)dx \)

\( =>g(x)=xsin(x)+c \) but \( g(0)=0=>c=0 \)

Therefore, the function \( g(x)=xsin(x) \)

(*) Find the potential function f.

\( F=\nabla f=>(M,N)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}) \), we have 

\( M=\frac{\partial f}{\partial x}=ysin(x)+xycos(x)+e^y ,(i) \)

\( N=\frac{\partial f}{\partial y}=xsin(x)+xe^y,(ii) \)

From \( (ii) \)  \( =>f(x,y)=xysin(x)+xe^y+c(x) \)

                 \( =>f_x(x,y)=xycos(x)+ysin(x)+e^y+c'(x),(iii) \)

From \( (i) \) and \( (iii) \) \( =>c'(x)=0=>c(x)=k,k\in R \)

so, the potential function is \( f(x,y)=xycos(x) +ysin(x)+e^y+k,k\in R \)

answer

Thus, The potential function is \( f(x,y)=xycos(x) +ysin(x)+e^y+k,k\in R \)