\section{Conservative Vector Fields}

\begin{enumerate}[1.]

\item
{\it Let ${\bf F}(x,y,z)=e^y \ii+ (xe^y-\cos z) \jj+y\sin z$ in $\R^3$. 
Prove that ${\bf F}$ conservative, and find a potential function $f$.}

\medskip

We look for a potential function such that ${\bf F}=\nabla f$:
\begin{align*}
\frac{\partial}{\partial x}f(x,y,z)& =F_1=e^y,\ \ f(x,y,z)=xe^y  + g(y,z),
\\
\frac{\partial}{\partial y}f(x,y,z)& =xe^y +\frac{\partial}{\partial y}g(y,z)= xe^y-\cos z,
\\
g(y,z)&= =-y\cos z +h(z),
\\
\frac{\partial}{\partial z}f(x,y,z)& = y\sin z +h'(z)= F_3=y\sin z,\ \ h(z)=c.
\end{align*}
The field is conservative, with $f(x,y,z)=xe^y-y\cos z+c$ for any constant $c$.

\bigskip





\item
{\it Consider the vector field $$\displaystyle {\bf G}=\frac{F_1(x)}{F_3(z)}\ii + \frac{F_2(y)}{F_3(z)}\jj + \kk,$$
where $F_1,F_2,F_3$ are smooth functions on a domain $D\subset \rr^3$ and $F_3$ is never zero on $D$. Does ${\bf G}$ 
have to be conservative on $D$? 
If yes, prove it. If no, give a counterexample.}

\medskip

The answer is no. Let $F_1(x) = 1$, $F_2(y) = 1$, $F_3(z) = z$. Then $$\displaystyle {\bf G} = G_1\ii + G_2 \jj + G_3 \kk = \frac{1}{z} \ii + \frac{1}{z} \jj + \kk,$$ is defined on the domain $z > 0$, but $\bf G$ does not satisfy the necessary condition of Proposition 1.5 for a conservative vector field in $\rr^3$. Namely, we have:
	\[\frac{\partial G_1}{\partial z} = - \frac{1}{z^2} \neq 0 = \frac{\partial G_3}{\partial x},\]
contradicting Proposition 2.2.5 in Section 2.2.
\bigskip



\item
{\it Give an example of two smooth vector fields ${\bf F}$ and ${\bf G}$ on $\rr^2$ such that
${\bf F}$ and ${\bf G}$ have the same field lines, ${\bf F}$ is conservative, and 
${\bf G}$ is not conservative. 

\smallskip\noindent
(Hint: Worked Problem 1 of this Section might be helpful in constructing a simple conservative field $\mathbf{F}$. Then, try modifying it to make a $\mathbf{G}$ that has the same field lines but is non-conservative.)}

\medskip

Take ${\bf F} = e^x{\bf i} $ and ${\bf G} = e^{x+y} {\bf i}$. For both fields, the vectors are always non-zero and pointing in the direction of ${\bf i}$, so the field lines are lines $y=C$, oriented in the direction of increasing $x$.
The field ${\bf F}$ is conservative, as a special case of Practice Problem 1.  On the other hand, 
${\bf G}$ is not conservative, since $\frac{\partial G_1}{\partial y} = \frac{\partial e^{x+y}}{\partial y} = e^{x+y}$ and
$\frac{\partial G_2}{\partial x}=0$.

\bigskip

\item {\it 
Find all functions $f(y)$ and $g(x)$, where $f(y)$ depends only on $y$ and $g(x)$ depends only on $x$,
such that the vector field ${\bf F}(x,y)=f(y)\ii+g(x)\jj$ is conservative on $\mathbb{R}^2$. For such $f$ 
and $g$, find the potential function for ${\bf F}$.}

\medskip

For ${\bf F}$ to be conservative, we must have
$$
\frac{\partial}{\partial y}f(y)=\frac{\partial}{\partial x}g(x).\ \ 
$$
But since the functions $f$ and $g$ depend only on one variable each, so do the functions
$\frac{\partial}{\partial y}f(y)=f'(y)$ and $\frac{\partial}{\partial x}g(x)=g'(x)$.
Therefore $f'(y)=g'(x)=A$ has to be constant. This means that $f(y)=Ay+B$ and $g(y)=Ax+C$,
where $B$ and $C$ are also constants. In that case, integrating in $x$ and $y$ we get 
$$
\phi(x,y)=Axy+Bx+Cy+D,
$$
where $A,B,C,D$ are constants, so that ${\bf F}$ is indeed a conservative field. 
(The constant $D$ is optional, since the potential function is defined
only up to a constant.) 


\end{enumerate}

