

\section{Path Independence}

\begin{enumerate}[1.]


\item {\it 
Let ${\bf F}(x,y,z)=e^{2y}{\bf i}+(axe^{2y}+be^z){\bf j}-3ye^z{\bf k}$.
Find the values of $a,b$ for which ${\bf F}$ is conservative, and the corresponding potential function.
For these values of $a$ and $b$, evaluate $\int_\mathcal{C} {\bf F}\cdot d{\bf r}$,
where $\mathcal{C}$ is a piecewise smooth curve from $(1,1,0)$ to $(0,1,3)$.}

\medskip

For ${\bf F}$ to be conservative, we must have
$$
\frac{\partial}{\partial y}e^{2y}=\frac{\partial}{\partial x}(axe^{2y}+be^z),\ \ 
2e^{2y}=ae^{2y},\ \ a=2,
$$
$$
\frac{\partial}{\partial z}(2xe^{2y}+be^{z})=\frac{\partial}{\partial y}(-3ye^z),\ \ 
be^z=-3e^z,\ \ b=-3.
$$
Let $a=2,b=-3$, and solve for $\phi$ such that ${\bf F}=\nabla\phi$:
$$
\frac{\partial \phi}{\partial x}=e^{2y},\ \ \phi=xe^{2y}+f(y,z)
$$
$$
\frac{\partial}{\partial y}(xe^{2y}+f(y,z))=2xe^{2y}+\frac{\partial f}{\partial y}=2xe^{2y}-3e^z,\ \ 
 f(y,z)=-3ye^z+g(z),
$$
$$
\frac{\partial}{\partial z}(xe^{2y}-3ye^z + g(z))=-3ye^z+g'(z)=-3ye^z,\ \ g(z)=C
$$

In particular, we may take $C=0$, so that 
$$\phi=xe^{2y}-3ye^z.$$

Finally,
 $$\int_\mathcal{C} {\bf F}\cdot d{\bf r}= \phi(0,1,3)-\phi(1,1,0)=(0-3e^3)-(e^2-3)=3-3e^3-e^2.$$
 
 
 \bigskip
 
 
\item {\it Find all values of the parameters $a,b,c$ 
such that the vector field 
$${\bf F}(x,y,z)=4xy^3 \,\ii+ (ax^2y^2+b\cos z)\jj+(y\sin z+c \cos z) \kk$$
 is conservative on $\mathbb{R}^3$. For all such $a,b,c$:
 
 \medskip
 
(a) find a potential function $f$ such that ${\bf F}=\nabla f$,

\medskip

(b) evaluate the integral $\int_C {\bf F}\cdot d{\bf r}$, if $C$ is the line segment from $(0,1,0)$ to $(1,1,\pi/2)$.}

\bigskip


(a) Suppose that ${\bf F}$ is conservative, with ${\bf F}=\nabla f$. Then we must have
$$
\frac{\partial}{\partial x}f(x,y,z)=F_1=4xy^3,$$
so that $f(x,y,z)=2x^2y^3  + g(y,z)$. We must also have 
$$
\frac{\partial}{\partial y}f(x,y,z)=6x^2y^2+\frac{\partial}{\partial y}g(y,z)= F_2 = ax^2y^2+b\cos z,
$$
so that $a=6$ and $g(y,z)=by\cos z +h(z)$. So far, we have $f(x,y,z)= 2x^2y^3+ by\cos z +h(z)$.
Finally,  
$$
\frac{\partial}{\partial z}f(x,y,z)= -by\sin z +h'(z)= F_3=y\sin z + c\cos z,
$$
so that $b=-1$ and we can take $h(z) =c\sin  z$, with no restrictions on $c$. The answer is $a=6$, $b=-1$, $c\in\rr$, and 
$f(x,y,z)= 2x^2y^3 - y\cos z +c\sin  z.$

\medskip

(b) With $f$ as above, we have
\begin{align*}
\int_C {\bf F}\cdot d{\bf r}&= f(1,1,\pi/2)-f(0,1,0)\\
&= (2-0+c)-(0-1+0) =3+c.
\end{align*}


\bigskip


\item { \it 
Let ${\bf F}:\rr^2\to\rr^2$ be a $C^1$ vector field. Assume that the following holds: for any $P,Q\in \rr^2$, and for any two curves $C_1,C_2$ from $P$ to $Q$ {\bf that are unions of at most 5 line segments each}, we have 
$\int_{C_1}{\bf F}\cdot d\bfr = \int_{C_2}{\bf F}\cdot d\bfr$. Does it follow that ${\bf F}$ is conservative? Prove your answer.
}

\medskip


The answer is yes, and we can prove it by modifying the argument from class just slightly. 
Assume that $\mathbf{F}$ has the path independence property on $\rr^2$, and define
$$
\phi (x,y) = \int_{C} \mathbf{F} \cdot d \mathbf{r},
$$
where $C$ is any path connecting the origin $O=(0,0)$ to $P =(x,y)$ and consisting of at most 5 line segments.

We need to prove that
$$
\frac{\partial \phi}{\partial x} = F_1, \, \frac{\partial \phi}{\partial y} = F_2.
$$

To prove the first identity, let $C$ be the path consisting of 2 line segments: $C_1$ from $(0,0)$ to $(0,y)$, and $C_2$ from $(0,y)$ to $(x,y)$. (This is a valid choice because the domain of ${\bf R}$ is all of $\rr^2$.) 
Then
$$
\phi(x,y) : = \int_{C_1} \mathbf{F} \cdot d \mathbf{r} +\int_{C_2} \mathbf{F} \cdot d \mathbf{r}.
$$
We now keep $y$ constant and differentiate in $x$. The integral on $C_1$ is independent of $x$, so the derivative is $0$. For $C_2$, we have
$$
\int_{C_2} \mathbf{F} \cdot d \mathbf{r} = \int_{0}^{x} F_1 (t,y) d t 
$$
so that 
$$
 \frac{\partial}{\partial x} \int_{C_2} \mathbf{F} \cdot d \mathbf{r} = F_1 (x, y)
$$
as claimed. The proof of the second part is identical but with $x$ and $y$ variables interchanged, so that we use a path consisting of the line segments from $(0,0)$ to $(x,0)$, and $C_2$ from $(x,0)$ to $(x,y)$. Path independence guarantees that both of the above paths get us the same function $\phi$.

Additional comments:

\begin{itemize}
\item We did not even need to use 5 line segments---paths consisting of at most 2 line segments would have been sufficient. 
\item It is also possible to solve this by approximating arbitrary piecewise smooth curves by paths consisting of increasing numbers of line segments. This is fine if done correctly.
\end{itemize}



\bigskip

\item { \it Let $f$ and $g$ be two $C^1$ functions on $D$. Let ${\bf F}=f\nabla g$ and ${\bf G}=g\nabla f$. Prove that 
${\bf F}$ has path-independent line integrals if and only if so does ${\bf G}$.}



\medskip


We have
$$
{\bf F}+{\bf G} = f\nabla g+ g\nabla f = \nabla (fg),
$$
so that ${\bf F}+{\bf G}$ has path independent line integrals.


Let $P_1,P_2\in D$, and let $C_1,C_2$ be  two piecewise smooth curves $C_1,C_2$ from $P_1$ to $P_2$. 
Then
\begin{align*}
& \int_{C_1}{\bf G}\cdot d{\bfr} - \int_{C_2}{\bf G}\cdot d{\bfr} 
\\
&\ \ = \left( \int_{C_1}({\bf F}+{\bf G})\cdot d{\bfr} - \int_{C_1}{\bf F}\cdot d{\bfr}\right)
-  \left( \int_{C_2}({\bf F}+{\bf G})\cdot d{\bfr} - \int_{C_2}{\bf F}\cdot d{\bfr}\right)
\\
&\ \ = \left( \int_{C_1}({\bf F}+{\bf G})\cdot d{\bfr}- \int_{C_2}({\bf F}+{\bf G})\cdot d{\bfr}\right)
-  \left( \int_{C_1}{\bf F}\cdot d{\bfr}
 - \int_{C_2}{\bf F}\cdot d{\bfr}\right)\\
&\ \ 
= 
\int_{C_2}{\bf F}\cdot d{\bfr}
 - \int_{C_1}{\bf F}\cdot d{\bfr}.
\end{align*}
Therefore the left side is always zero if and only if the right side is always zero, and the conclusion follows.








 
 
 
 
\end{enumerate}

