IMPORTANT: This is a problem I created for fun, so if it is ill-posed in any way, please let me know in the comments. Also, I think my title is a bit vague, so please suggest any improvements that come to mind.
Let's consider the vector field
$\textbf{F}(x,y)=x\textbf{i}-y\textbf{j}$. Out of all the smooth
curves lying in $(0,\infty)^2$, is there one connecting the points
$\left(\frac{1}{2},2\right)$, $\left(2,\frac{1}{2}\right)$ that
maximizes the value of $\int_{C}\textbf{F}\cdot d\textbf{r}$? Here, a
curve is said to be smooth if it admits a parameterization
$\textbf{r}$ with a continuous, non-vanishing derivative.
I've spent some time thinking about this problem, and am beginning to suspect that the curve, if it exists, is the hyperbola $y=\frac{1}{x}$, where $\frac{1}{2}\leq x\leq 2$. Here's how I arrived at this conclusion:
If we interpret the line integral $\int_{C}\textbf{F}\cdot d\textbf{r}$ as the work done by $\textbf{F}$ on a particle moving along the "track" drawn out by $\textbf{r}(t)=x(t)\textbf{i}+y(t)\textbf{j}$, then my intuition of work suggests that it will be maximized when the track "flows" with $\textbf{F}$ (if it didn't, then $\textbf{F}$ would act against the particle at certain times). More precisely, I think that $\textbf{r}'(t)=\textbf{F}\left(\textbf{r}(t)\right)$ should be true for all $t$ in the domain of $\textbf{r}$, as this condition will ensure that the particle points in the direction of the force field at all times.
Using the fact that $\textbf{r}'(t)=x'(t)\textbf{i}+y'(t)\textbf{j}$ and $\textbf{F}\left(\textbf{r}(t)\right)=\textbf{F}\left(x(t),y(t)\right)=x(t)\textbf{i}-y(t)\textbf{j}$, we can transform the equation $\textbf{r}'(t)=\textbf{F}\left(\textbf{r}(t)\right)$ into the system
\begin{align*}
x'(t) &= x(t)\\
y'(t) &= -y(t)
\end{align*}
This gives $x(t)=Ce^t$ and $y(t)=Ke^{-t}=\frac{K}{e^t}$ (clearly $K,C\neq 0$). We deduce that the Cartesian equation of the curve is
$$y=\frac{K}{\frac{x}{C}}$$
or
$$y=\frac{CK}{x}$$
This curve passes through $\left(\frac{1}{2},2\right)$ and $\left(2,\frac{1}{2}\right)$, so we end up with the following system
$$2=\frac{CK}{\frac{1}{2}}\iff CK=1$$
$$\frac{1}{2}=\frac{CK}{2}\iff CK=1$$
They're both the same equation, so we solve for one of the constants, say $K$.
$$K=\frac{1}{C}$$
Substituting this into the equation of the curve gives
$$y=\frac{C\cdot\frac{1}{C}}{x}=\frac{1}{x}$$
which is the section of the hyperbola I talked about (I found the domain of $x$ with a graph).
This seems reasonable, but I have no idea how to prove that the hyperbola actually maximizes the value of the line integral $\int_{C}\textbf{F}\cdot d\textbf{r}$. Heck, I don't even know if there exists a curve that maximizes the integral, let alone how to prove its existence. Any help is appreciated.
Note: the line integral of $\textbf{F}$ over this hyperbola is easily determined to be $\frac{15}{4}$, so if you find a curve where the line integral is greater than this amount, please share it with me.
Best Answer
Consider $$\phi=\frac12(x^2-y^2)$$ Then $$\vec F=\nabla \phi$$ Therefore the force is conservative, and your integral depends only on the initial and final points. See the gradient theorem. No matter the path, your integral is the same.