## Problem on finding a potential function of a vector field

Is $\mathbf{F}(x,y) = \frac{1}{x+y} \mathbf{i} + \frac{1}{x+y} \mathbf{j}$ conservative?
If so, find a $\phi(x,y)$ such that $\mathbf{F}(x,y) = \nabla \phi(x,y)$?
• ## Solution

#### Determining if $\mathbf{F}(x,y)$ is conservative

We recall how to determine if a vector field is conservative:
We identify $u(x,y) = \frac{1}{x+y}$ and $v(x,y) = \frac{1}{x+y}$, from which we can compute the partial derivatives
\begin{align}
\partial_y u(x,y) &= -\frac{1}{(x+y)^2}\\
\partial_x v(x,y) &= - \frac{1}{(x+y)^2}
\end{align}
These are equal, so $\mathbf{F}(x,y)$ is conservative.

#### Finding the potential $\phi(x,y)$

We recall the definition of a conservative vector field:
Further:
Writing down these components:
\begin{align}
\frac{1}{x+y} &= \partial_x \phi(x,y) \tag{1}\\
\frac{1}{x+y} &= \partial_y \phi(x,y) \tag{2}\\
\end{align}
Integrating (1) with respect to $x$ and (2) with respect to $y$ gives
\begin{align}
\phi(x,y) &= \log(x+y) + g(y)\\
\phi(x,y) &= \log(x+y) + h(x)
\end{align}
for any $g(y)$ and $h(x)$.
These relations are both satisfied by $$\phi(x,y) = \log(x+y) + c$$ for any constant $c$.