Conservative Vector Fields Examples 2

# Conservative Vector Fields Examples 2

Recall from the Conservative Vector Fields page that a vector field $\mathbf{F}$ is said to be conservative on the domain $D$ if there exists a function $\phi$ known as a potential function such that $\mathbf{F} = \nabla \phi$ on $D$.

We also proved a necessary condition for a vector field on $\mathbb{R}^2$ and a vector field on $\mathbb{R}^3$ to possess. Recall that if $\mathbf{F}(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a conservative vector field on $D$ then we must have that for all points in $D$ that::

(1)
\begin{align} \quad \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \end{align}

Furthermore, recall that if $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a conservative vector field on $D$ then we must have that for all points in $D$ that:

(2)
\begin{align} \quad \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \quad , \quad \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x} \quad , \quad \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y} \end{align}

We will now look at some examples of determining whether a vector field is conservative or not.

# Example 1

Determine whether the vector field $\mathbf{F}(x, y) = 2x \ln \mid y \mid \vec{i} + \frac{x^2}{y} \vec{j}$ is a conservative vector field.

Let's first check the necessary condition for a vector field to be conservative. Let $P(x, y) = 2x \ln \mid y \mid$ and let $Q(x, y) = \frac{x^2}{y}$. We first compute the partial derivative of $P$ with respect to $y$ to get that:

(3)
\begin{align} \quad \frac{\partial P}{\partial y} = \frac{2x}{y} \end{align}

We then compute the partial derivative of $Q$ with respect to $x$ to get that:

(4)
\begin{align} \quad \frac{\partial Q}{\partial x} = \frac{2x}{y} \end{align}

Therefore we have that $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, so $\mathbf{F}$ may be a conservative vector field. Let's try to find a potential function for $\mathbf{F}$.

We note that if a potential function $\phi$ exists, then we will have that $\mathbf{F} = \nabla \phi$ and so $\frac{\partial \phi}{\partial x} = P(x, y)$ and $\frac{\partial \phi}{\partial y} = Q(x, y)$.

We will start by integrating both sides of $\frac{\partial \phi}{\partial x} = P(x, y)$ with respect to $x$ to get that:

(5)
\begin{align} \quad \int \frac{\partial \phi}{\partial x} \: dx = \phi (x, y) = \int 2x \ln \mid y \mid \: dx = x^2 \ln \mid y \mid + h(y) \end{align}

We will now take the partial derivative with respect to $y$ of our result above to get that:

(6)
\begin{align} \quad \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} \left ( x^2 \ln \mid y \mid + h(y) \right ) = \frac{x^2}{y} + h'(y) \end{align}

Now we already have that $\frac{\partial \phi}{\partial y} = \frac{x^2}{y}$, which implies that $h'(y) = 0$ and so $h(y) = C$. Therefore, for $C$ as a constant we have found a set of potential functions:

(7)
\begin{align} \quad \phi (x, y) = x^2 \ln \mid y \mid + C \end{align}

So $\mathbf{F}$ is a conservative vector for $(x, y) \in \mathbb{R}^2 \setminus \{ 0, 0 \}$.

## Example 2

Determine whether the vector field $\mathbf{F}(x, y) = \frac{1}{x + y} \vec{i} + \frac{1}{x+y} \vec{j}$ is a conservative vector field.

Let's first check the necessary condition for a vector field to be conservative. Let $P(x, y) = \frac{1}{x+y}$ and $Q(x, y) = \frac{1}{x + y}$. We then compute the partial derivative of $P$ with respect to $y$ to get that:

(8)
\begin{align} \quad \frac{\partial P}{\partial y} = -\frac{1}{(x + y)^2} \end{align}

We then compute the partial derivative of $Q$ with respect to $x$ to get that:

(9)
\begin{align} \quad \frac{\partial Q}{\partial x} = -\frac{1}{(x + y)^2} \end{align}

So $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. Once again, $\mathbf{F}$ may be a conservative vector field, so let's try to find a potential function. If a potential function $\phi$ exists, then we will have that $\frac{\partial \phi}{\partial x} = P(x, y)$ and $\frac{\partial \phi}{\partial y} = Q(x, y)$. Using the first equation and integrating with respect to $x$ and we have that:

(10)
\begin{align} \quad \int \frac{\partial \phi}{\partial x} \: dx = \phi (x, y) = \int \frac{1}{x + y} \: dx = \ln \mid x + y \mid + h(y) \end{align}

We will now partial differentiate both sides with respect to $y$ and get that:

(11)
\begin{align} \quad \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} \left ( \ln \mid x + y \mid + h(y) \right ) = \frac{1}{x + y} + h'(y) \end{align}

Therefore we have that $h'(y) = 0$ so $h(y) = C$ and so:

(12)
\begin{align} \quad \phi (x, y) = \ln \mid x + y \mid + C \end{align}

Thus $\mathbf{F}$ is a conservative vector field.