Conservative Vector Fields Examples 1

Conservative Vector Fields Examples 1

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

We will first see if we can verify the necessary condition for a vector field to be conservative. We note that $P(x, y) = \frac{x}{x^2 + y^2}$ and $Q(x, y) = \frac{y}{x^2 + y^2}$.

First let's take the partial derivative of $P$ with respect to $y$:

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

Now let's take the partial derivative of $Q$ with respect to $x$:

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

Therefore we have verified the necessary condition for a vector field to be conservative with $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. This does not guarantee us that $\mathbf{F} = P\vec{i} + Q \vec{j}$ is conservative though, but now we have reason to suspect it could be.

Now suppose that a potential function exists, that is suppose that $\mathbf{F} (x, y) = \nabla \phi (x, y)$. Then we must have that:

(5)
\begin{align} \quad \frac{\partial \phi}{\partial x} = \frac{x}{x^2 + y^2} \\ \quad \frac{\partial \phi}{\partial y} = \frac{y}{x^2 + y^2} \end{align}

Take the first equation and partial integrate with respect to $x$ to get that:

(6)
\begin{align} \quad \int \frac{\partial \phi}{\partial x} \: dx = \phi(x, y) = \int \frac{x}{x^2 + y^2} \: dx \end{align}

Using substitution, we let $u = x^2 + y^2$. Then $du = 2x \: dx$ and so $\frac{1}{2} du = x \: dx$ and so:

(7)
\begin{align} \quad = \frac{1}{2} \int \frac{1}{u} \: du = \frac{1}{2} \ln \mid u \mid = \frac{1}{2} \ln (x^2 + y^2) + h(y) \end{align}

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

(8)
\begin{align} \quad \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} \left ( \frac{1}{2} \ln (x^2 + y^2) + h(y) \right ) = \frac{1}{2} \left ( \frac{2y}{x^2 + y^2} \right ) + h(y) = \frac{y}{x^2 + y^2} + h'(y) \end{align}

Comparing this with the fact that we're given $\frac{\partial \phi}{\partial y}$ tells us that $h'(y) = 0$, and so $h(y) = C$.

Therefore, for any constant $C$ we have that $\phi (x, y) = \frac{1}{2} \ln (x^2 + y^2) + C$ is a potential function.

Therefore $\mathbf{F}$ is a conservative vector field on all of $\mathbb{R}^2$ except for at the origin.

Example 2

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

Let's first check the necessary condition for a vector field to be conservative. Let $P(x, y) = x^2y$ and $Q(x, y) = 2xy^2$. Let's calculate the partial derivative of $P$ with respect to $y$:

(9)
\begin{align} \quad \frac{\partial P}{\partial y} = x^2 \end{align}

Now let's calculate the partial derivative of $Q$ with respect to $x$:

(10)
\begin{align} \quad \frac{\partial Q}{\partial x} = 2y^2 \end{align}

We note that $\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$ and so $\mathbf{F}$ is not a conservative vector field.