How to find the potential function of a conservative vector field

Potential function of a conservative vector field blog post.jpeg

Let’s look at the formulas we’ll use for the potential function

A vector field $F$ is called conservative if it’s the gradient of some scalar function, that is, if there exists a function $f$ such that

$\bold{F}=\nabla f$

In this situation $f$ is called a potential function for $F$ .

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Showing that a vector field is conservative

Given a vector field

$\bold{F}(x,y)=P(x,y)\bold{i}+Q(x,y)\bold{j}$

$F$ is conservative if

The domain of $F$ is open and simply-connected, and
The scalar curl of $F$ is $0$

Open and simply-connected

The domain of $F$ is open and simply-connected if $F$ is defined on the entire plane $R^2$ . If the domain of $F$ is $R^2$ , then the domain of $F$ is open, such that it doesn’t contain any of its boundary points, and the domain of $F$ is simply-connected, such that it’s connected and contains no holes.

Scalar curl

The scalar curl of $F$ is $0$ , which means that

1. $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$

This also implies that

$\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}=0$

and

$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0$

2. Because $\partial P/\partial y$ must be equal to $\partial Q/\partial x$ , we should be able to subtract $P$ from $Q$ or vice versa and get $0$ . Because the partial derivatives are interchangeable, you’ll sometimes see them written as

$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$

but with this notation it’s important to remember that we have to take the partial derivative with respect to $x$ of the function $Q$ , and the partial derivative with respect to $y$ of the function $P$ .

Whichever notation we use,

$\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\quad=\quad\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\quad=\quad\frac{\partial F_1}{\partial y}-\frac{\partial F_2}{\partial x}\quad=\quad\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\quad=\quad0$

is called the scalar curl of the vector field $F$ .

Line integrals of conservative vector fields

The value of the line integral over the curve $C$ inside a conservative vector field is always the same, regardless of the path of the curve $C$ . This means that the value of the line integral only depends on the initial and terminal points of $C$ .

This means that we can evaluate the line integral of a conservative vector field using only the endpoints of the curve $C$ , because the line integral of $\nabla f$ is just the net change in $f$ .

So the theorem that defines the line integral of a conservative vector field says:

Assume $C$ is a smooth curve defined by the vector function $r(t)$ , with $a\leq t\leq b$ . If $f$ is a differentiable function whose gradient vector $\nabla f$ is continuous on $C$ , then

$\int_C\bold{F}\cdot d\bold{r}=\int_C\nabla f\cdot d\bold{r}$

$=f(\bold{r}(b))-f(\bold{r}(a))$

$=f(x_1,y_1)-f(x_2,y_2)$

The theorem shows us that, in order to find the value of the line integral of a conservative vector field, we just follow these steps:

Show that $F$ is conservative
If $F$ is conservative, find its potential function $f$
Evaluate $f$ over the interval $[a,b]$ , and the answer is the value of the line integral