Finding the potential function for path-independent vector fields

You are given a vector field, say,

You calculate

$${\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial }}{{\partial x}}}$$

$${\frac{{\partial F_1}}{{\partial y}}} {\frac{{\partial }}{{\partial y}}}$$

so that its curl is zero, i.e.,

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

F is defined on all of R², so there are no tricks to worry about. The vector field F is path-independent.

Since F is path-independent, we know there exists some potential function f so that $\nabla$f = F. How do we go about finding f?

Well, $\nabla$f = F means that

$$\left(\vphantom{\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}}\right. {\frac{{\partial f}}{{\partial x}}} {\frac{{\partial f}}{{\partial y}}} \left.\vphantom{\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}}\right)$$

So we need to find a function f (x, y) that satisfies the two conditions

$${\frac{{\partial f}}{{\partial x}}}$$ (1)

and

$${\frac{{\partial f}}{{\partial y}}}$$ (2)

Let's take these conditions one by one and see if we can find an f (x, y) that satisfies both of them. (We know this is possible since F is path-independent. If F were path-dependent, the procedure that follows would hit a snag somewhere.)

Let's start with condition (1). We can take the equation

$${\frac{{\partial f}}{{\partial x}}}$$

and treat y as though it were a number. In other words, we pretend that the equation is

$${\frac{{\mathrm{d} f}}{{ \mathrm{d} x}}}$$

for some number a. We can integrate the equation with respect to x and obtain that

But, then we have to remember that a really was the variable y so that

But actually, that's not right yet either. Since we were viewing y as a constant, really C could be a function of y and it wouldn't make a difference. We can replace C with any function of y, say g(y), and condition (1) will be satisfied

f (x, y) = y sin x + y²x + g(y). (3)

If you are still skeptical, try taking the partial derivative with respect to x of f (x, y) defined by equation (3). Since g(y) does not depend on x, we can conclude that $${\frac{{\partial }}{{\partial x}}}$$g(y) = 0. Indeed, condition (1) is satisfied for the f (x, y) of equation (3).

Now, we need to satisfy condition (2). We can take the f (x, y) of equation (3) (so we know that condition (1) will be satisfied) and take its partial derivative with respect to y, obtaining

$${\frac{{\partial f}}{{\partial y}}} {\frac{{\partial }}{{\partial y}}} \left(\vphantom{ y \sin x + y^2x +g(y)}\right. \left.\vphantom{ y \sin x + y^2x +g(y)}\right) {\frac{{\mathrm{d} g}}{{ \mathrm{d} y}}}$$

Comparing this to condition (2), we are in luck. We can easily make this f (x, y) satisfy condition (2) as long as

$${\frac{{\mathrm{d} g}}{{ \mathrm{d} y}}}$$

(If the vector field F had been path-dependent, we would have found it impossible to satisfy both condition (1) and condition (2), and would have run into trouble at this point.)

If we let

for some constant k, then

$${\frac{{\partial f}}{{\partial y}}}$$

and we have satisfied both conditions.

Combining this definition of g(y) with equation (3), we conclude that the function

is a potential function for F. You can verify that indeed

$$\nabla$$

With this in hand, calculating the integral

$$\int_{{C}}^{}$$

is simple, no matter what path C is. The integral is simply f (q) - f (p), where p is the beginning point and q is the ending point of C. (For this reason, if C is a closed curve, the integral is zero.)

We might like to give a problem such as find

$$\int_{{C}}^{}$$

where C is the curve given by the following graph.

The answer is simply

$$\int_{{C}}^{} \pi \pi$$

$$\pi {\frac{{\pi}}{{2}}} \pi \pi$$

$$\pi {\frac{{9\pi}}{{2}}} {\frac{{9\pi}}{{2}}}$$

(The constant k is always guaranteed to cancel, so you could just set k = 0.)

If the curve C is complicated, it's a good hint that F is probably path-independent. It's always a good idea to check if F is path-independent before computing its line integral

$$\int_{{C}}^{}$$

You might save yourself a lot of work.