17.2 Conservative vector fields and potential functions

Conservative Vector Fields and Potential Functions

Defining Conservative Vector Fields

A conservative vector field is one where the work done moving between two points doesn't depend on the path you take. Only the endpoints matter. This is a powerful property because it lets you skip painful line integral computations entirely.

Three equivalent ways to recognize a conservative field:

• The line integral between any two points is the same regardless of path (path independence)
• The field can be written as the gradient of some scalar function, called a potential function
• The line integral around every closed loop is zero: $\oint_C \vec{F} \cdot d\vec{r} = 0$

These three conditions are all saying the same thing from different angles. If one holds (on a simply connected domain), the others follow.

Path Independence and Potential Functions

Path independence means that $\int_C \vec{F} \cdot d\vec{r}$ depends only on the starting point $A$ and ending point $B$, not on the curve $C$ connecting them.

A potential function (or scalar potential) is a scalar function $f$ whose gradient recovers the vector field:

$\vec{F} = \nabla f = \left(\frac{\partial f}{\partial x},\; \frac{\partial f}{\partial y},\; \frac{\partial f}{\partial z}\right)$

If you can find such an $f$, the field is conservative. The potential function is unique up to an additive constant, just like antiderivatives in single-variable calculus.

Exact Differentials and Conservativeness

When $\vec{F} = (M, N, P)$ is conservative, the expression $M\,dx + N\,dy + P\,dz$ is an exact differential, meaning it equals $df$ for some function $f$. That's just another way of saying $M = \frac{\partial f}{\partial x}$, $N = \frac{\partial f}{\partial y}$, and $P = \frac{\partial f}{\partial z}$.

The necessary and sufficient condition for conservativeness on a simply connected domain is that the mixed partials match up (which, as you'll see below, is equivalent to the curl being zero).

Gradient and Curl

Gradient and Its Properties

The gradient of a scalar function $f(x, y, z)$ is the vector field:

$\nabla f = \left(\frac{\partial f}{\partial x},\; \frac{\partial f}{\partial y},\; \frac{\partial f}{\partial z}\right)$

• $\nabla f$ points in the direction of greatest increase of $f$, and its magnitude $\|\nabla f\|$ gives the rate of change in that direction.
• $\nabla f$ is always perpendicular to the level surfaces of $f$ (the surfaces where $f = \text{const}$).

These properties matter here because a conservative field $\vec{F} = \nabla f$ inherits all of them. The field vectors are always perpendicular to the equipotential surfaces of $f$.

Curl and Its Interpretation

The curl of $\vec{F} = (P, Q, R)$ is the vector field:

$\nabla \times \vec{F} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z},\; \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x},\; \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$

Curl measures the tendency of the field to rotate around a point. A field with zero curl everywhere is called irrotational.

The key test for conservativeness: if $\nabla \times \vec{F} = \vec{0}$ throughout a simply connected domain, then $\vec{F}$ is conservative. The "simply connected" condition is critical. On domains with holes, a zero-curl field can still fail to be conservative (the classic example is $\vec{F} = \frac{1}{x^2+y^2}(-y, x)$ on $\mathbb{R}^2 \setminus \{0\}$).

Why does this work? If $\vec{F} = \nabla f$, then by Clairaut's theorem the mixed partials of $f$ are equal, which forces every component of $\nabla \times \vec{F}$ to vanish. The converse (zero curl implies conservative) requires the domain to be simply connected.

Finding a Potential Function

When you've confirmed $\nabla \times \vec{F} = \vec{0}$, here's how to find $f$:

1. Integrate $M = \frac{\partial f}{\partial x}$ with respect to $x$. You'll get $f = \int M\,dx + g(y,z)$, where $g(y,z)$ is an unknown function (it plays the role of the "constant" of integration, but it can depend on the other variables).
2. Differentiate your result with respect to $y$ and set it equal to $N = \frac{\partial f}{\partial y}$. Solve for $\frac{\partial g}{\partial y}$, then integrate to get $g(y,z) = \ldots + h(z)$.
3. Differentiate with respect to $z$ and set equal to $P = \frac{\partial f}{\partial z}$. Solve for $h(z)$ and integrate.
4. Combine everything. The result is your potential function $f$, determined up to a constant.

Example: Suppose $\vec{F} = (2xy + z,\; x^2,\; x)$. You can verify $\nabla \times \vec{F} = \vec{0}$. Then:

1. $f = \int (2xy + z)\,dx = x^2 y + xz + g(y,z)$
2. $\frac{\partial f}{\partial y} = x^2 + \frac{\partial g}{\partial y} = x^2 \implies \frac{\partial g}{\partial y} = 0 \implies g(y,z) = h(z)$
3. $\frac{\partial f}{\partial z} = x + h'(z) = x \implies h'(z) = 0 \implies h(z) = C$
4. $f(x,y,z) = x^2 y + xz + C$

Line Integrals and the Fundamental Theorem

Line Integrals and Their Computation

The line integral of $\vec{F}$ along a curve $C$ is written $\int_C \vec{F} \cdot d\vec{r}$ and represents the work done by the field along that path.

To compute it in general, you parameterize $C$ as $\vec{r}(t)$ for $t \in [a, b]$, then evaluate:

$\int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}\,'(t)\,dt$

This can be tedious. For conservative fields, you don't need to do any of this.

Fundamental Theorem of Line Integrals

This is the payoff. If $\vec{F} = \nabla f$ is conservative with potential function $f$, then for any smooth curve $C$ from point $A$ to point $B$:

$\int_C \vec{F} \cdot d\vec{r} = f(B) - f(A)$

Just evaluate the potential function at the two endpoints and subtract. The curve itself doesn't matter at all.

This is a direct generalization of the single-variable Fundamental Theorem of Calculus ($\int_a^b f'(x)\,dx = f(b) - f(a)$) to line integrals in higher dimensions.

Using the earlier example: To find the work done by $\vec{F} = (2xy+z,\; x^2,\; x)$ along any path from $(1,0,0)$ to $(2,1,3)$, just compute:

$f(2,1,3) - f(1,0,0) = (4 \cdot 1 + 2 \cdot 3) - (1 \cdot 0 + 1 \cdot 0) = 10 - 0 = 10$

No parameterization needed.