Gradient Fields
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.1: “Vector Fields” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-1-vector-fields |
| Textbook used in class | Stewart, Calculus, Section 16.1: “Vector Fields” (Examples 4, 5) |
Quick Reference
A vector field $\mathbf{F}$ is a gradient field (or conservative field) if there exists a scalar function $f$ such that $\mathbf{F} = \nabla f$. The function $f$ is called a potential function for $\mathbf{F}$.
In 2D: $\mathbf{F} = \langle P, Q\rangle = \nabla f$ means $P = f_x$ and $Q = f_y$.
Necessary condition: If $\mathbf{F} = \nabla f$ and the partials are continuous, then $\partial P/\partial y = \partial Q/\partial x$ (since both equal $f_{xy} = f_{yx}$).
Motivation
Not every vector field is a gradient field. Gravity, electric fields, and spring forces are gradient fields (they have potential energy functions). A rotating fluid vortex is not. The distinction matters greatly in physics: gradient fields are path-independent (the work done depends only on start and end points, not the path taken), while non-conservative fields are path-dependent.
Key Concepts
1. Finding a Potential Function
Given $\mathbf{F} = \langle P, Q\rangle$, find $f$ such that $f_x = P$ and $f_y = Q$.
Strategy:
- Integrate $P$ with respect to $x$: $f(x,y) = \int P\,dx + g(y)$.
- Differentiate $f$ with respect to $y$ and set equal to $Q$: solve for $g'(y)$, then integrate.
2. The Test for Conservative Fields
For $\mathbf{F} = \langle P, Q\rangle$ on a simply-connected region: $\mathbf{F} = \nabla f$ for some $f$ if and only if $\partial P/\partial y = \partial Q/\partial x$.
Worked Example
Find a potential function for $\mathbf{F}(x,y) = \langle 2xy, x^2 + 3y^2\rangle$, or show that no potential function exists. (Stewart 16.1.)
Check: $\partial P/\partial y = 2x$ and $\partial Q/\partial x = 2x$. Equal, so $\mathbf{F}$ may be conservative.
Find $f$: $f_x = 2xy \implies f(x,y) = x^2 y + g(y)$.
Differentiate: $f_y = x^2 + g'(y) = x^2 + 3y^2 \implies g'(y) = 3y^2 \implies g(y) = y^3 + C$.
Potential function: $f(x,y) = x^2 y + y^3$.
Verify: $\nabla f = \langle 2xy, x^2 + 3y^2\rangle = \mathbf{F}$. Correct.
every vector field has a potential function. Only conservative (gradient) fields have potential functions. A simple test: if $\partial P/\partial y \neq \partial Q/\partial x$, no potential function exists. For example, $\mathbf{F} = \langle y, 0\rangle$ fails the test ($\partial P/\partial y = 1 \neq 0 = \partial Q/\partial x$), so it is not a gradient field.
Common Misconceptions
on a non-simply-connected domain, the condition $\partial P/\partial y = \partial Q/\partial x$ is sufficient to guarantee a potential function exists.
This is the concept-image-conflicts-definition error. The mixed-partial test guarantees a potential function only on a simply-connected domain (one with no holes). On a domain with holes, such as $\mathbb{R}^2\setminus\{(0,0)\}$, a field can satisfy the partial-derivative test everywhere yet fail to be conservative. The standard example is $\mathbf{F} = \langle -y/(x^2+y^2),\, x/(x^2+y^2)\rangle$: its mixed partials are equal wherever it is defined, but its line integral around the origin equals $2\pi \neq 0$, so no global potential function exists.
Leveled Practice
Problem 1. Determine if $\mathbf{F}(x,y) = \langle y\cos x, \sin x\rangle$ is conservative. If so, find a potential function.
Show answer
$\partial P/\partial y = \cos x$, $\partial Q/\partial x = \cos x$. Equal, so $\mathbf{F}$ is conservative.
$f_x = y\cos x \implies f = y\sin x + g(y)$.
$f_y = \sin x + g'(y) = \sin x \implies g'(y) = 0 \implies g = C$.
$f(x,y) = y\sin x$.