---
title: "Conservative Fields and Path Independence"
description: "Some fields are gradients of a potential function. For those, the line integral depends only on the endpoints, and there is a test to recognise them."
canonical: https://lightmysky.com/learn/mathematics/conservative-fields-and-path-independence-mt_5djKpI5F33
source: https://lightmysky.com/learn/mathematics/conservative-fields-and-path-independence-mt_5djKpI5F33.md
retrieved: 2026-09-12
---

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# Conservative Fields and Path Independence

Some fields are gradients of a potential function. For those, the line integral depends only on the endpoints, and there is a test to recognise them.

Subject: Mathematics · Area: Calculus & Analysis · Ages 20 to 21
Page: https://lightmysky.com/learn/mathematics/conservative-fields-and-path-independence-mt_5djKpI5F33

## Ready when they can

- Test a field for the mixed-partial condition on a simply connected region
- Recover a potential function from a conservative field
- Explain why the integral around any closed loop of a conservative field is zero

## Lesson: Fields that only care about endpoints

A conservative field is a gradient in disguise: the whole field equals grad f for some potential function f. Such fields have path independent line integrals, so only the endpoints matter and the route between them does not. Three views say the same thing: the integral forgets the path, the flow around any closed loop sums to zero, and a potential function exists. Gravity is the famous example, which is why path independence feels like energy conservation.

**Example.** To test a field F = (P, Q) on a nice region, compare the cross partials: check whether dQ/dx equals dP/dy. Equal cross partials signal a conservative field, and differing ones mean it is not. Try F = (2xy, x squared + 1): dQ/dx is 2x and dP/dy is 2x, so the test passes. A field like (minus y, x) fails it, so you stop there and never hunt for a potential.

Once the test passes, recover the potential by integrating P with respect to x. That gives f up to a leftover function of y alone. Differentiate your result with respect to y and match it against Q to pin down the leftover piece. Then the fundamental theorem for line integrals finishes the job: the integral along any curve from A to B is just f(B) minus f(A).

**Tip.** Always certify the field before finding the potential, since a non conservative field has none at all. If the cross partials differ, stop: the shortcut is unavailable and you must integrate along the path directly.

**Recap.** Test the cross partials, recover the potential, and trade any path integral for a subtraction at the endpoints.

## Practice

14 questions on this page, each with its working shown.

## Needs first

- [Directional Derivatives and the Gradient](https://lightmysky.com/learn/mathematics/directional-derivatives-and-the-gradient-mt_LW_KckY5Ad)
- [Vector Fields and Line Integrals](https://lightmysky.com/learn/mathematics/vector-fields-and-line-integrals-mt_VZxKEmsVgO)

## Opens up

- [Cauchy's Theorem and Deforming a Contour](https://lightmysky.com/learn/mathematics/cauchys-theorem-and-deforming-a-contour-mt_aacYFVN-HH)
- [Parallel Transport, Christoffel Symbols and the Geodesic Equation](https://lightmysky.com/learn/science/parallel-transport-christoffel-symbols-and-the-geodesic-equation-mt_bUEL2R91Vy)
- [Green's Theorem in the Plane](https://lightmysky.com/learn/mathematics/greens-theorem-in-the-plane-mt_UDgivmPVEJ)
