Conservative Fields and Potentials

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Endpoint information

A field is conservative on a domain if $\mathbf F=\nabla\phi$ for a single-valued potential there. Along a smooth path, the chain rule gives $d\phi(\mathbf r(t))/dt=\nabla\phi\cdot\mathbf r'$ . Integrating produces

\int_C\mathbf F\cdot d\mathbf r=\phi(B)-\phi(A).

Every closed loop then has zero work. Conversely, path independence on a path-connected domain allows a potential to be defined by work from a fixed starting point.

Visual guide

VISUAL GUIDECurl can vanish away from a missing point

Loop surrounding the hole

For F = (−y/r², x/r²), arrows circle the excluded origin. The unit loop has circulation 2π despite zero curl everywhere in the punctured plane. The hole prevents the local derivative test from proving a global potential.

Find a potential

For $\mathbf F=(P,Q)$ , integrate $P$ with respect to $x$ , adding an unknown function of $y$ . Differentiate that candidate with respect to $y$ and match $Q$ . In three dimensions repeat with the remaining component. Always check every partial derivative at the end.

Worked example: reconstructing energy

For $\mathbf F=(2xy,x^2+2y)$ , integration gives $\phi=x^2y+g(y)$ . Matching $\phi_y=x^2+g'(y)$ yields $g'=2y$ , so $\phi=x^2y+y^2+C$ . Work from $(0,0)$ to $(1,2)$ is $2+4=6$ , on any path within the plane.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

If F=∇φ along a path from A to B, what is ∫ F·dr?

φ(A)−φ(B)

Always zero

φ(B)−φ(A)

Hint 1 · Find a starting point

A potential makes work depend only on endpoints.

Hint 2 · Take the next step

Use final potential minus initial potential.

Show the reasoning

Answer: φ(B)−φ(A)

The fundamental theorem for line integrals gives φ(B)−φ(A); a closed path has equal endpoints.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a hole matters

On the punctured plane consider $\mathbf F=(-y/(x^2+y^2),x/(x^2+y^2))$ . Here $Q_x-P_y=0$ everywhere the field is defined. But its integral around the unit circle is $2\pi$ , not zero. A globally single-valued angle potential cannot be chosen around the hole.

Conditions behind the shortcut

For a continuously differentiable field on an open simply connected planar domain, $Q_x=P_y$ implies conservativeness. Curl zero is always necessary for a smooth gradient, but domain topology matters for the converse. Do not silently include excluded points inside a loop.

Practice

Find a potential for $(3x^2,2y)$ .
Use it to find work from $(1,1)$ to $(2,3)$ .
Is $(-y,x)$ conservative on the plane?

Show worked solutions

$\phi=x^3+y^2+C$ ; differentiating verifies both components.
$(8+9)-(1+1)=15$ .
No: $Q_x-P_y=1-(-1)=2$ . Its nonzero curl rules out a potential even before evaluating a path.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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