Multivariable-Calculus · Unit 18 · Video 3 · Interactive Practice

Path Independence and Why the Rotation Field Fails It

IKey Formulas

Formula	Name	What it says
$\displaystyle\int_{C_1} \nabla f \cdot d\vec r = \int_{C_2} \nabla f \cdot d\vec r = f(P_1) - f(P_0)$	Path independence	$C_1$ , $C_2$ both run from $P_0$ to $P_1$ ; no formula for $f$ needed
$\displaystyle\oint_C \vec F \cdot d\vec r = 0$ for every closed $C$	Conservative	For $\nabla f$ : $f(P) - f(P) = 0$
$\displaystyle\oint_{x^2+y^2=1} (-y\,dx + x\,dy) = 2\pi$	Rotation field, counterclockwise	$\vec F \cdot \hat T = 1$ on a curve of length $2\pi$
$\pi$ (upper half) vs $-\pi$ (lower half)	From $(1, 0)$ to $(-1, 0)$	Same endpoints, different work: not a gradient

Key Insight: one loop with nonzero work is enough to show a field is not a gradient; one loop with zero work proves nothing.

IISame Endpoints, Same Work?

From $(1, 0)$ to $(-1, 0)$ , a gradient field gives every path the same work; the rotation field does not.

Field Path

IIIOne Zero Loop Proves Nothing

The rotation field does zero work around some closed curves, yet it is not conservative.

Loop fraction travelled 1

IVQuiz Questions

Problem 1 · A Second Path

Given: $\vec F$ is the gradient of some function (no formula given), and along a curve $C_1$ from $P$ to $Q$ , $\int_{C_1} \vec F \cdot d\vec r = 5$ — find the work along a different curve $C_2$ from $P$ to $Q$ .

✅ Correct! Both integrals equal

f(Q) - f(P)

, whatever

f

is.

❌ The formula for $f$ is not needed. Both works equal

f(Q) - f(P)

, so they are equal to each other.

❌ $C_2$ runs the same way, from $P$ to $Q$ . Only a path from

Q

back to

P

would give

-5

❌ Zero is for closed curves.

C_2

starts at

P

and ends at

Q

, a different point.

Show solution

If $\vec F = \nabla f$ , the fundamental theorem gives

\int_{C_1} \vec F \cdot d\vec r = f(Q) - f(P) = \int_{C_2} \vec F \cdot d\vec r,

so the work along $C_2$ is also $5$ .

Problem 2 · One Loop with Zero Work

Given: a field $\vec F$ with $\oint_C \vec F \cdot d\vec r = 0$ for $C$ the unit circle — find what this tells you about $\vec F$ .

✅ Correct! The rotation field itself gives zero around a figure eight, and it is not conservative.

❌ One loop is not every loop. Conservative means zero work around every closed curve.

❌ Nothing rules it out either. A gradient field also gives zero here; one loop decides nothing in either direction.

❌ That would require zero work around every closed curve. A single loop cannot establish it.

Show solution

$\vec F$ is conservative only if $\oint_C \vec F \cdot d\vec r = 0$ for every closed curve. A single loop with zero work is consistent with both possibilities: every gradient field gives zero there, and so do some non-conservative fields on particular loops (the rotation field gives $0$ around a figure eight made of two equal circles traversed in opposite directions).

A single loop with nonzero work, on the other hand, does prove that $\vec F$ is not conservative.

Problem 3 · The Lower Half Circle

Given: $\vec F = -y\,\hat\imath + x\,\hat\jmath$ and $C$ the lower half of the circle of radius $2$ about the origin, from $(2, 0)$ to $(-2, 0)$ — find $\int_C \vec F \cdot d\vec r$ .

✅ Correct!

\vec F \cdot \hat T = -|\vec F| = -2

on a half circle of length

2\pi

❌ Check the direction. From

(2, 0)

(-2, 0)

along the bottom is clockwise, against the counterclockwise field.

❌ $|\vec F| = 2$ on this circle, not $1$ . The work is

-2

times the length

2\pi

❌ The endpoints differ, and the field is not a gradient.

\vec F

is tangent to the circle the whole way, so the work is not zero.

Show solution

On the circle of radius $2$ , $\vec F$ is tangent with $|\vec F| = \sqrt{x^2 + y^2} = 2$ , pointing counterclockwise. The lower half from $(2, 0)$ to $(-2, 0)$ is traversed clockwise, so $\hat T = -\vec F/|\vec F|$ and $\vec F \cdot \hat T = -2$ . The half circle has length $2\pi$ :

\int_C \vec F \cdot d\vec r = \int_C (-2)\,ds = -2 \cdot 2\pi = -4\pi.

The upper half gives $+4\pi$ between the same endpoints.

Problem 4 · Perpendicular Is Not Enough

Given: $f = x^2 + y^2$ and a field $\vec G$ that points radially outward (perpendicular to every circle about the origin), with $\vec G = 5\nabla f$ on the positive $x$ -axis and $\vec G = \nabla f$ on the negative $x$ -axis. The loop $C$ goes from $(1, 0)$ to $(2, 0)$ along the axis, around the upper half of the circle of radius $2$ to $(-2, 0)$ , along the axis to $(-1, 0)$ , and back around the upper half of the unit circle to $(1, 0)$ — find $\oint_C \vec G \cdot d\vec r$ .

✅ Correct!

15 + 0 - 3 + 0 = 12 \ne 0

, so

\vec G

is not conservative and not a gradient.

❌ Being perpendicular to level curves does not make $\vec G$ a gradient. Compute the two axis pieces: they are

15

and

-3

❌ Check the sign of the inward piece. From

(-2, 0)

(-1, 0)

f

drops from

4

1

, so that piece is

-3

❌ The inward piece also does work. On the negative axis

\vec G = \nabla f

, and

f

changes from

4

1

there:

-3

Show solution

Out along the positive axis: $\vec G = 5\nabla f = \langle 10x, 0\rangle$ , so the work is $\int_1^2 10x\,dx = 5\,(4 - 1) = 15$ .

Both half circles: $\vec G$ points radially, perpendicular to the path, so $\vec G \cdot \hat T = 0$ and the work is $0$ .

In along the negative axis: $\vec G = \nabla f = \langle 2x, 0\rangle$ , so the work is $\int_{-2}^{-1} 2x\,dx = 1 - 4 = -3$ .

\oint_C \vec G \cdot d\vec r = 15 + 0 - 3 + 0 = 12.

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