Multivariable-Calculus · Unit 18 · Video 4 · Interactive Practice

Conservative Fields: Four Equivalent Properties

IKey Formulas

Statement	Property	Meaning
$\oint_C \vec F \cdot d\vec r = 0$ for every closed $C$	(1) Conservative	No net work around any loop
$\int_{C_1} \vec F \cdot d\vec r = \int_{C_2} \vec F \cdot d\vec r$	(2) Path independent	Same endpoints, same work
$\vec F = \nabla f = f_x\,\hat\imath + f_y\,\hat\jmath$	(3) Gradient field	$M = f_x$ , $N = f_y$
$M\,dx + N\,dy = df$	(4) Exact differential	(3) written in differential notation

Key Insight: $\oint_{C_1 - C_2} = \int_{C_1} - \int_{C_2}$ links (1) and (2), and $f(x, y) = \int_{(0,0)}^{(x,y)} \vec F \cdot d\vec r$ builds the potential from path independence.

IITwo Paths Make a Loop

Out along $C_1$ and back along $C_2$ reversed is the closed curve $C_1 - C_2$ , with work $\int_{C_1} - \int_{C_2}$ .

Field position on C₁ − C₂ end

IIIThe Potential as Work from the Origin

$f(x, y)$ is the work from $(0, 0)$ to $(x, y)$ ; it is well defined only if every route gives the same number.

Field

IVQuiz Questions

Problem 1 · Kinetic Energy from the Potential

Given: the only force on a particle is $\vec F = \nabla f$ with $f = x^2 + y^2$ (our plus sign), and the particle moves from $(1, 0)$ to $(0, 2)$ — find the change in its kinetic energy.

✅ Correct! Change in kinetic energy = work =

f(0, 2) - f(1, 0) = 4 - 1 = 3

❌ The order is reversed. Work is

f(P_1) - f(P_0)

, end minus start.

❌ That adds the two values of $f$ . The kinetic energy changes by the difference.

❌ For a gradient field it does not. The work, and so the change in kinetic energy, is

f(P_1) - f(P_0)

on every path.

Show solution

The change in kinetic energy equals the work done by the force. For a gradient field the work is the change in the potential:

\Delta(\text{KE}) = \int_C \nabla f \cdot d\vec r = f(0, 2) - f(1, 0) = 4 - 1 = 3.

Problem 2 · Two Paths, Two Answers

Given: two curves $C_1$ , $C_2$ from $P$ to $Q$ with $\int_{C_1} \vec F \cdot d\vec r = 7$ and $\int_{C_2} \vec F \cdot d\vec r = 4$ — find $\oint_{C_1 - C_2} \vec F \cdot d\vec r$ and what it shows.

✅ Correct!

7 - 4 = 3 \ne 0

: one loop with nonzero work, so all four properties fail.

❌ $C_2$ is traversed backwards. Reversing it flips the sign of its work:

7 - 4

, not

7 + 4

❌ A gradient field is path independent. Different work on two paths with the same endpoints rules a gradient out.

❌ That holds only for conservative fields. Here the loop work is

\int_{C_1} - \int_{C_2} = 3

Show solution

The closed curve $C_1 - C_2$ runs out along $C_1$ and back along $C_2$ reversed, so

\oint_{C_1 - C_2} \vec F \cdot d\vec r = \int_{C_1} \vec F \cdot d\vec r - \int_{C_2} \vec F \cdot d\vec r = 7 - 4 = 3.

A loop with nonzero work means $\vec F$ is not conservative; by the equivalence it is not path independent, not a gradient, and $M\,dx + N\,dy$ is not exact.

Problem 3 · Building the Potential

Given: $\vec F = (2x + y)\,\hat\imath + x\,\hat\jmath$ is a gradient field and $f(0, 0) = 0$ — find $f(2, 3)$ as the work from $(0, 0)$ to $(2, 3)$ , going along the $x$ -axis to $(2, 0)$ and then straight up.

✅ Correct!

4 + 6 = 10

, and indeed

f = x^2 + xy

gives

f(2, 3) = 10

❌ That is only the horizontal leg. The vertical leg adds

\int_0^3 N(2, y)\,dy = \int_0^3 2\,dy = 6

❌ That is only the vertical leg. The horizontal leg adds

\int_0^2 2x\,dx = 4

❌ On the vertical leg only $N$ counts. There

dx = 0

, so the integrand is

N\,dy = 2\,dy

, giving

6

, not

9

Show solution

Horizontal leg $(0, 0) \to (2, 0)$ : $y = 0$ , $dy = 0$ , so $\vec F \cdot d\vec r = 2x\,dx$ and $\int_0^2 2x\,dx = 4$ .

Vertical leg $(2, 0) \to (2, 3)$ : $x = 2$ , $dx = 0$ , so $\vec F \cdot d\vec r = 2\,dy$ and $\int_0^3 2\,dy = 6$ .

f(2, 3) = 4 + 6 = 10.

Check: $f = x^2 + xy$ has $f_x = 2x + y$ , $f_y = x$ , and $f(2, 3) = 4 + 6 = 10$ .

Problem 4 · The Physicists' Sign

Given: in the physics convention $\vec F = -\nabla \varphi$ , where $\varphi$ is the potential energy; a particle moves from $P_0$ with $\varphi = 5$ to $P_1$ with $\varphi = 2$ — find the work done by $\vec F$ and the change in kinetic energy.

✅ Correct! Work

= \varphi(P_0) - \varphi(P_1) = 3

: potential energy lost is kinetic energy gained.

❌ That uses our plus sign. With

\vec F = -\nabla\varphi

the work is

-(\varphi(P_1) - \varphi(P_0)) = 3

❌ The work and the kinetic energy change agree. Positive work speeds the particle up.

❌ That adds the potential energies. Only the change,

5 - 2

, matters.

Show solution

With $\vec F = \nabla(-\varphi)$ , the fundamental theorem applied to $-\varphi$ gives

\int_C \vec F \cdot d\vec r = -\varphi(P_1) + \varphi(P_0) = 5 - 2 = 3.

The change in kinetic energy equals the work, $+3$ , so kinetic plus potential energy stays constant: $\text{KE}$ rises by $3$ while $\varphi$ falls by $3$ .

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