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Conservative Fields and Path Independence · Tutorial 797 of 999

Workshop: Planar Potential Fields

Practice a complete planar-field workflow: inspect the domain, test cross-partials, find a potential when one exists, and compute work along specified paths.

Intermediate 10 min read

What You'll Learn

  • Check whether a planar field is defined and continuously differentiable throughout its domain.
  • Use matching or mismatched cross-partials to assess whether a potential can exist.
  • Recover a potential by integrating one component and determining the remaining function.
  • Apply domain conditions before concluding that matching cross-partials guarantee a potential.
  • Calculate work along a parameterized path and verify it using an endpoint potential when available.

A Reliable Workflow for Planar Fields

A planar vector field assigns a vector to each point in part of the plane. For a field \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\), work along a curve depends on both components and the direction of travel. When the field has a potential, that calculation simplifies: work depends only on the potential's values at the endpoints.

The useful question is not just whether the cross-partials match. First check where the field is defined, then apply the cross-partial test with the domain conditions in mind. If a potential exists, recover it and use it to evaluate work. The examples below illustrate each part of that process.

Definition: A potential for \(\mathbf{F}=\langle P,Q\rangle\) on a domain \(D\) is a scalar function \(f:D\to\mathbb{R}\) such that \(f_x=P\) and \(f_y=Q\) throughout \(D\). A field with a potential is called conservative.

The Cross-Partial Test and Its Conditions

Suppose \(P\) and \(Q\) have continuous first partial derivatives on an open domain \(D\). If \(\mathbf{F}\) has a potential \(f\) with continuous second partial derivatives, then \(P=f_x\) and \(Q=f_y\). Equality of the mixed partial derivatives gives the necessary condition \(P_y=Q_x\).

On a simply connected open domain, the condition \(P_y=Q_x\) is also sufficient for a potential to exist. Here, simply connected means that every closed curve in the domain can be continuously contracted to a point while staying in the domain. On a domain that is not simply connected, matching cross-partials alone may not be enough.

Test: For a continuously differentiable planar field, \(P_y=Q_x\) is necessary for a potential. It is sufficient if the domain is also simply connected. Check that the field and its partial derivatives are defined throughout the domain before using this conclusion.

The domain matters because the test concerns derivatives at points where the field is defined, while the conclusion concerns a potential defined throughout the domain. A missing point or curve can obstruct a single-valued potential even if the cross-partials agree everywhere the field exists.

Recovering a Potential

When the test applies, integrate one component with respect to its variable. If \(f_x=P\), then integration with respect to \(x\) gives \(f(x,y)=\int P(x,y)\,dx+g(y)\), where \(g(y)\) is an unknown function. Differentiating this expression with respect to \(y\), then matching \(f_y\) to \(Q\), determines \(g\) up to an additive constant.

This extra function is essential: an antiderivative with respect to \(x\) may omit any term depending only on \(y\), because such a term has zero \(x\)-derivative. The recovered function must ultimately pass both checks, \(f_x=P\) and \(f_y=Q\).

1
Inspect the domain.
Identify where each component and its first partial derivatives are defined. Confirm the domain condition needed for the cross-partial test.
2
Compare cross-partials.
Compute \(P_y\) and \(Q_x\). If they differ at any point, the field cannot have a potential on a domain containing that point.
3
Recover and verify a potential.
Integrate one component, determine the remaining single-variable function, and check both partial derivatives.
4
Evaluate the work.
If a potential exists, subtract its value at the starting point from its value at the ending point. Otherwise, compute the line integral along the specified path.

Worked Example: Recover a Potential and Find Work

Worked Example: Recover a Potential and Find Work

Let \(\mathbf{F}(x,y)=\langle 2xy+3,\ x^2+4y\rangle\) on \(D=\mathbb{R}^2\). Find a potential and calculate the work from \((0,0)\) to \((1,2)\) along \(\gamma(t)=(t,2t)\), \(0\leq t\leq 1\).

1. Check the domain and cross-partials. Both components are polynomials, so the field and its first partial derivatives are continuous throughout \(\mathbb{R}^2\). The plane is simply connected. With \(P=2xy+3\) and \(Q=x^2+4y\),

$$P_y=2x,\qquad Q_x=2x.$$

The cross-partials agree, so a potential exists on this domain.

2. Recover the potential. Integrating \(P\) with respect to \(x\) gives

$$f(x,y)=\int(2xy+3)\,dx=x^2y+3x+g(y).$$

Differentiate with respect to \(y\) and match the result to \(Q\):

$$f_y=x^2+g'(y)=x^2+4y,\qquad g'(y)=4y.$$

Thus \(g(y)=2y^2+C\), where \(C\) is an arbitrary constant. Choose \(C=0\), so \(f(x,y)=x^2y+3x+2y^2\). Check both components:

$$f_x=2xy+3=P,\qquad f_y=x^2+4y=Q.$$

3. Compute work from endpoint values. The starting value is \(f(0,0)=0\). The ending value is

$$f(1,2)=(1)^2(2)+3(1)+2(2)^2=2+3+8=13.$$

Therefore, the work is \(f(1,2)-f(0,0)=13\).

4. Check by direct integration along the specified path. The path has \(x=t\), \(y=2t\), \(dx=dt\), and \(dy=2\,dt\). Along it, \(P=4t^2+3\) and \(Q=t^2+8t\). Hence

$$\int_\gamma \mathbf{F}\cdot d\mathbf{r}=\int_0^1\bigl(P\,dx/dt+Q\,dy/dt\bigr)\,dt=\int_0^1\bigl((4t^2+3)+2(t^2+8t)\bigr)\,dt$$
$$=\int_0^1(6t^2+16t+3)\,dt=\left[2t^3+8t^2+3t\right]_0^1=2+8+3=13.$$

The direct calculation agrees with the endpoint-potential result.

Why the Endpoint Formula Works

The endpoint formula follows from the chain rule and the fundamental theorem of calculus. Suppose \(\mathbf{F}=\nabla f=\langle f_x,f_y\rangle\), and let \(\gamma(t)=(x(t),y(t))\), \(a\leq t\leq b\), be a piecewise smooth path. On each smooth piece, the chain rule gives

$$\frac{d}{dt}f(x(t),y(t))=f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t).$$

Since \(f_x=P\) and \(f_y=Q\), the right side is \(P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)\), the line-integral integrand. Integrating on each smooth piece and adding the results cancels the values at shared endpoints. The fundamental theorem then gives

$$\int_\gamma \mathbf{F}\cdot d\mathbf{r}=f(x(b),y(b))-f(x(a),y(a)).$$

This argument explains both why a potential makes work independent of the path and why the direction of travel matters: reversing the endpoints reverses the difference.

Worked Example: Matching Cross-Partials but No Global Potential

Worked Example: Matching Cross-Partials but No Global Potential

Consider \(\mathbf{F}(x,y)=\left\langle-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right\rangle\) on the punctured plane \(D=\mathbb{R}^2\setminus\{(0,0)\}\). Decide whether the cross-partial test guarantees a potential, and compare work along two paths from \((1,0)\) to \((-1,0)\).

1. Check the domain. The denominator is zero only at the origin, which is excluded. The field is continuously differentiable on \(D\). However, \(D\) is not simply connected, so matching cross-partials will not by itself guarantee a global potential.

2. Compare cross-partials. Write \(P=-y/(x^2+y^2)\) and \(Q=x/(x^2+y^2)\). Differentiating gives

$$P_y=\frac{y^2-x^2}{(x^2+y^2)^2},\qquad Q_x=\frac{y^2-x^2}{(x^2+y^2)^2}.$$

They agree at every point of \(D\). Because the domain condition for sufficiency fails, this does not establish a potential on all of \(D\).

3. Compute work along the upper semicircle. Set \(\gamma_1(t)=(\cos t,\sin t)\) for \(0\leq t\leq\pi\). Then \(\gamma_1'(t)=(-\sin t,\cos t)\). On the unit circle, \(\mathbf{F}(\gamma_1(t))=(-\sin t,\cos t)\), so

$$\int_{\gamma_1}\mathbf{F}\cdot d\mathbf{r}=\int_0^\pi\bigl(\sin^2t+\cos^2t\bigr)\,dt=\int_0^\pi 1\,dt=\pi.$$

4. Compute work along the lower semicircle. Set \(\gamma_2(t)=(\cos t,-\sin t)\) for \(0\leq t\leq\pi\). Its derivative is \((-\sin t,-\cos t)\), while \(\mathbf{F}(\gamma_2(t))=(\sin t,\cos t)\). Therefore,

$$\int_{\gamma_2}\mathbf{F}\cdot d\mathbf{r}=\int_0^\pi\bigl(-\sin^2t-\cos^2t\bigr)\,dt=\int_0^\pi(-1)\,dt=-\pi.$$

Both paths stay in the domain and have the same endpoints, but the work values differ. Thus there is no global potential on the punctured plane. The example shows why domain conditions cannot be skipped.

Worked Example: Unequal Cross-Partials and Path-Dependent Work

Worked Example: Unequal Cross-Partials and Path-Dependent Work

Let \(\mathbf{F}(x,y)=\langle y,0\rangle\) on \(\mathbb{R}^2\). Test for a potential and calculate work from \((0,0)\) to \((1,1)\) along \(\gamma(t)=(t,t^2)\), \(0\leq t\leq 1\).

Here \(P=y\), \(Q=0\), and the domain is the simply connected plane. The cross-partials are \(P_y=1\) and \(Q_x=0\), which are unequal. A potential cannot exist on \(\mathbb{R}^2\).

For the specified path, \(x=t\), \(y=t^2\), \(dx=dt\), and \(dy=2t\,dt\). Along the path \(P=t^2\) and \(Q=0\). Thus

$$\int_\gamma \mathbf{F}\cdot d\mathbf{r}=\int_0^1\bigl(t^2(1)+0(2t)\bigr)\,dt=\left[\frac{t^3}{3}\right]_0^1=\frac{1}{3}.$$

As a check of path dependence, travel first along the \(x\)-axis from \((0,0)\) to \((1,0)\), then vertically to \((1,1)\). On the horizontal segment \(y=0\), so the work is zero. On the vertical segment \(dx=0\) and \(Q=0\), so the work is also zero. The total is \(0\), different from \(1/3\) along \(\gamma\), as expected for a field without a potential.

Common Pitfall: Treating the Test as an Unconditional Guarantee

A common error is to calculate \(P_y=Q_x\) and immediately declare that a potential exists. For the punctured-plane field in the previous example, that calculation is correct, but the conclusion is not: the domain is not simply connected, and the two path integrals differ.

The correction is to separate what the test establishes from what the domain allows. Unequal cross-partials rule out a potential wherever the test's differentiability assumptions hold. Matching cross-partials establish a potential when the domain is simply connected; on a domain with a hole, further reasoning is needed. In either case, a proposed potential should be checked by differentiating it with respect to both variables.

Check Your Understanding

For each question, identify the domain conditions as well as the calculation needed.

  1. For \(\mathbf{F}=\langle 3x^2+y,\ x+2y\rangle\) on \(\mathbb{R}^2\), compare the cross-partials and recover a potential.
  2. If \(P_y\neq Q_x\) at one point of a domain, what does that imply about a potential on that domain?
  3. Why do matching cross-partials on the punctured plane not by themselves guarantee a global potential?
  4. If \(f\) is a potential for \(\mathbf{F}\), what endpoint values give the work along a path from \(A\) to \(B\)?
  5. For \(\mathbf{F}=\langle y,0\rangle\), compare the work from \((0,0)\) to \((1,1)\) along the diagonal line segment with the work along the two axis-parallel segments.

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