+ - * / ^. Functions: sin, cos, tan, exp, log/ln, sqrt, abs, pow(a,b), min(a,b), max(a,b). Constants: pi, e.
| Example | M(x,y) | N(x,y) | Exactness condition | One potential function F(x,y) | Implicit solution |
|---|---|---|---|---|---|
| A | 2*x*y + y^2 | x^2 + 2*x*y | ∂M/∂y = 2x+2y, ∂N/∂x = 2x+2y | F = x^2*y + x*y^2 | x^2*y + x*y^2 = C |
- Exactness: ∂M/∂y = ∂N/∂x on a region where M and N are smooth.
- Potential: If exact, there exists F(x,y) with dF = M dx + N dy, so F(x,y) = C.
- This tool: Central differences approximate partials, and Simpson’s rule estimates path integrals to compute F(P1) − F(P0).
- Type M(x,y) and N(x,y) with explicit multiplication.
- Set region bounds where you want to test exactness.
- Choose (x0,y0) and (x1,y1) for the path integral.
- Adjust h, samples, tolerance, and integration steps if needed.
- Press Submit to display results above the form.
- Use CSV or PDF buttons to export the latest report.
Exactness testing metrics
The calculator analyzes differential forms M(x,y)dx + N(x,y)dy by estimating the mismatch Δ = ∂M/∂y − ∂N/∂x. It reports maximum |Δ| and assigns a grade. As a rule of thumb, values near 1e-6 indicate strong agreement, 1e-4 is often acceptable for smooth inputs, and larger values suggest non‑exactness or domain issues.
Sampling density and coverage
Grid sampling converts a theoretical condition into evidence. With S samples on each axis, the tool evaluates S² points; S=9 means 81 checks, while S=21 means 441. If your functions change rapidly, increase S so peaks in |Δ| are not missed. For speed, confirm with one slightly larger grid. Use bounds that avoid zero denominators and roots.
Step size and derivative noise
Central differences approximate partial derivatives using a step h. Large h blurs curvature and inflates truncation error; tiny h can amplify floating‑point noise, especially near steep gradients. For many classroom expressions, h between 1e-4 and 1e-3 is stable. If the heatmap looks speckled, increase h slightly or avoid singular bounds.
Path integrals for verification
Exactness implies path independence for line integrals of the form ∫(M dx + N dy). This tool compares two rectangle paths between (x0,y0) and (x1,y1): first along x then y, and first along y then x. The displayed path difference quantifies consistency. If the difference shrinks when you increase integration steps, numerical error was the main cause.
Reading the potential estimate
When the form is exact, there exists a potential F such that dF = M dx + N dy and the implicit solution is F(x,y)=C. The calculator estimates F(x1,y1)−F(x0,y0) via the path integral. Use it to validate a derived F or to compare two candidate solutions. If the form looks non‑exact, treat the value as diagnostic, not a solution.
Exporting results responsibly
The CSV export captures inputs, bounds, step size, tolerance, and computed metrics for reproducible records. The PDF export summarizes the same information in a compact report suitable for notes or assignments. For best practice, save multiple runs when you tweak S, h, or bounds, then compare max |Δ| and path difference to ensure conclusions remain stable.
1) What condition makes a form exact?
On a smooth region, exactness requires ∂M/∂y = ∂N/∂x. Then a potential F exists with dF = M dx + N dy, giving F(x,y) = C.
2) Why does the tool show a heatmap?
The heatmap visualizes |∂M/∂y − ∂N/∂x| across your bounds. Uniformly low values support exactness, while spikes highlight locations where the condition fails or becomes unstable.
3) What if my expression has divisions or roots?
Choose bounds that keep denominators nonzero and radicands nonnegative. If the region crosses singular points, derivative estimates may explode, and the exactness conclusion can become misleading.
4) How do I tune h, samples, and steps?
Start with h=5e-4, samples=9, steps=200. If results change noticeably, increase samples and steps, then slightly adjust h upward or downward to reduce noise.
5) Why compare two path integrals?
Exact forms are path independent, so both rectangle paths should match closely. A small difference supports exactness; a persistent difference suggests non‑exactness or insufficient numeric resolution.
6) Does the potential value give the full solution?
It gives an estimate of F(x1,y1) − F(x0,y0). To write the full implicit solution, you still construct F(x,y) analytically and set F(x,y) = C.