Parameterize work along competing planar paths
Use the line integral to distinguish a selected interaction, a net force and a conservative field.
Learning outcome: Compare A=1,B=2 with A=B=2. Independently parameterize both legs, then reverse one path to build a closed-loop argument. Retain a unit audit and explain the role of unspecified forces.
Starting recap
Derive the integral or derivative, preserve the system boundary, and reject physically inaccessible continuations.
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Try
A=1,B=2 gives curved work 40/3 J, horizontal-first work 16 J and vertical-first work 8 J. For A=B=2 all are 16 J.
Open the experimentPredict
What makes these three path works agree within the supplied field family?
Experiment
Compare A=1,B=2 with A=B=2. Independently parameterize both legs, then reverse one path to build a closed-loop argument. Retain a unit audit and explain the role of unspecified forces.
Compare work along specified planar paths
Worked teaching model · synthetic values · Selected field F=(Ay,Bx). Paths connect (0,0) to (2,4) m: parabola y=x²/(1 m), or coordinate-aligned legs. Other path-maintaining interactions are unspecified; selected-field work alone is not necessarily net work.
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Parabolic line-integral integrand
Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.
Parabolic line-integral integrand: numerical coordinates
| parameter u (m) | dW/du (N) |
|---|---|
| 0 | 0 |
| 0.05 | 0.0125 |
| 0.1 | 0.05 |
| 0.15 | 0.1125 |
| 0.2 | 0.2 |
| 0.25 | 0.3125 |
| 0.3 | 0.45 |
| 0.35 | 0.6125 |
| 0.4 | 0.8 |
| 0.45 | 1.0125 |
| 0.5 | 1.25 |
| 0.55 | 1.5125 |
| 0.6 | 1.8 |
| 0.65 | 2.1125 |
| 0.7 | 2.45 |
| 0.75 | 2.8125 |
| 0.8 | 3.2 |
| 0.85 | 3.6125 |
| 0.9 | 4.05 |
| 0.95 | 4.5125 |
| 1 | 5 |
| 1.05 | 5.5125 |
| 1.1 | 6.05 |
| 1.15 | 6.6125 |
| 1.2 | 7.2 |
| 1.25 | 7.8125 |
| 1.3 | 8.45 |
| 1.35 | 9.1125 |
| 1.4 | 9.8 |
| 1.45 | 10.5125 |
| 1.5 | 11.25 |
| 1.55 | 12.0125 |
| 1.6 | 12.8 |
| 1.65 | 13.6125 |
| 1.7 | 14.45 |
| 1.75 | 15.3125 |
| 1.8 | 16.2 |
| 1.85 | 17.1125 |
| 1.9 | 18.05 |
| 1.95 | 19.0125 |
| 2 | 20 |
| Quantity | Value | Unit |
|---|---|---|
| Parabolic-path selected work | 13.3333 | J |
| Horizontal-then-vertical work | 16 | J |
| Vertical-then-horizontal work | 8 | J |
| Leg-path work difference | -8 | J |
| Conservative within this field family | 0 | flag |
Parameterize both position and its differential before taking the dot product.
On r(u)=(u,u²), dr=(1,2u)du and F·dr=(A+2B)u²du in the stipulated SI scaling. When A=B, U=−Axy has this force as its negative gradient. For unequal coefficients the paths give different work; no single position-only potential describes this field.
Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.
Parameterize both position and its differential before taking the dot product.
On r(u)=(u,u²), dr=(1,2u)du and F·dr=(A+2B)u²du in the stipulated SI scaling. When A=B, U=−Axy has this force as its negative gradient. For unequal coefficients the paths give different work; no single position-only potential describes this field.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
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Public practice checks a specific claim. It does not establish independent mastery.
Why does path dependence prevent a single potential difference from replacing this selected-force integral?
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