Curve parametrizations and line integrals
Distinguish accumulated scalar density from oriented work and test when path independence is justified.
On this page
The question for today
What you’ll understand
- Compute speed and arc-length weighting from a curve parametrization.
- Evaluate an oriented vector line integral and track reversal.
- Use potentials carefully while recognizing a domain with a hole.
A wire can carry a varying density, while a force field can perform work on an object moving along the same wire. Both calculations involve a curve, but their integrals respond differently to orientation. A scalar density is weighted by distance traveled. Work pairs the field with an oriented displacement. Parametrization provides a common language for both while revealing exactly where speed, direction, and the curve's domain enter the calculation.
A parameter describes motion; speed supplies length
Let r(t), for a≤t≤b, be a continuously differentiable regular curve, meaning r′(t) never vanishes. Its speed is the norm of r′, and its length is the integral of that speed. Piecewise regular curves are handled by adding the contributions of their regular pieces. A scalar line integral of density ρ along the curve inserts ρ(r(t)) before the speed factor. This computes mass when ρ is mass per unit length. It counts the parametrized traversal, so tracing a segment twice counts its contribution twice.
An increasing differentiable reparametrization with nonzero derivative changes the rate of travel but preserves the scalar integral through the ordinary substitution rule. Reversing orientation also preserves it because speed is a norm. That invariance does not mean that any formula with the same endpoints describes the same curve: different paths can pass through different densities and have different lengths. Nor can the speed factor be omitted unless the parameter happens to be arc length. A unit-speed parametrization is a useful special case, not a default assumption.
Work retains orientation and can detect circulation
For a continuous vector field F, the work integral pairs its value with the infinitesimal oriented displacement. Reversing a curve changes the sign. Only the tangential field component contributes; a field normal to the instantaneous velocity does no work along that motion. On a closed curve, this integral is often called circulation. A nonzero circulation records a failure of global path independence, even when a plot of the field looks locally smooth. The actual parametrized traversal, including repeated turns, remains part of the data.
If F is the gradient of a C¹ potential φ on a neighborhood of the curve, the chain rule turns the integrand into the derivative of φ(r(t)). The fundamental theorem then gives only the endpoint potential difference, so every closed curve has zero work. This argument proves path independence for gradient fields without requiring the path to be straight. It also explains the sign convention: work by the gradient is final potential minus initial potential; physical force is sometimes defined as the negative gradient of potential energy, changing that interpretation.
Curl-free needs a domain qualification
In the plane, a C¹ field (P,Q) with Qx−Py=0 on a simply connected open domain has a potential. A hole can obstruct this conclusion. The field (−y/(x²+y²),x/(x²+y²)) is curl-free on the punctured plane, but its counterclockwise unit-circle work is 2π. The field is not defined at the origin, so the usual no-hole hypothesis is absent. This is a topological obstruction to a single-valued global potential, not an arithmetic contradiction in the local derivative test.
A practical line-integral solution begins by declaring whether the integral is scalar or vector and whether orientation matters. Then check the curve's parameter range, regular pieces, and any excluded field points. Search for a potential only after checking its domain; otherwise use the defining parameter integral. Endpoints alone are enough precisely when path independence has been established. A closed-curve result of zero for one chosen loop is not proof that every loop has zero circulation, just as one successful numerical test does not establish a global theorem.
Work it through
One circle, two different kinds of accumulation
On the counterclockwise unit circle r(t)=(cos t,sin t), 0≤t≤2π, compute the scalar integral of ρ=2+x and the work of F=(−y,x).
The derivative is (−sin t,cos t), whose norm is one. The scalar integral is ∫₀²π(2+cos t)dt=4π; the cosine term contributes zero.
Along this curve, F(r(t))=(−sin t,cos t), identical to the velocity. Their dot product is one, so the vector integral is 2π.
Reversing the parametrization leaves speed and scalar mass unchanged but reverses the oriented velocity. The scalar result remains 4π and the work becomes −2π.
The same geometric circle carries an unoriented scalar accumulation and an oriented circulation. Their numerical values and reversal rules differ.
The living notebook
Explore the model
Interactive lab · fields
Separate radial and tangential field components
Reset, set source strength q=0 and rotation Ω=1. Predict that arrows on a counterclockwise circle are tangential and contribute positive work. For a fresh comparison, Reset, set q=1 and Ω=0: arrows are radial and have zero tangential work on that circle. The source is softened with denominator r²+0.2; this field diagram does not numerically integrate a chosen path.
- Fx at (1, 0)
- 0.833
- Fy at (1, 0)
- 0.5
- Field magnitude there
- 0.972
The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forWork uses the field component along oriented motion, while a radial component contributes no work on a centered circle.
Pause & reason
Check your understanding
Your turn
Try it yourself
For F=(2xy,x²), compute the work from (0,1) to (2,3) along any piecewise C¹ path in the plane. Give a potential and explain why no path equation is needed.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Integrate the first field component with respect to x, then check the y derivative of the candidate.
Compare with the solution
The function φ(x,y)=x²y has gradient (2xy,x²) everywhere. It is C¹ on the entire plane, so the work equals φ(2,3)−φ(0,1)=12. A curve formula is unnecessary because the global potential establishes path independence. If a constant is added to φ, it cancels from the endpoint difference; potentials are not unique, but the work is.
Keep the idea
Take it with you
- Scalar line integrals use speed; vector line integrals use oriented velocity.
- A global potential reduces work to an endpoint difference.
- A curl-free test needs domain hypotheses before it implies global path independence.
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