Vector Fields and Line Integrals
Attach a vector to every point of a region, then integrate along a curve to accumulate work done by the field or mass along a wire.
What a learner can do afterwards
- Sketch a simple vector field and describe its flow
- Parameterise a curve and evaluate the work integral along it
- Distinguish integrating a scalar along a curve from integrating a field along it
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A vector field assigns a vector to each point. For F(x, y) = (P(x, y), Q(x, y)), evaluate both components at the point. The first is horizontal; the second is vertical. On standard axes, positive x points right and positive y points up. Draw sampled vectors with their tails at their points, using one arrow scale. Magnitude gives speed for a velocity field and force strength for a force field.
A parametrisation r(t) = (x(t), y(t)), a <= t <= b, lists points in travel order. Check its endpoints and curve equation. For r(t) = (2t, t²), t = 1/4 gives (1/2, 1/16). Differentiate each component: constants give 0, ct gives c, and ct² gives 2ct. Here r'(t) = (2, 2t). A segment from A to B can use r(t) = A + t(B - A), 0 <= t <= 1.
For continuous F on a piecewise smooth oriented curve r(t), a <= t <= b, use ∫C F dot dr = ∫a^b F(r(t)) dot r'(t) dt. Here dr = r'(t) dt includes displacement direction and length. In two dimensions, the dot product is P x' + Q y'. For a force field, this integral is work. Aligned force contributes positively, opposing force negatively, and perpendicular force zero. Integrating |F| ds loses this signed tangential contribution.
For continuous f along a piecewise smooth curve, ∫C f ds = ∫a^b f(r(t)) |r'(t)| dt. Speed |r'| is the square root of the sum of squared derivative components. For nonnegative mass per unit length f, this gives wire mass. With r(t) = (0, 3t), 0 <= t <= 1, and f = 1 + y, speed is 3 and the integrand is 3 + 9t. Its antiderivative is 3t + 9t²/2, giving mass 15/2.
Reverse the same traversal while keeping the field or scalar function fixed. Each displacement reverses, so ∫C F dot dr is multiplied by -1. In ∫C f ds, the function values and positive arc lengths stay the same, so the result is unchanged. These statements assume the integrals exist and no part of the traversal is added or removed.
Let F(x, y) = (2y, x) and r(t) = (t, t²), 0 <= t <= 2. Substitution gives F(r(t)) = (2t², t). Differentiating gives r'(t) = (1, 2t). Their dot product is 2t² × 1 + t × 2t = 4t². An antiderivative is 4t³/3. Evaluate at the parameter endpoints: 4 × 2³/3 - 0 = 32/3. This is the work if F represents force.
In a velocity-field sketch, arrows pointing outward around a point suggest a source pattern; inward arrows suggest a sink pattern. Nonzero arrows tangent to circles around a point show a rotational pattern. These describe the sampled directions. A zero vector has no direction. Arrow lengths compare speeds only when the drawing uses a common scale.
Evaluate the field on the curve. For force-field work, integrate F dot dr. For wire mass, integrate linear density against arc length ds.
This lesson, practice questions and flashcards were written for this dataset.
Additional readings and practice questions were written for this dataset. Linked videos are Khan Academy's "Vector fields, introduction" and "Line integrals and vector fields". Their reuse terms have not been verified.
More to read
Converting wire density into mass
Linear density is mass per unit length. A short wire piece with nearly constant density has mass approximately equal to density times length. Adding these estimates over smaller pieces leads to ∫C f ds.
A parameter interval need not measure length. For a wire described by r(t), use ds = |r'(t)| dt. Thus the mass calculation is ∫a^b f(r(t)) |r'(t)| dt. Evaluate the density on the wire and multiply by the magnitude of the curve derivative.
For example, r(t) = (6t, 8t), 0 <= t <= 1, has speed 10. If its linear density is 2 + y, the density on the wire is 2 + 8t. The mass integrand is 20 + 80t, and its integral is [20t + 40t²] from 0 to 1 = 60.
Reversing the same traversal preserves this scalar integral. Traversing part of the wire twice would count that part twice, so check that the parametrisation describes the intended wire once.
From a field formula to sampled arrows
The location supplied to a vector field and the vector it returns have different jobs. The location tells you where to place the arrow's tail. The returned components tell you the arrow's direction and magnitude.
Take F(x, y) = (y + 1, -2x). At (1, 2), its value is (3, -2): the arrow points right and down. At (-1, 2), its value is (3, 2): it points right and up. At (0, -1), the returned vector is zero, which has no direction.
Repeat this calculation at a grid of sample points. Choose one scale for all arrow lengths if you want their lengths to compare magnitudes. A picture using equal arrow lengths or colours needs an explanation of its convention.
Interpret the vectors using the quantity being modelled. A velocity arrow gives motion direction and speed. A force arrow gives force direction and strength; it does not by itself give the object's velocity.
Keeping direction in a work calculation
For a force field, work accumulates force dotted with displacement. On a parametrised path, displacement is r'(t) dt. Evaluate the field at r(t), take its dot product with r'(t), and integrate over the stated parameter interval.
Suppose F(x, y) = (y, -2x) and r(t) = (2t, t), 0 <= t <= 1. The field on the path is (t, -4t), and the derivative is (2, 1). Their dot product is 2t - 4t = -2t. Its integral is [-t²] from 0 to 1 = -1. The negative result means the field does net work against this motion.
The notation f ds is not restricted to wire mass. Work can also be written as ∫C (F dot T) ds, where T is the unit tangent in the travel direction. The scalar function F dot T contains the signed tangential force. Replacing it with |F| discards that information.
A tangential integral of a velocity field uses the same mathematics, but its value is not physical work.
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