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Parametric Arc Length

You’ll be able to

Length of a parametric path

The arc length of a parametric curve from t = a to t = b sums the tiny hypotenuses √(dx² + dy²), now with both dx and dy expressed through t: L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt. This is the same Pythagorean idea as the y = f(x) arc-length formula, but symmetric in x and y because neither is privileged as the independent variable. It is literally the integral of the particle’s speed.

Parametric arc length
L = ∫ₐᵇ √( (dx/dt)² + (dy/dt)² ) dt
Identical to the integral of speed over time, so arc length = total distance traveled by the particle.

Arc length is distance traveled, not displacement

Because the integrand is the speed (always nonnegative), the parametric arc-length integral gives the total distance traveled along the path, even if the particle doubles back. This differs from displacement, the straight-line vector from start ⟨x(a), y(a)⟩ to end ⟨x(b), y(b)⟩. A particle that returns to its starting point has zero displacement but a positive arc length equal to the full path it walked.

Worked example

Set up and evaluate the arc length of x = cos(t), y = sin(t) for t from 0 to π.

  1. 1.Differentiate: dx/dt = −sin(t), dy/dt = cos(t).
  2. 2.Form the integrand: √((−sin t)² + (cos t)²) = √(sin²t + cos²t) = √1 = 1.
  3. 3.The arc length integral becomes L = ∫₀^π 1 dt.
  4. 4.Evaluate: [t]₀^π = π.
Answer: L = π. The curve is the unit circle traced at constant speed 1, so over t ∈ [0, π] it covers half the circumference, length π.
Tip

Whenever the speed simplifies to a constant (as with a circle traced at unit speed), the arc-length integral collapses to speed × time. Always simplify the radicand before integrating — a Pythagorean identity often reduces it dramatically.

Checkpoint

The arc length of a parametric curve from t = a to t = b is given by:

Checkpoint

A particle travels along a parametric path and returns exactly to its starting point. Which statement is true?

On the exam

Parametric arc length and "total distance traveled" are the same integral, ∫√((dx/dt)² + (dy/dt)²) dt. If a free-response question asks for the distance a particle travels in the plane, this is the formula — not the magnitude of the displacement vector.

Answer the 2 checkpoints as you read.

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