Parametric Arc Length- Complete Tutorial

What Is Parametric Arc Length?

Arc length measures the distance along a curve between two points. When a curve is defined parametrically—meaning x and y are expressed as functions of a third variable t—you need a different formula than the standard y = f(x) approach.

Parametric curves show up everywhere: projectile motion, robotics paths, computer graphics, and anywhere you describe motion with a parameter. If you're working with these, you'll need this formula.

The Parametric Arc Length Formula

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

S = ∫_a^b √((dx/dt)² + (dy/dt)²) dt

That's it. The integrand is the magnitude of the velocity vector. Think of it as the instantaneous speed of a point moving along the curve, integrated over time.

Why This Formula Works

You can derive this from the Pythagorean theorem. Over an infinitesimally small change dt:

• dx = (dx/dt) dt
• dy = (dy/dt) dt
• ds = √(dx² + dy²) = √((dx/dt)² + (dy/dt)²) dt

Integrate ds from a to b and you get the total distance traveled.

Step-by-Step Calculation

Example 1: Line Segment

Let's start simple. Given x = 2t + 1 and y = 3t - 2, find the arc length from t = 0 to t = 3.

Step 1: Find the derivatives

• dx/dt = 2
• dy/dt = 3

Step 2: Plug into the formula

S = ∫₀³ √(2² + 3²) dt = ∫₀³ √13 dt

Step 3: Evaluate

S = √13 · (3 - 0) = 3√13 ≈ 10.82

This checks out. The line goes from (1, -2) to (7, 7). Distance = √((7-1)² + (7-(-2))²) = √(36 + 81) = √117 = 3√13. Same answer.

Example 2: Circle

Parametric equation of a circle: x = cos(t), y = sin(t), from t = 0 to t = 2π.

Step 1: Derivatives

• dx/dt = -sin(t)
• dy/dt = cos(t)

Step 2: Plug in

S = ∫₀^2π √(sin²t + cos²t) dt = ∫₀^2π √1 dt = ∫₀^2π 1 dt

Step 3: Evaluate

S = 2π

That's the circumference of a unit circle. The formula works exactly as expected.

Example 3: Semicubical Parabola

Curve: x = t², y = t³ from t = 0 to t = 2.

Step 1: Derivatives

• dx/dt = 2t
• dy/dt = 3t²

Step 2: Set up the integral

S = ∫₀² √(4t² + 9t⁴) dt = ∫₀² t√(4 + 9t²) dt

Step 3: Solve using substitution

Let u = 4 + 9t², du = 18t dt, so t dt = du/18

When t = 0: u = 4

When t = 2: u = 4 + 36 = 40

S = ∫₄⁴⁰ √u · (1/18) du = (1/18) · [2/3 · u^3/2]₄⁴⁰

S = (1/27) · (40^3/2 - 4^3/2) = (1/27) · (40√40 - 8)

S ≈ (1/27) · (252.98 - 8) ≈ 9.07

Common Parametric Curves Reference

How to Actually Solve These Problems

The Method

1. Find dx/dt and dy/dt — differentiate each parametric equation with respect to t
2. Square them — (dx/dt)² and (dy/dt)²
3. Add and take the square root — this is your integrand
4. Set the limits — use the given t-values, not x or y values
5. Integrate — evaluate the definite integral

Common Integrals You'll See

• √(a² + b²t²) — use hyperbolic sine substitution or trig if a² is positive
• √(a²t² - b²) — trig substitution with secant
• √(t² ± constant) — complete the square, then trig or log substitution

Where Students Screw Up

1. Using x-limits instead of t-limits. Your integral is in terms of t. Always. The bounds come from the parameter.

2. Forgetting to square the derivatives. (dx/dt)² + (dy/dt)², not dx/dt + dy/dt. That's a common and costly mistake.

3. Not simplifying the integrand. √(4t² + 4t⁴) = 2|t|√(1 + t²). For positive t, this becomes 2t√(1 + t²). Don't leave it unsimplified.

4. Wrong substitution. When you do u-substitution, remember to change the limits. Or integrate with respect to u and switch back. Pick one approach and be consistent.

Arc Length vs. Speed

The integrand √((dx/dt)² + (dy/dt)²) is the speed of a particle moving along the curve. If you integrate speed over time, you get distance traveled.

This connects to physics directly. If you know the velocity vector v(t) = ⟨dx/dt, dy/dt⟩, the arc length is ∫|v(t)| dt.

For 3D curves with z = f(t), just add (dz/dt)² inside the square root:

S = ∫_a^b √((dx/dt)² + (dy/dt)² + (dz/dt)²) dt

Quick Practice Problem

Find the arc length of the curve x = t - sin(t), y = 1 - cos(t) from t = 0 to t = 2π.

Answer:

dx/dt = 1 - cos(t)

dy/dt = sin(t)

S = ∫₀^2π √((1 - cos t)² + sin²t) dt = ∫₀^2π √(2 - 2cos t) dt

Using 1 - cos t = 2sin²(t/2):

S = ∫₀^2π √(4sin²(t/2)) dt = ∫₀^2π 2|sin(t/2)| dt

From 0 to 2π, sin(t/2) is positive, so:

S = 2 · [-2cos(t/2)]₀^2π = -4(-1 - 1) = 8

This is the length of one arch of a cycloid. The answer is 8.

When There's No Closed Form

Some integrals don't have elementary antiderivatives. For example, the ellipse arc length requires elliptic functions—special functions with no simple formula.

In those cases:

• Leave the answer as an integral
• Use numerical integration (Simpson's rule, trapezoidal rule)
• Use software for approximations

Being able to set up the integral correctly is often the actual goal. The evaluation is sometimes just arithmetic.