Arc Length of Parametric Curves

The arc length of a parametric curve x = x(t), y = y(t) from t = a to t = b is L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt. The expression under the integral is the speed of a point moving along the curve, so the arc length is the distance the point travels if it traces the curve once.

Before this lesson: Derivatives of Parametric Equations and Arc Length of a Curve

The idea: add up tiny straight pieces

Zoom in on a curve and a tiny piece of it looks straight. If x changes by Δx and y changes by Δy, that piece has length √(Δx² + Δy²), by the Pythagorean theorem.

Over a short time Δt, Δx ≈ (dx/dt)Δt and Δy ≈ (dy/dt)Δt. So each piece is about √((dx/dt)² + (dy/dt)²) Δt long. Adding the pieces and letting Δt shrink turns the sum into an integral.

The square root is the speed: how fast the point moves along its path. Distance equals speed times time, and the integral adds it up when the speed keeps changing.

Arc length formula for parametric curves

L = \int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}\,dt

1. Find dx/dt and dy/dt.
2. Square each one and add them.
3. Simplify. The sum is often a perfect square, or it simplifies with sin² t + cos² t = 1.
4. Take the square root and integrate from t = a to t = b.
5. Check that the curve is traced only once on [a, b]. If it repeats, the integral counts the repeat too.

If the integral can’t be done by hand, use a calculator’s numerical integration. Many curves, such as most ellipses, have no simple formula for their length.

Arc length of parametric curves examples

Example 1: A line segment

Find the length of x = 3t, y = 4t for 0 ≤ t ≤ 2.

\dfrac{dx}{dt} = 3, \quad \dfrac{dy}{dt} = 4

3^2 + 4^2 = 25

\sqrt{25} = 5

L = \int_0^2 5\,dt = 5(2) - 5(0) = 10

Answer: L = 10. Check: the segment runs from (0, 0) to (6, 8), and the distance formula gives √(36 + 64) = 10.

Example 2: A circle

Find the length of x = 5 cos t, y = 5 sin t for 0 ≤ t ≤ 2π.

\dfrac{dx}{dt} = -5\sin t, \quad \dfrac{dy}{dt} = 5\cos t

25\sin^2 t + 25\cos^2 t

25(\sin^2 t + \cos^2 t) = 25

L = \int_0^{2\pi} 5\,dt = 10\pi

Answer: L = 10π ≈ 31.42, the circumference of a circle with radius 5, as expected.

Example 3: A curve that needs u-substitution

Find the length of x = t², y = (2/3)t³ for 0 ≤ t ≤ √3.

\dfrac{dx}{dt} = 2t, \quad \dfrac{dy}{dt} = 2t^2

(2t)^2 + (2t^2)^2 = 4t^2 + 4t^4

4t^2 + 4t^4 = 4t^2(1 + t^2)

\sqrt{4t^2(1 + t^2)} = 2t\sqrt{1 + t^2}

L = \int_0^{\sqrt{3}} 2t\sqrt{1 + t^2}\,dt

L = \int_1^{4} u^{1/2}\,du

L = \left[\tfrac{2}{3}u^{3/2}\right]_1^4

L = \tfrac{2}{3}(8) - \tfrac{2}{3}(1) = \dfrac{14}{3}

Answer: L = 14/3 ≈ 4.667.

Example 4: A spiral

Find the length of x = eᵗ cos t, y = eᵗ sin t for 0 ≤ t ≤ 1.

\dfrac{dx}{dt} = e^t(\cos t - \sin t)

\dfrac{dy}{dt} = e^t(\sin t + \cos t)

e^{2t}\left[(\cos t - \sin t)^2 + (\sin t + \cos t)^2\right]

e^{2t}\left[2\cos^2 t + 2\sin^2 t\right] = 2e^{2t}

\sqrt{2e^{2t}} = \sqrt{2}\,e^t

L = \int_0^1 \sqrt{2}\,e^t\,dt = \sqrt{2}(e - 1)

Answer: L = √2(e − 1) ≈ 2.430.

Example 5: A point on a rolling wheel (word problem)

A reflector sits on the rim of a bike wheel with radius 1 foot. As the wheel rolls, the reflector follows the cycloid x = t − sin t, y = 1 − cos t. How far does it travel during one full turn of the wheel, 0 ≤ t ≤ 2π?

\dfrac{dx}{dt} = 1 - \cos t, \quad \dfrac{dy}{dt} = \sin t

(1 - \cos t)^2 + \sin^2 t

1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t

2 - 2\cos t = 4\sin^2\dfrac{t}{2}

\sqrt{4\sin^2\dfrac{t}{2}} = 2\sin\dfrac{t}{2}

L = \int_0^{2\pi} 2\sin\dfrac{t}{2}\,dt = \left[-4\cos\dfrac{t}{2}\right]_0^{2\pi}

L = -4\cos\pi + 4\cos 0 = 4 + 4 = 8

Answer: 8 feet. The wheel itself only rolls forward 2π ≈ 6.28 feet, but the reflector travels farther because it also goes up and down.

Common mistakes

Related methods

Arc length of a graph

For a graph y = f(x), let x = t and y = f(t). The parametric formula becomes ∫ √(1 + (dy/dx)²) dx, the formula from arc length of a curve.

Speed and distance traveled

The integrand is speed. Integrating speed gives total distance, just as in displacement and total distance for motion on a line. Vector-Valued Functions, later in this unit, write the same speed as the length of the velocity vector.

Surface area and polar curves

The same arc-length piece appears in surface area of revolution. Arc Length of Polar Curves, at the end of this unit, uses it with x = r cos θ and y = r sin θ.

Practice problems

1. Find the length of x = 6t, y = 8t for 0 ≤ t ≤ 1.
2. Find the length of x = 2 cos t, y = 2 sin t for 0 ≤ t ≤ π.
3. Find the length of x = 3t², y = 2t³ for 0 ≤ t ≤ 1.
4. Find the length of the astroid piece x = cos³ t, y = sin³ t for 0 ≤ t ≤ π/2.
5. Find the length of x = t, y = (2/3)t^(3/2) for 0 ≤ t ≤ 3.
6. Find the length of x = eᵗ + e⁻ᵗ, y = 2t for 0 ≤ t ≤ 1.
7. For x = cos 2t, y = sin 2t on 0 ≤ t ≤ 2π, find the integral of the speed. Is it the length of the circle?
8. Find the length of half an arch of the cycloid, x = t − sin t, y = 1 − cos t for 0 ≤ t ≤ π.
9. A particle moves along x = 2t, y = t² for 0 ≤ t ≤ 2. How far does it travel?
10. Use a calculator to estimate the length of the ellipse x = 3 cos t, y = 2 sin t, 0 ≤ t ≤ 2π.

\sqrt{6^2 + 8^2} = \sqrt{100} = 10

L = \int_0^1 10\,dt = 10

4\sin^2 t + 4\cos^2 t = 4

L = \int_0^{\pi} 2\,dt = 2\pi

(6t)^2 + (6t^2)^2 = 36t^2(1 + t^2)

\sqrt{36t^2(1 + t^2)} = 6t\sqrt{1 + t^2}

L = \int_1^2 3u^{1/2}\,du

L = \left[2u^{3/2}\right]_1^2 = 2(2\sqrt{2}) - 2

L = 4\sqrt{2} - 2 \approx 3.657

\dfrac{dx}{dt} = -3\cos^2 t\sin t, \quad \dfrac{dy}{dt} = 3\sin^2 t\cos t

9\cos^4 t\sin^2 t + 9\sin^4 t\cos^2 t = 9\sin^2 t\cos^2 t

\sqrt{9\sin^2 t\cos^2 t} = 3\sin t\cos t

L = \left[\tfrac{3}{2}\sin^2 t\right]_0^{\pi/2} = \tfrac{3}{2}

1 + \left(t^{1/2}\right)^2 = 1 + t

L = \int_0^3 (1 + t)^{1/2}\,dt

L = \left[\tfrac{2}{3}(1 + t)^{3/2}\right]_0^3 = \tfrac{2}{3}(8 - 1) = \dfrac{14}{3}

(e^t - e^{-t})^2 + 2^2 = e^{2t} - 2 + e^{-2t} + 4

e^{2t} + 2 + e^{-2t} = (e^t + e^{-t})^2

L = \int_0^1 (e^t + e^{-t})\,dt = \left[e^t - e^{-t}\right]_0^1

L = e - e^{-1} \approx 2.350

4\sin^2 2t + 4\cos^2 2t = 4

\int_0^{2\pi} 2\,dt = 4\pi

L = \int_0^{\pi} 2\sin\dfrac{t}{2}\,dt

L = \left[-4\cos\dfrac{t}{2}\right]_0^{\pi} = 0 + 4 = 4

2^2 + (2t)^2 = 4(1 + t^2)

L = \int_0^2 2\sqrt{1 + t^2}\,dt

L = \left[t\sqrt{1 + t^2} + \ln\left(t + \sqrt{1 + t^2}\right)\right]_0^2

L = 2\sqrt{5} + \ln(2 + \sqrt{5}) \approx 5.916

L = \int_0^{2\pi} \sqrt{9\sin^2 t + 4\cos^2 t}\,dt

L \approx 15.865

Parametric arc length FAQ

What is the arc length formula for parametric equations?

L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt, where the curve is traced once as t goes from a to b.

How is it different from the arc length formula for y = f(x)?

The y = f(x) formula integrates √(1 + (dy/dx)²) with respect to x. The parametric formula integrates with respect to t. The first is a special case of the second, with x = t.

Is arc length the same as distance traveled?

Only if the curve is traced exactly once. If the point goes over part of the path again, the integral of speed counts that part twice, so it gives distance traveled, not the curve’s length.

Why can’t some arc length integrals be done by hand?

The square root often has no elementary antiderivative, as for most ellipses. Then a numerical method, like Simpson’s rule or a calculator, gives the length.

Summary

• Parametric arc length: L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt.
• The integrand is the speed along the curve.
• Square and add the derivatives, then look for a perfect square or sin² + cos² = 1.
• The limits are values of t, not x.
• Make sure the curve is traced only once on the interval.
• When there is no antiderivative, estimate the integral numerically.

Previous lesson: Derivatives of Parametric Equations

Next lesson: Vector-Valued Functions

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