Math Core

How far does a particle travel as it moves along a curved path? For a straight segment you'd use the distance formula. For a curve, you chop it into tiny nearly straight pieces, use the distance formula on each, and add them up with an integral. Parametric equations make this especially natural.

From the distance formula to an integral

Suppose a particle is at $(x(t), y(t))$ at time $t$ . Over a short time $\Delta t$ , it moves about $\Delta x$ horizontally and $\Delta y$ vertically, so it covers a distance of roughly

\sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{\left(\frac{\Delta x}{\Delta t}\right)^2 + \left(\frac{\Delta y}{\Delta t}\right)^2}\;\Delta t.

As $\Delta t \to 0$ , the ratios become derivatives, and adding up all the pieces becomes an integral.

Arc length of a parametric curve

If $x'(t)$ and $y'(t)$ are continuous and the curve is traced exactly once as $t$ goes from $a$ to $b$ , its length is

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

The integrand $\sqrt{(x')^2 + (y')^2}$ is the speed of the particle, and integrating speed over time gives distance. You'll use exactly this idea again in the lesson on motion in the plane.

A check with a circle

Worked example: Half of a circle

Find the length of $x = 3\cos t$ , $y = 3\sin t$ for $0 \le t \le \pi$ .

$\dfrac{dx}{dt} = -3\sin t$ and $\dfrac{dy}{dt} = 3\cos t$ , so

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

The length is $\displaystyle\int_0^{\pi} 3\, dt = 3\pi$ . That's half the circumference of a circle of radius 3, as it should be.

Integrals you can do by hand

Most arc length integrals can't be done in closed form, so exam problems that ask for exact answers are carefully built so the square root simplifies. The usual trick is that the expression under the root has a common factor you can pull out, leaving something a $u$ -substitution can handle.

Worked example: Factoring under the radical

Find the length of $x = t^2$ , $y = \dfrac{2}{3}t^3$ for $0 \le t \le \sqrt{3}$ .

$\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 2t^2$ . Under the root:

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

Since $t \ge 0$ , $\sqrt{4t^2(1 + t^2)} = 2t\sqrt{1 + t^2}$ . Let $u = 1 + t^2$ , so $du = 2t\,dt$ ; $u$ runs from $1$ to $4$ :

L = \int_0^{\sqrt{3}} 2t\sqrt{1 + t^2}\, dt = \int_1^4 u^{1/2}\, du = \frac{2}{3}u^{3/2}\Big|_1^4 = \frac{2}{3}(8 - 1) = \frac{14}{3}.

Worked example: A slightly messier root

Find the length of $x = 1 + 3t^2$ , $y = 4 + 2t^3$ for $0 \le t \le 1$ .

$\dfrac{dx}{dt} = 6t$ and $\dfrac{dy}{dt} = 6t^2$ , so the speed is $\sqrt{36t^2 + 36t^4} = 6t\sqrt{1 + t^2}$ for $t \ge 0$ . With $u = 1 + t^2$ :

L = \int_0^1 6t\sqrt{1 + t^2}\, dt = 3\int_1^2 u^{1/2}\, du = 2u^{3/2}\Big|_1^2 = 2(2\sqrt{2} - 1) = 4\sqrt{2} - 2.

That's about $3.657$ . The constants $1$ and $4$ in $x$ and $y$ don't matter: shifting a curve doesn't change its length.

Common mistake

$\sqrt{a^2 + b^2}$ is not $a + b$ . You can't split the square root over the sum. Simplify inside the root first (factor, or use an identity like $\sin^2 t + \cos^2 t = 1$ ), then take the root.

Calculator arc length

On the calculator-active part of the AP exam, you set up the integral and let the calculator evaluate it. Your written work should show the integral with the correct integrand and limits; the number alone doesn't earn full credit.

Worked example: Length of the looping curve

Find the length of the curve $x = t^2$ , $y = t^3 - 3t$ for $-2 \le t \le 2$ .

$\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2 - 3$ , so

L = \int_{-2}^{2} \sqrt{(2t)^2 + (3t^2 - 3)^2}\, dt \approx 15.209.

This includes the loop and both tails. There's no nice antiderivative, so the calculator is the right tool.

The whole curve x = t², y = t³ − 3t for −2 ≤ t ≤ 2 has length about 15.209.Open in grapher →

Tip

Arc length depends on the interval of $t$ , not just the shape. If the parameter interval makes the particle go around a closed curve twice, the integral counts the length twice. For example, $x = \cos t$ , $y = \sin t$ on $0 \le t \le 4\pi$ gives $4\pi$ , twice the circumference. Always check that the curve is traced once if the question asks for the length of the curve itself.