Arc Length of Parametric Curves

How long is a curved path? If you walk along a winding trail, the distance on your step counter is longer than the straight-line distance between the start and the finish. For a curve given by parametric equations, one integral measures that curved length. It’s also exactly how you find the total distance travelled by a particle moving in the plane.

Key ideas

Where the formula comes from

Chop the curve into many tiny pieces. Each piece is almost a straight segment, the hypotenuse of a right triangle with legs $\Delta x$ and $\Delta y$:

$$\Delta s \approx \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{\left( \frac{\Delta x}{\Delta t} \right)^2 + \left( \frac{\Delta y}{\Delta t} \right)^2}\ \Delta t$$

Adding up all the pieces and letting $\Delta t \to 0$ turns the sum into an integral.

Each short chord has length $\sqrt{(\Delta x)^2 + (\Delta y)^2}$. More, shorter chords give a better estimate of the curve’s length.

The arc length formula

If $x(t)$ and $y(t)$ have continuous derivatives 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 is always $\ge 0$, so the length is never negative. (For a graph $y = f(x)$, the same idea gives $\int_a^b \sqrt{1 + (f'(x))^2}\, dx$; see arc length.)

Distance travelled vs displacement

If $(x(t), y(t))$ is the position of a particle at time $t$, then $\sqrt{(x'(t))^2 + (y'(t))^2}$ is its speed. So the same integral is the total distance travelled:

	What it measures	How to find it
Total distance travelled	every bit of the path, as a positive length	$\displaystyle\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\, dt$
Displacement	the change in position, from start to finish	the vector $\langle x(b) - x(a),\ y(b) - y(a) \rangle$

The straight-line distance between the start and the finish is the length of the displacement vector, and it’s never more than the distance travelled. If the particle goes around a curve more than once, the distance travelled counts every lap, even though the curve itself has the same length.

Exact or calculator?

The square root makes most arc length integrals impossible to find by hand. On the AP exam, arc length almost always appears on the calculator-active section: write the integral, then evaluate it numerically (in radian mode) and give $3$ decimal places. A few curves are designed so that $(x')^2 + (y')^2$ is a perfect square, or so that a u-substitution works. Those can be done exactly.

Worked examples

Example 1: A line segment, checked two ways

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

Solution. $\dfrac{dx}{dt} = 3$ and $\dfrac{dy}{dt} = 4$, so

$$L = \int_0^2 \sqrt{3^2 + 4^2}\, dt = \int_0^2 5\, dt = 10$$

Check: the curve is the segment from $(0, 0)$ to $(6, 8)$, and the distance formula gives $\sqrt{36 + 64} = 10$.

Example 2: A circle, once and twice

Find the length of $x = 2\cos t$, $y = 2\sin t$ for $0 \le t \le 2\pi$. What changes if $0 \le t \le 4\pi$?

Solution. $\left( \dfrac{dx}{dt} \right)^2 + \left( \dfrac{dy}{dt} \right)^2 = 4\sin^2 t + 4\cos^2 t = 4$, so

$$L = \int_0^{2\pi} \sqrt{4}\, dt = 2(2\pi) = 4\pi$$

That’s the circumference of a circle of radius $2$, as it should be.

For $0 \le t \le 4\pi$, the point goes around the circle twice. The integral gives $8\pi$: that’s the distance travelled, but the curve itself is still only $4\pi$ long. The arc length formula measures the curve only when it is traced once.

Example 3: An exact answer by u-substitution

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

Solution. $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 2t^2$, so

$$\sqrt{4t^2 + 4t^4} = \sqrt{4t^2(1 + t^2)} = 2t\sqrt{1 + t^2} \qquad (t \ge 0)$$

Let $u = 1 + t^2$, $du = 2t\, dt$. When $t = 0$, $u = 1$; when $t = \sqrt{3}$, $u = 4$.

$$L = \int_0^{\sqrt{3}} 2t\sqrt{1 + t^2}\, dt = \int_1^4 \sqrt{u}\, du = \left[ \frac{2}{3}u^{3/2} \right]_1^4 = \frac{2}{3}(8 - 1) = \frac{14}{3}$$

Check: the straight line from $(0, 0)$ to $(3, 2\sqrt{3})$ has length $\sqrt{9 + 12} = \sqrt{21} \approx 4.583$, a little less than $\dfrac{14}{3} \approx 4.667$. Good: the curve is slightly longer than the chord.

Example 4: Calculator active

A particle moves in the plane with $x(t) = t^2 - 4t$ and $y(t) = 2\sin(1.5t)$ for $0 \le t \le 4$ (calculator in radian mode). Find the total distance travelled, and the particle’s displacement.

Solution.

$$\text{distance} = \int_0^4 \sqrt{(2t - 4)^2 + \big( 3\cos(1.5t) \big)^2}\ dt \approx 11.878$$

Displacement: the particle starts at $(0, 0)$ and ends at $(16 - 16,\ 2\sin 6) \approx (0, -0.559)$. The displacement vector is $\langle 0,\ 2\sin 6 \rangle \approx \langle 0, -0.559 \rangle$, so the particle ends up only about $0.559$ units from where it started, even though it travelled almost $12$ units.

Common mistakes

Forgetting to square the derivatives. The integrand is $\sqrt{(x')^2 + (y')^2}$, not $\sqrt{x' + y'}$ and not $x' + y'$.

Splitting the square root. $\sqrt{a^2 + b^2}$ is not $a + b$. In Example 1, $\sqrt{9 + 16} = 5$, not $3 + 4 = 7$.

Using the wrong interval when the curve is retraced. If the curve is traced more than once on $[a, b]$, the integral gives the distance travelled, not the length of the curve. For a curve’s length, integrate over one trip only.

Confusing distance with displacement. Distance travelled is an integral of speed and is always positive. Displacement is the change in position, $\langle x(b) - x(a),\ y(b) - y(a) \rangle$. AP questions often ask for both, so read each part carefully.

Not showing the integral on calculator questions. AP graders need to see the setup, such as $\int_0^4 \sqrt{(x'(t))^2 + (y'(t))^2}\, dt$, before the decimal answer. A bare number usually loses the setup point.

Calculator in degree mode. With trig functions in $x(t)$ or $y(t)$, degree mode gives a wrong value with no warning.

Practice

1. (Warm-up) Find the length of $x = 1 + 6t$, $y = 2 - 8t$ for $0 \le t \le 1$.

Solution

$$L = \int_0^1 \sqrt{6^2 + (-8)^2}\, dt = \int_0^1 10\, dt = 10$$

Check: from $(1, 2)$ to $(7, -6)$ is $\sqrt{36 + 64} = 10$.

2. (Warm-up) Find the length of $x = 3\cos t$, $y = 3\sin t$ for $0 \le t \le \pi$. What shape is this?

Solution

$(x')^2 + (y')^2 = 9\sin^2 t + 9\cos^2 t = 9$, so

$$L = \int_0^{\pi} 3\, dt = 3\pi$$

It’s the top half of a circle of radius $3$: half of the circumference $6\pi$.

3. (Warm-up) Write, but do not evaluate by hand, the integral for the length of $x = t^2$, $y = \sin t$ for $0 \le t \le \pi$. Then use a calculator to evaluate it.

Solution

$$L = \int_0^{\pi} \sqrt{4t^2 + \cos^2 t}\ dt \approx 10.354$$

4. (Core) Find the exact length of $x = t^2$, $y = t^3$ for $0 \le t \le 2$.

Solution

$\sqrt{4t^2 + 9t^4} = t\sqrt{4 + 9t^2}$ for $t \ge 0$. Let $u = 4 + 9t^2$, $du = 18t\, dt$; $u$ runs from $4$ to $40$.

$$L = \int_0^2 t\sqrt{4 + 9t^2}\, dt = \frac{1}{18}\int_4^{40} \sqrt{u}\, du = \frac{1}{27}\Big[ u^{3/2} \Big]_4^{40} = \frac{40\sqrt{40} - 8}{27} = \frac{80\sqrt{10} - 8}{27} \approx 9.073$$

5. (Core) Find the exact length of $x = e^t\cos t$, $y = e^t\sin t$ for $0 \le t \le 1$.

Solution

By the product rule, $x' = e^t(\cos t - \sin t)$ and $y' = e^t(\sin t + \cos t)$.

$$(x')^2 + (y')^2 = e^{2t}\big[ (\cos t - \sin t)^2 + (\sin t + \cos t)^2 \big] = e^{2t}(2) = 2e^{2t}$$

(The cross terms $\mp 2\sin t\cos t$ cancel, and $\sin^2 t + \cos^2 t = 1$ in each bracket.)

$$L = \int_0^1 \sqrt{2}\, e^t\, dt = \sqrt{2}(e - 1) \approx 2.430$$

6. (Core) Find the exact length of $x = t^3 - 3t$, $y = 3t^2$ for $0 \le t \le 2$.

Solution

$x' = 3t^2 - 3$ and $y' = 6t$:

$$(x')^2 + (y')^2 = 9t^4 - 18t^2 + 9 + 36t^2 = 9t^4 + 18t^2 + 9 = 9(t^2 + 1)^2$$

A perfect square, so the root is $3(t^2 + 1)$:

$$L = \int_0^2 3(t^2 + 1)\, dt = \Big[ t^3 + 3t \Big]_0^2 = 8 + 6 = 14$$

7. (Core) (Calculator active.) A particle moves with $x(t) = t^2 - 3t$ and $y(t) = \ln(1 + t^2)$ for $0 \le t \le 3$. Find the total distance travelled, and the distance between the particle’s starting and ending points.

Solution

$x'(t) = 2t - 3$ and $y'(t) = \dfrac{2t}{1 + t^2}$.

$$\text{distance travelled} = \int_0^3 \sqrt{(2t - 3)^2 + \left( \frac{2t}{1 + t^2} \right)^2}\ dt \approx 5.377$$

The particle starts at $(0, 0)$ and ends at $(9 - 9,\ \ln 10) = (0, \ln 10)$. The distance between them is $\ln 10 \approx 2.303$, much less than the distance travelled, because the particle moves left and then back right.

8. (Challenge) Find the length of one arch of the cycloid $x = t - \sin t$, $y = 1 - \cos t$, $0 \le t \le 2\pi$. (Hint: $1 - \cos t = 2\sin^2\frac{t}{2}$.)

Solution

$$(x')^2 + (y')^2 = (1 - \cos t)^2 + \sin^2 t = 1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t$$

Using the hint, $2 - 2\cos t = 4\sin^2\dfrac{t}{2}$, so the root is $2\left\lvert \sin\dfrac{t}{2} \right\rvert = 2\sin\dfrac{t}{2}$ (it’s non-negative for $0 \le t \le 2\pi$).

$$L = \int_0^{2\pi} 2\sin\frac{t}{2}\, dt = \left[ -4\cos\frac{t}{2} \right]_0^{2\pi} = -4(-1) + 4(1) = 8$$

9. (Challenge) The astroid $x = \cos^3 t$, $y = \sin^3 t$, $0 \le t \le 2\pi$, is a four-pointed star. Find its total length. (Watch the square root carefully.)

Solution

$x' = -3\cos^2 t\sin t$ and $y' = 3\sin^2 t\cos t$:

$$(x')^2 + (y')^2 = 9\sin^2 t\cos^2 t(\cos^2 t + \sin^2 t) = 9\sin^2 t\cos^2 t$$

So the integrand is $3\lvert \sin t\cos t \rvert$, with absolute value. Integrating $3\sin t\cos t$ over $[0, 2\pi]$ would wrongly give $0$.

By symmetry, the four quarters have equal length. On $\left[ 0, \tfrac{\pi}{2} \right]$, $\sin t\cos t \ge 0$:

$$L = 4\int_0^{\pi/2} 3\sin t\cos t\, dt = 12\left[ \frac{\sin^2 t}{2} \right]_0^{\pi/2} = 12 \cdot \frac{1}{2} = 6$$