Arc Length

How long is a curve? A string laid along the graph of $y = \sin x$ from $0$ to $\pi$ and then pulled straight would have some length, but you can’t measure it with a ruler. Calculus finds it the same way it finds area: chop the curve into tiny pieces, approximate each piece with something simple (here, a straight line), and add them up with an integral. Arc length is a BC-only application of integration, and on the AP exam it’s usually calculator active.

Key ideas

Where the formula comes from

Approximate the curve by short straight chords. Over a small step $\Delta x$, the curve rises by $\Delta y$, and the Pythagorean theorem gives the chord’s length:

$$\sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2}\,\Delta x$$

As the chords get shorter, $\dfrac{\Delta y}{\Delta x}$ becomes the derivative $f'(x)$, and the sum of the chords becomes an integral.

Four chords give a length of about $3.790$; the true length of $y = \sin x$ on $[0, \pi]$ is about $3.820$.

The arc length formula

If $f'$ is continuous on $[a, b]$, the length of the curve $y = f(x)$ from $x = a$ to $x = b$ is

$$L = \int_a^b \sqrt{1 + \big(f'(x)\big)^2}\,dx$$

For a curve written as $x = g(y)$, from $y = c$ to $y = d$, swap the roles of the variables:

$$L = \int_c^d \sqrt{1 + \big(g'(y)\big)^2}\,dy$$

Use the $dy$ version when the curve is easier to write as $x$ in terms of $y$, or when $\dfrac{dy}{dx}$ is undefined somewhere on the interval (a vertical tangent).

Setting up, then evaluating

The square root makes most arc length integrals impossible to do by hand. So on the AP exam, arc length almost always appears in the calculator-active section: write the integral with the correct derivative and limits, then let your calculator evaluate it and report 3 decimal places. A few special curves are built so that $1 + \big(f'(x)\big)^2$ is a perfect square, and those can be done exactly.

Two quick checks: arc length is always positive, and it is always at least the straight-line distance between the endpoints.

Curves given by parametric equations (and the total distance travelled by a moving particle) use a related formula; see parametric arc length.

Worked examples

Example 1: An exact answer

Find the length of $y = \tfrac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$.

Solution. $f'(x) = \tfrac{2}{3} \cdot \tfrac{3}{2}x^{1/2} = \sqrt{x}$, so $1 + \big(f'(x)\big)^2 = 1 + x$.

$$L = \int_0^3 \sqrt{1 + x}\,dx = \Big[\tfrac{2}{3}(1 + x)^{3/2}\Big]_0^3 = \tfrac{2}{3}\left(4^{3/2} - 1\right) = \tfrac{2}{3}(8 - 1) = \frac{14}{3}$$

Check: the endpoints are $(0, 0)$ and $(3, 2\sqrt{3})$, about $4.583$ apart in a straight line, and $\tfrac{14}{3} \approx 4.667$ is a bit longer. ✓

Example 2: A parabola (calculator)

Find the length of $y = x^2$ from $x = 0$ to $x = 1$.

Solution. $f'(x) = 2x$, so

$$L = \int_0^1 \sqrt{1 + 4x^2}\,dx \approx 1.479$$

(by calculator). The straight-line distance from $(0, 0)$ to $(1, 1)$ is $\sqrt{2} \approx 1.414$, a little less. ✓

Example 3: One arch of a sine curve

Find the length of $y = \sin x$ from $x = 0$ to $x = \pi$. (Radians, as always in calculus.)

Solution. $f'(x) = \cos x$, so

$$L = \int_0^\pi \sqrt{1 + \cos^2 x}\,dx \approx 3.820$$

This matches the figure: the four chords add up to about $3.790$, just a little shorter than the curve, as chords always are.

Example 4: Integrating with respect to y

Find the length of the curve $x = y^2$ from $y = 0$ to $y = 2$.

Solution. Here $g(y) = y^2$ and $g'(y) = 2y$:

$$L = \int_0^2 \sqrt{1 + 4y^2}\,dy \approx 4.647$$

This is the same curve as $y = \sqrt{x}$ for $0 \le x \le 4$. But $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$ is undefined at $x = 0$ (the curve has a vertical tangent there), so the $dy$ version avoids an improper integral.

Common mistakes

Forgetting to square the derivative. The integrand is $\sqrt{1 + \big(f'(x)\big)^2}$. Writing $\sqrt{1 + f'(x)}$ is a common slip that costs the setup point.

Using f(x) instead of f’(x). The formula uses the derivative. For $y = x^2$, the integrand is $\sqrt{1 + 4x^2}$, not $\sqrt{1 + x^4}$.

Simplifying the square root wrongly. $\sqrt{1 + 4x^2}$ is not $1 + 2x$. A square root of a sum doesn’t split. Only simplify when the inside is truly a perfect square.

Mismatched limits. In the $dy$ version, the limits must be $y$-values. In Example 4, the limits are $y = 0$ to $y = 2$, not $x = 0$ to $x = 4$.

Rounding too early or too little. Let the calculator evaluate the whole integral and give 3 decimal places. Don’t round the derivative first.

Degree mode on a trig curve. With the calculator in degrees, $\displaystyle\int_0^\pi \sqrt{1 + \cos^2 x}\,dx$ comes out wrong. Use radians.

Practice

1. (Warm-up) Use the arc length formula to find the length of the line $y = 2x + 1$ from $x = 0$ to $x = 3$. Check with the distance formula.

Solution

$f'(x) = 2$, so

$$L = \int_0^3 \sqrt{1 + 4}\,dx = 3\sqrt{5} \approx 6.708$$

Check: the endpoints are $(0, 1)$ and $(3, 7)$, and $\sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$. ✓

2. (Warm-up) Write an integral for the length of $y = x^3$ from $x = 0$ to $x = 2$, then evaluate it with a calculator.

Solution

$f'(x) = 3x^2$, so $\big(f'(x)\big)^2 = 9x^4$:

$$L = \int_0^2 \sqrt{1 + 9x^4}\,dx \approx 8.630$$

3. (Core) Find the exact length of $y = \tfrac{4}{3}x^{3/2}$ from $x = 0$ to $x = 2$.

Solution

$f'(x) = 2x^{1/2}$, so $1 + \big(f'(x)\big)^2 = 1 + 4x$. With $u = 1 + 4x$, $dx = \tfrac{1}{4}\,du$:

$$L = \int_0^2 \sqrt{1 + 4x}\,dx = \Big[\tfrac{1}{6}(1 + 4x)^{3/2}\Big]_0^2 = \tfrac{1}{6}(27 - 1) = \frac{13}{3}$$

4. (Core) Find the length of $y = e^x$ from $x = 0$ to $x = 1$, to 3 decimal places.

Solution

$f'(x) = e^x$, so

$$L = \int_0^1 \sqrt{1 + e^{2x}}\,dx \approx 2.003$$

5. (Core) Find the length of the curve $x = \tfrac{1}{2}y^2$ from $y = 0$ to $y = 1$, to 3 decimal places.

Solution

$g'(y) = y$, so

$$L = \int_0^1 \sqrt{1 + y^2}\,dy \approx 1.148$$

6. (Core) The curve $y = \sqrt{4 - x^2}$ is part of a circle of radius $2$. Find the exact length of the curve from $x = 0$ to $x = 1$, and check your answer with geometry.

Solution

$f'(x) = \dfrac{-x}{\sqrt{4 - x^2}}$, so

$$1 + \big(f'(x)\big)^2 = 1 + \frac{x^2}{4 - x^2} = \frac{4}{4 - x^2}$$$$L = \int_0^1 \frac{2}{\sqrt{4 - x^2}}\,dx = \Big[2\arcsin\frac{x}{2}\Big]_0^1 = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \approx 1.047$$

Geometry check: from $(0, 2)$ to $(1, \sqrt{3})$ the radius turns through $\tfrac{\pi}{6}$ radians (from $90^\circ$ to $60^\circ$), and arc length is radius × angle $= 2 \cdot \tfrac{\pi}{6} = \tfrac{\pi}{3}$. ✓

7. (Core) A suspension bridge cable hangs in the shape $y = 0.05x^2$, where $x$ and $y$ are in metres, between towers at $x = -10$ and $x = 10$. Find the length of the cable to 3 decimal places.

Solution

$f'(x) = 0.1x$, so

$$L = \int_{-10}^{10} \sqrt{1 + 0.01x^2}\,dx \approx 22.956 \text{ m}$$

That’s a bit more than the $20$ m straight span between the towers, as it should be.

8. (Challenge) Find the exact length of $y = \dfrac{x^3}{6} + \dfrac{1}{2x}$ from $x = 1$ to $x = 2$.

Solution

$f'(x) = \dfrac{x^2}{2} - \dfrac{1}{2x^2}$. Then

$$\begin{aligned} 1 + \big(f'(x)\big)^2 &= 1 + \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4} \\ &= \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2 \end{aligned}$$

a perfect square. So

$$L = \int_1^2 \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)dx = \Big[\frac{x^3}{6} - \frac{1}{2x}\Big]_1^2 = \left(\frac{4}{3} - \frac{1}{4}\right) - \left(\frac{1}{6} - \frac{1}{2}\right) = \frac{17}{12}$$

9. (Challenge) Let $s(x) = \displaystyle\int_0^x \sqrt{1 + 4t^2}\,dt$ be the length of $y = t^2$ from $t = 0$ to $t = x$.

• (a) Find $s'(x)$ and explain why $s$ is increasing.
• (b) Use a calculator to find the value $b$ for which the curve from $0$ to $b$ has length $3$. Give $b$ to 3 decimal places.

Solution

(a) By the Fundamental Theorem of Calculus, $s'(x) = \sqrt{1 + 4x^2}$. This is always positive (at least $1$), so $s$ is increasing: the length keeps growing as you go farther along the curve.

(b) Solve $\displaystyle\int_0^b \sqrt{1 + 4t^2}\,dt = 3$ with the calculator’s solver (or by graphing $y = s(x)$ and $y = 3$). This gives $b \approx 1.554$.