Arc Length – AP Calculus

The arc length formula allows us to calculate the length of a curve between two points. By summing up infinitely many tiny straight line segments (using the Pythagorean theorem), we derive an integral that gives the exact length. This is a BC Calculus topic.

🎯 In this section you will learn

The arc length formula: $L = \int_a^b \sqrt{1 + [f'(x)]^2} dx$
How to derive the formula from the distance formula
How to find arc length for functions given as $y = f(x)$
Arc length for functions given as $x = g(y)$

📌 The Arc Length Formula

$ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx $$ Length of the curve $y = f(x)$ from $x=a$ to $x=b

💡 Derivation

On a small interval $dx$, the change in $y$ is $dy = f'(x) dx$. The small arc length $ds$ is approximately $\sqrt{dx^2 + dy^2} = \sqrt{1 + (dy/dx)^2} dx$. Summing (integrating) gives the total length.

📊 Visualizing Arc Length

Interactive graph below. The curve length is approximated by the sum of small straight segments:

Example 1Arc Length of a Linear Function

Find the arc length of $f(x) = 2x + 1$ from $x=0$ to $x=3$.

①

Find $f'(x)$

$f'(x) = 2$

②

Set up the integral

$L = \int_0^3 \sqrt{1 + (2)^2} dx = \int_0^3 \sqrt{5} dx$

③

Evaluate

$L = \sqrt{5} \cdot 3 = 3\sqrt{5} \approx 6.708$

Example 2Arc Length of a Parabola

Find the arc length of $f(x) = x^2$ from $x=0$ to $x=1$.

①

Find $f'(x)$

$f'(x) = 2x$

②

Set up the integral

$L = \int_0^1 \sqrt{1 + (2x)^2} dx = \int_0^1 \sqrt{1 + 4x^2} dx$

③

Evaluate (using formula or trig substitution)

$\int \sqrt{1+4x^2} dx = \frac{x}{2}\sqrt{1+4x^2} + \frac{1}{4}\ln\left|2x + \sqrt{1+4x^2}\right| + C$
$L = \left[\frac{x}{2}\sqrt{1+4x^2} + \frac{1}{4}\ln\left(2x + \sqrt{1+4x^2}\right)\right]_0^1$
$= \frac{1}{2}\sqrt{5} + \frac{1}{4}\ln(2+\sqrt{5}) - 0 \approx 1.4789$

Example 3Arc Length with $x = g(y)$

Find the arc length of $f(x) = x^{2/3}$ from $x=0$ to $x=8$.

①

Find $f'(x)$

$f'(x) = \frac{2}{3}x^{-1/3}$

②

Set up the integral

$L = \int_0^8 \sqrt{1 + \frac{4}{9}x^{-2/3}} dx$ — this integral is messy in terms of $x$.

③

Rewrite in terms of $y$

$y = x^{2/3}$ ⇒ $x = y^{3/2}$, $dx = \frac{3}{2} y^{1/2} dy$
Limits: when $x=0$, $y=0$; when $x=8$, $y=4$

④

Arc length formula in terms of $y$

$L = \int_c^d \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy = \int_0^4 \sqrt{1 + \left(\frac{3}{2} y^{1/2}\right)^2} dy = \int_0^4 \sqrt{1 + \frac{9}{4} y} dy$

⑤

Evaluate

Let $u = 1 + \frac{9}{4} y$, $du = \frac{9}{4} dy$ ⇒ $dy = \frac{4}{9} du$
When $y=0$, $u=1$; when $y=4$, $u=10$
$L = \int_1^{10} \sqrt{u} \cdot \frac{4}{9} du = \frac{4}{9} \cdot \frac{2}{3} [u^{3/2}]_1^{10} = \frac{8}{27}(10\sqrt{10} - 1)$

Example 4Arc Length of Sine

Find the arc length of $f(x) = \sin x$ from $x=0$ to $x=2\pi$.

$f'(x) = \cos x$
$L = \int_0^{2\pi} \sqrt{1 + \cos^2 x} dx$
This integral cannot be expressed in elementary functions — it's an elliptic integral. On the AP exam, you would set up the integral but not evaluate it.

⚠️ Common Mistakes

Forgetting the square root: The formula is $\sqrt{1 + (f')^2}$, not $1 + (f')^2$.
Confusing with surface area: Arc length is $\int \sqrt{1 + (f')^2} dx$; surface area of revolution is $2\pi \int f(x) \sqrt{1 + (f')^2} dx$.
Incorrect limits: Make sure the limits of integration match the variable of integration.
Not simplifying when possible: Sometimes rewriting in terms of $y$ makes the integral easier.

🔍 Key Takeaways

Arc length formula: $L = \int_a^b \sqrt{1 + [f'(x)]^2} dx$.
For $x = g(y)$, $L = \int_c^d \sqrt{1 + [g'(y)]^2} dy$.
The formula comes from the Pythagorean theorem: $ds = \sqrt{dx^2 + dy^2}$.
Many arc length integrals require trigonometric substitution or other techniques.
On the AP exam, you may only need to set up the integral, not evaluate it.

← Back to Module Page

📏 Arc Length (BC)

📌 The Arc Length Formula

📊 Visualizing Arc Length