Arc Length Calculator

Q: What is the arc length formula?

For y = f(x) on [a,b]: L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx. For a parametric curve (x(t), y(t)): L = \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt.

Q: What is the arc length of a circle?

Use the parametric mode with x = r\cos t, y = r\sin t, 0 \le t \le 2\pi: L = \int_0^{2\pi} r\,dt = 2\pi r. Try it with r = 1 to get 2\pi.

Q: How do I find arc length in polar coordinates?

L = \int_\alpha^\beta \sqrt{r^2 + (dr/d\theta)^2}\,d\theta. Equivalently, use parametric mode with x = r(\theta)\cos\theta and y = r(\theta)\sin\theta.

Measure the length of a curve between two points, for y = f(x) or a parametric curve. You get the √(1 + f′²) integral, the exact value when one exists, and a polygon that closes in on the true length.

Curve y = f(x)

Reads as

x^{\frac{3}{2}}

Try:

x fromto

Polygon segments: 4

Arc length

L = 9.07341529

4-segment polygon: 9.04531

Pink polygon: straight pieces whose total length approaches L as you add segments.

Step by step

Differentiate
$f'(x) = \frac{3}{2}\sqrt{x}$
Set up the arc length integral
L = ∫ √(1 + f′(x)²) dx.
$L = \int_{0}^{4} \sqrt{\frac{9}{4}x + 1}\,dx$
Antiderivative
$\int \sqrt{\frac{9}{4}x + 1}\,dx = \frac{8}{27}\left(\frac{9}{4}x + 1\right)^{\frac{3}{2}}$
Evaluate
$L = 9.07341529$

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An AI tutor reads the problem and steps above and explains them in plain words. It can make mistakes, so check it against the worked steps.

Why the formula has a square root

Zoom in on a tiny piece of the curve: it is almost a straight segment with horizontal run $dx$ and vertical rise $dy = f'(x)\,dx$ . By Pythagoras its length is $\sqrt{dx^2 + dy^2} = \sqrt{1 + f'(x)^2}\,dx$ . Adding every piece is an integral, and the polygon on the graph is that sum with a finite number of pieces.

L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx

Worked example

For $y = x^{3/2}$ on $[0, 4]$ : $y' = \tfrac32 x^{1/2}$ , so $1 + y'^2 = 1 + \tfrac94 x$ and $L = \int_0^4 \sqrt{1 + \tfrac94 x}\,dx = \tfrac{8}{27}\left(10^{3/2} - 1\right) \approx 9.073$ . This is the default example above.

The derivative step uses the derivative calculator, and the integral the integral calculator.

Questions students ask

What is the arc length formula?

For $y = f(x)$ on $[a,b]$ : $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$ . For a parametric curve $(x(t), y(t))$ : $L = \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt$ .

Where does the formula come from?

Chop the curve into tiny straight pieces. A piece with horizontal change $dx$ and vertical change $dy$ has length $\sqrt{dx^2 + dy^2} = \sqrt{1 + (dy/dx)^2}\,dx$ by Pythagoras. Adding them up is the integral. The graph shows the polygon approximation converging.

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

The square root rarely simplifies. Textbook examples are chosen so that $1 + f'^2$ is a perfect square (like $y = x^{3/2}$ or $y = \cosh x$ ). For most curves, even $y = x^2$ , the calculator falls back to a precise numerical value.

What is the arc length of a circle?

Use the parametric mode with $x = r\cos t$ , $y = r\sin t$ , $0 \le t \le 2\pi$ : $L = \int_0^{2\pi} r\,dt = 2\pi r$ . Try it with $r = 1$ to get $2\pi$ .

How do I find arc length in polar coordinates?

$L = \int_\alpha^\beta \sqrt{r^2 + (dr/d\theta)^2}\,d\theta$ . Equivalently, use parametric mode with $x = r(\theta)\cos\theta$ and $y = r(\theta)\sin\theta$ .