Length of a Curve (Arc Length)

The length of a curve—also called arc length—is an application of definite integrals. In Calculus BC, you will learn how to compute the length of curves written in different forms: parametric, polar, and Cartesian (y = f(x)).

What Is Arc Length?

Imagine a curve on a graph. You want to measure its length, just like measuring a piece of string. Since the curve is not straight, you approximate it using many tiny line segments.

As the number of segments increases, the approximation gets closer to the true length. In calculus, this leads to an integral formula for exact length.

Arc Length Formula for y = f(x)

If a curve is defined by a differentiable function

y=f(x) on the interval [a,b]

the length of the curve is:

$$L=\int_{a}^{b}\sqrt{1+(f^\prime(x))^{2}}$$​

Why this formula?

Because for a tiny change in x:

• Horizontal change: dx

• Vertical change: dy=f′(x) dx

The tiny segment length is:

$$ds=\sqrt{dx^{2}+dy^{2}}dx=\sqrt{1+\left(\frac{dy^{2}}{dx^{2}}\right)}dx=\sqrt{1+(f^\prime(x))^{2}}dx$$

Integrate from a to b is total length.

Arc Length for Parametric Curves

If a curve is given by:

$$x=x(t) , y=y(t) a\leq{t}\leq{b}$$

then the length is:

$$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}​

careful, don’t divide by $$\frac{dx}{dt}$$ inside square root

Arc Length for Polar Curves

For a polar function:

$$r=f(\theta)$$ on $$\alpha\leq\theta\leq\beta$$,

the arc length is:

$$L=\int_{\alpha}^{beta}\sqrt{(f(\theta))^{2}+(f^\prime(\theta))^{2}}$$​

Reason: A small arc in polar coordinates has length:

$$ds=\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}d\theta$$

Example 1 (Cartesian)

Find the length of the curve

$$y=\frac{x^{3}}{3} , 0\leq{x}\leq1$$

Step 1 find derivative:

$$f^\prime(x)=x^{2}$$

Step 2:

$$L=\int_{0}^{1}\sqrt{1+x^{4}}dx$$

Example 2 (Parametric)

$$x(t)=t^{2} , y(t)=t^{3} , 0\leq{t}\leq1$$ then

$$\frac{dx}{dt}=2t , \frac{dy}{dt}=3t^{2}$$

length:

$$=\int_{0}^{1}\sqrt{(2t)^{2}+(3t^{2})^{2})dt=\int_{0}^{1}\sqrt{4t^{2}+9t^{4}}$$

Example 3 (Polar)

Find the arc length of

$$r-2\theta , 0\leq\theta\leq\pi$$

$$f(\theta)=2\theta , f^\prime(\theta)=2$$ then

$$L=\int_{0}^{\pi}\sqrt{(2\theta)^{2}+(2^{2})}d\theta=\int_{0}^{\pi}\sqrt{4\theta^{2}+4}d\theta$$