# 5.3: The Length of a Plane Curve

**At Grade**Created by: CK-12

## Learning Objectives

A student will be able to:

- Learn how to find the length of a plane curve for a given function.

In this section will consider the problem of finding the length of a plane curve. Formulas for finding the arcs of circles appeared in early historical records and they were known to many civilizations. However, very little was known about finding the lengths of general curves, such as the length of the curve \begin{align*}y = x^2\end{align*} in the interval \begin{align*}[0, 2],\end{align*} until the discovery of calculus in the seventeenth century.

In calculus, we define an ** arc length** as the length of a plane curve \begin{align*}y = f(x)\end{align*} over an interval \begin{align*}[a, b]\end{align*} (Figure 17). When the curve \begin{align*}f(x)\end{align*} has a continuous first derivative \begin{align*}f'\end{align*} on \begin{align*}[a, b],\end{align*} we say that \begin{align*}f\end{align*} is a smooth function (or smooth curve) on \begin{align*}[a, b].\end{align*}

**The Arc Length Problem**

If \begin{align*}y = f(x)\end{align*} is a smooth curve on the interval \begin{align*}[a, b],\end{align*} then the arc length \begin{align*}L\end{align*} of this curve is defined as

\begin{align*} L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx = \int_{a}^{b} \sqrt{1 + \left (\frac{dy} {dx}\right )^2} dx.\end{align*}

**Example 1:**

Find the arc length of the curve \begin{align*}y = x^{\frac{3}{2}}\end{align*} on \begin{align*}[1, 3]\end{align*} (Figure 18).

*Solution:*

Since \begin{align*}y = x^3/2,\end{align*}

\begin{align*} \frac{dy} {dx} = \frac{3} {2} x^{1/2}.\end{align*}

Using the formula above, we get

\begin{align*}\int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx & = \int_{1}^{3} \sqrt{1 + \left [\frac{3} {2} x^{1/2}\right ]^2} dx\\ & = \int_{1}^{3} \sqrt{1 + \frac{9} {4}x} dx.\end{align*}

Using \begin{align*}u-\end{align*}substitution by letting \begin{align*} u = 1 + \frac{9} {4}x\end{align*}, then \begin{align*} du = \frac{9} {4} dx.\end{align*} Substituting, and remembering to change the limits of integration,

\begin{align*}L & = \frac{4} {9} \int_{13/4}^{31/4} \sqrt{u} du\\ & = \frac{8} {27}\left [u^{3/2}\right ]^{31/4}_{13/4}\\ & \approx 4.65.\end{align*}

## Multimedia Links

The formula you just used to find the length of a curve was derived by using line segments to approximate the curve. The derivation of that formula can be found at Wikipedia Entry on Arc Length. In the following applet you can explore this further. Experiment with various curves and change the number of segments to see how changing the number of segments is related to approximating the arc length. Arc Length Applet.

For video presentations showing how to obtain the arc length using parametric curves **(16.0)**, see Just Math Tutoring, Arc Length Using Parametric Curves, Example 1 (8:17)

and Just Math Tutoring, Arc Length Using Parametric Curves, Example 2 (7:27).

## Review Questions

- Find the arc length of the curve \begin{align*} y = \frac{(x^2 + 2)^{3/2}} {3}\end{align*} on \begin{align*}[0, 3].\end{align*}
- Find the arc length of the curve \begin{align*} x = \frac{1} {6} y^3 + \frac{1} {2y}\end{align*} on \begin{align*}y \in [1, 2].\end{align*}
- Integrate \begin{align*} x = \int_{0}^{y} \sqrt{\sec^4 t - 1} dt,\ - \frac{\pi} {4} \le y \le \frac{\pi} {4}.\end{align*}
- Find the length of the curve shown in the figure below. The shape of the graph is called the
*astroid*because it looks like a star. The equation of its graph is \begin{align*}x^{2/3} + y^{2/3} = 1.\end{align*} - The figure below shows a suspension bridge. The cable has the shape of a parabola with equation \begin{align*}kx^2 = y.\end{align*} The suspension bridge has a total length of \begin{align*}2S\end{align*} and the height of the cable is \begin{align*}h\end{align*} at each end. Show that the total length of the cable is \begin{align*} L = 2 \int_{0}^{s} \sqrt{1 + \frac{4h^2} {S^4} x^2} dx.\end{align*}

## Review Answers

- \begin{align*}12\end{align*}
- \begin{align*}(17/12)\end{align*}
- \begin{align*}2\end{align*}
- \begin{align*}6\end{align*}
- \begin{align*} L = \int_{0}^{s} \sqrt{1 + \frac{4h^2} {S^4} x^2} dx.\end{align*}

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |