Length of an arc

This article needs to be cleaned up to conform to a higher standard of quality.
See How to Edit and Style and How-to for help, or this article's talk page.

Determining the length of an irregular arc segment was historically difficult. Although many methods were used it wasn't until calculus that a universal formula became available.

Contents
1 Modern Methods
2 Historical Methods

2.1 Ancient
2.2 Fermat

3 External links

Modern Methods

Consider a function y = f(x) on a 2-dimensional Cartesian plane. The length L of the arc bounded by a and b is found as follows:

L = \int_{a}^{b} \sqrt { 1 + (f'(x))^2 } dx = \int_{a}^{b} \sqrt { 1 + \left ( \frac{dy}{dx} \right )^2 } dx

Where y = f(x) and y = f'(x) are continuous on [a,b].

Historical Methods

Ancient

For much of the history of mathematics, even the greatest thinkers considered it impossible to compute the length of an irregular arc. Although Archimedes had pioneered a rectangular approximation for finding the area beneath a curve with his method of exhaustion, few believed it was even possible for curves to have definite lengths, as do straight lines. The first ground was broken in this field, as it often has been in calculus, by approximation. People began to inscribe polygons within the curves and compute the length of the sides for a somewhat accurate measurement of the length. By using more segments, and by decreasing the length of each segment, they were able to obtain a more and more accurate approximation.

Fermat

Fermat was the first to propose a formula for calculating arc lengths. Building on his previous work with tangents, he used the curve

y = x^{3 \over 2}

whose tangent at x=a had a slope of

{3 \over 2} a^{1 \over 2}

so the tangent line would have the equation

y = {3 \over 2} {a^{1 \over 2}} * (x - a) + b

Next, he increased a by a small amount to a + c, making segment AC a relatively good approximation for the length of the curve from A to D. To find the length of \overrightarrow{A C}, he used the Pythagorean Theorem:

AC^2 = AB^2 + BC^2 = {3 \over 2} a^{1 \over 2} + e^2 = e^2 \left (1 + {9 \over 4} a \right )

which, when solved, yields

AC = e * \sqrt { 1 + {9 \over 4} a }

In order to approximate the length, Fermat would sum up a sequence of short ACs.

External links

See also: Length of an arc, Approximation, Archimedes, Calculus, Continuous function, Fermat, Function (mathematics), History of mathematics, Method of exhaustion