Bring the ARC back to Warcraft

Determining the length of an irregular arc segment—also called rectification of a curve—was historically difficult. Although many methods were used for specific curves, the advent of calculus led to a general formula that provides closed-form solutions in some cases. [edit] Precise definition Choose a finite number of points along a curve and connect each point to the next with a straight line. The sum of the lengths of such line segments is the length of a "polygonal path". Definition: The length of the curve is the smallest number that such lengths of polygonal paths can never exceed, no matter how close together the discretely placed endpoints of line segments are. In the language of mathematicians, the arc length is the supremum of all lengths of such polygonal paths. This definition does not require the curve to be "smooth"; it need not be either the graph or the image of a differentiable function. [edit] Modern methods Consider a function f(x) \, such that f(x) \, and f'(x) \, (its derivative with respect to x) are continuous on [a, b]. The length s of the part of the graph of f between x = a and x = b is found by the formula s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx. which is derived from the distance formula approximating the arc length with many small lines. As the number of line segments increases (to infinity by use of the integral) this approximation becomes an exact value. If a curve is defined parametrically by x=X(t)\, and y=Y(t)\,, then its arc length between t = a and t = b is s = \int_{a}^{b} \sqrt { [X'(t)]^2 + [Y'(t)]^2 }\, dt. This is more clearly a consequence of the distance formula where instead of a Δx and Δy, we take the limit. A useful mnemonic is s = \lim \sum_a^b \sqrt { \Delta x^2 + \Delta y^2 } = \int_{a}^{b} \sqrt { dx^2 + dy^2 } = \int_{a}^{b} \sqrt { \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 }\,dt. If function is defined in polar coordinates by r = f(θ) then the arclength is defined by: s = \int_a^b \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} \, d\theta. In most cases, including even simple curves, there is no closed-form solutions of arc length and numerical integration is necessary. Curves with closed-form solution for arc length include the catenary, circle, cycloid, logarithmic spiral, parabola, semicubical parabola and (mathematically, a curve) straight line. The lack of closed form solution for the arc length of an elliptic arc led to the development of the elliptic integrals. [edit] Derivation A representative linear element of the function A representative linear element of the function \begin{cases} y = t^5 \\ x = t^3 \end{cases} In order to approximate the arc length of the curve, it is split into many linear segments. To make the value exact, and not an approximation, infinitely many linear elements are needed. This means that each element is infinitely small. This fact manifests itself later on when an integral is used. Begin by looking at a representative linear segment (see image) and observe that its length (element of the arc length) will be the differential ds. We will call the horizontal element of this distance dx, and the vertical element dy. The distance formula tells us that ds = \sqrt{dx^2 + dy^2}.\, Since the function is defined in time, segments (ds) are added up across infintesimally small intervals of time (dt) yielding the integral \int_a^b \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}\,dt, which is the arc length from t = a to t = b of the parametric function f(t). For example, the curve in this figure is defined by \begin{cases} y = t^5, \\ x = t^3 \end{cases}. Subsequently, the arc length integral for values of t from −1 to 1 is \int_{-1}^1 \sqrt{(3t^2)^2 + (5t^4)^2}\,dt = \int_{-1}^1 \sqrt{9t^4 + 25t^8}\,dt. Using computational approximations, we can obtain a very accurate (but still approximate) arc length of 2.905. [edit] Historical methods [edit] 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. [edit] 1600s In the 1600s, the method of exhaustion led to the rectification by geometrical methods of several transcendental curves: the logarithmic spiral by Evangelista Torricelli in 1645 (some sources say John Wallis in the 1650s), the cycloid by Christopher Wren in 1658, and the catenary by Gottfried Leibniz in 1691. In 1659, Wallis credited William Neile's discovery of the first rectification of a nontrivial algebraic curve, the semicubical parabola. [edit] Integral form Before the full formal development of the calculus, the basis for the modern integral form for arc length was independently discovered by Hendrik van Heuraet and Pierre Fermat. In 1659 van Heuraet published a construction showing that arc length could be interpreted as the area under a curve - this integral, in effect - and applied it to the parabola. In 1660, Fermat published a more general theory containing the same result in his De linearum curvarum cum lineis rectis comparatione dissertatio geometrica. Fermat's Method of determining arc length Fermat's Method of determining arc length Building on his previous work with tangents, Fermat 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) + f(a). Next, he increased a by a small amount to a + e, making segment AC a relatively good approximation for the length of the curve from A to D. To find the length of the segment AC, 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 segments.