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Arclength

April 13th, 2009 |

Author: Merrin

There are many types of possible curves you can draw. If the curves have certain properties then we can use calculus to find their length or so called arclength. If the curves are continuous and can be expressed as a function then the arclength can be expressed in terms of an integral. For a curve with these properties, it can be subdivided into segments which are approximately straight lines. In the limit that the number of segments goes to infinity this statement becomes exact. Each segment corresponds to a hypotenuse of width $latex \Delta x$ and a height $latex \Delta y$

$\Delta s=\sqrt{(\Delta x)^{2} +(\Delta y)^{2} } =\Delta x\sqrt{1+\cfrac{(\Delta y)^{2} }{(\Delta x)^{2} } }$

In the limit that $\Delta x$ is infinitesimal we obtain a differential. The quantity under the right radical is not the second derivative, but becomes the first derivative squared.

${ds=\sqrt{(dx)^{2} +(dy)^{2} } }$

${ds=dx\sqrt{1+(y')^{2} } =dy\sqrt{1+(x')^{2} } }$

If you approximate the length of a curve as the sum of the lengths of many infinitesimal segments then you get the arc length formula.

Theorem 1. Arclength formula

If y(x) is a continuous function on the interval $latex a\le x\le b$ then the arc length of y(x) is given by

$s=\int _{a}^{b}\sqrt{1+(y')^{2} } dx$

Arclength can also be calculated if x(y) is given for $latex c \le y \le d$ then

$s=\int _{c}^{d}\sqrt{1+(x')^{2} } dy$

It is often hard to evaluate arc length integrals in terms of elementary functions because of the radical in the formula. I will list some elementary functions for which there is a result in terms of elementary functions. Some functions which work are $latex \ln x, a^x, e^x, x^2, \cosh x, mx+b$. Arc length is fairly straightforward. You set up the integral in one line and then it is just a matter of integration skill to solve it.

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