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How to Find the Arc Length of an Archimedean Spiral: Calculus Integration Tutorial

Three rotations of an Archimedes spiral with the equation r(θ) = θ, where θ is measured in radians. Image via Wikimedia Commons.
Three rotations of an Archimedes spiral with the equation r(θ) = θ, where θ is measured in radians. Image via Wikimedia Commons.

An Archimedes spiral, also called an arithmetic spiral, is a plane curve with the property that any ray drawn from the origin and passing through the spiral will intersect the curve at equidistant points. If you look at the spiral in just one quadrant, it appears as if it is a ring of concentric and equally spaced circles, although this is actually not the case. The general form of an Archimedean spiral in polar coordinates is

r(θ) = p + qθ

where p and q are some real numbers. The parameter q determines the spacing between successive turnings of the spiral, while p determines the starting position of the spiral near the origin. For an Archimedes spiral defined by r(θ) = p + qθ, the distance between spirals is 2πq. The simplest example of an Archimedes spiral is

r(θ) = θ

where θ is measured in radians. Here, p = 0 and q = 1. The spacing between consecutive rotations is 2π. You can use the polar arc length integral to compute the length of the spiral's curve over a given interval. The general integral formula for any polar arc length is

∫ sqrt(r(θ)^2 + r'(θ)^2) dθ

and the limits of integration are the starting and ending angles of rotation. Since r(θ) = p + qθ and r'(θ) = q in this tutorial, the integral simplifies to

∫ sqrt(p^2 + q^2 + 2pqθ + (q^2)θ^2) dθ

Setting up and Evaluating the Arc Length Integral of the Spiral

For the example of r(θ) = θ, the first derivative of this function with respect to θ is r'(θ) = 1. The integral that must be evaluated is then

sqrt(θ^2 + 1) dθ

In this example, we will find the length of a spiral from θ = 0 to θ = 3π, which is one and a half turns. First we must find the antiderivative of the function sqrt(θ^2 + 1). Using either a table of integrals or integration software, we find the antiderivative is

0.5θ*sqrt(θ^2 + 1) + 0.5*ln(θ + sqrt(θ^2 + 1))

Evaluating this expression at the limits of the integral gives us the arc length

0.5*3π*sqrt(9π^2 + 1) + 0.5*ln(3π + sqrt(9π^2 + 1)) - 0 - 0

= 46.1321661307

Approximation Formula

For an Archimedean spiral r(θ) = p + qθ, where θ ranges from 0 to T, the exact arc length involves square roots and natural logarithms. Therefore, given an exact length L and the values of the parameters p and q, it is not possible to algebraically solve for the exact value of T.

However, there is a simpler approximation formula for the arc length L in terms of T, p, and q:

L ≈ (1/2)[ q*T^2 + (sqrt(p^2 + q^2)+p)*T ]

This approximation formula can then be solved for T in terms of p, q, and L using the quadratic formula.

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Comments 5 comments

Brian 3 years ago

In the example, the result 46.13 is what you get integrating to 3*pi, not 6*pi.


calculus-geometry profile image

calculus-geometry 3 years ago from Germany Author

Thanks for catching that typo! Fixed it now.


benblo profile image

benblo 2 years ago from Montpellier, France

Thanks for your post!

I've successfully implemented the general version equation (with arbitrary p & q, not just p = 0 & q = 1 as in your closing example) in C# with the help of Wolfram: http://www.wolframalpha.com/share/clip?f=d41d8cd98... (I've replaced q with a and p with b).

However now I want to do the opposite: given a known length (and the same known p & q), what's the angle? ~ number of turns needed to roll a length of rope.

Your solution gives exact results, however it is very cumbersome and I have no idea how to reverse it (if that's even the word): do you?

For anyone interested, I found an approximate solution of the length problem (http://www.intmath.com/blog/length-of-an-archimedi... that approximates to "the circumference of a circle of average radius": l = ax²/2 + bx

This one I was able to solve to: x = (-b + sqrt(b² - 2ac)) / a (it's actually supposed to have 2 solutions but I just always use the + one).


calculus-geometry profile image

calculus-geometry 2 years ago from Germany Author

Since the exact length formula involves both square roots and logarithms, it is not possible to solve it exactly for the angle. But if you use a quadratic approximation for the length, then you can solve for the angle.


benblo profile image

benblo 2 years ago from Montpellier, France

Ah, that's a pity... never mind then, I'll stick with the approximation.

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