Special Plane Curves
Archimedes Spiral
Michael Liu and Tim Myers
18 May 2005
Multivariable Calculus
Instructor: David Arnold
The Definition and Description of the Curve
The curve is defined by the polar equation r = a*θ, where θ≥0. If we let a=1, we
will begin at the origin with θ = 0 and r = 0. The curve is traced out counterclockwise as
the radius of the circle and the angle of rotation both run from 0 to 2pi. The unique
property of the Archimedes Spiral is that the radius of the spiral and the rotation angle are
linearly proportional, meaning that r, the distance from the origin to any point on the
curve, increases at the rate of a*θ, which is θ when a = 1. Because a is a constant of
proportionality, the only effect a has on the graph is to change the ratio at which the
position vector r increases. As a changes, the distance between the windings of the spiral
changes, but is always equal to a*θ, and the distance between any two successive
windings is the same.
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8
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2
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B
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A
D
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Archimedes gives this definition in his work On Spirals, “If a straight line drawn
in a plane revolves uniformly any number of times about a fixed extremity until it returns
to its original position, and if, at the same time as the line revolves, a point moves
uniformly along the straight line beginning at the fixed extremity, the point will describe
a spiral in the plane.”
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We must also understand that since the distance r from the origin is increasing,
the curvature of the spiral should be increasing as well. Below, we constructed
Archimedes spiral using Geometer’s SketchPad, along with its osculating circle to show
the relative magnitude of the curvature on each point of the spiral.
Osculating Circle
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Construction of the Osculating Circle
Let r = [ aθcos(θ), aθsin(θ) ] be a position vector from the origin to an arbitrary point on
the Spiral Curve of Archimedes.
Differentiating r with respect to θ and dividing by its magnitude to obtain a unit tangent
vector T(t) in the direction of r:
r(θ) = [ a θ cos(θ), a θ sin(θ) ]
( position vector )
dr/dθ = [ a cos(θ) – a θ sin(θ), a sin(θ) + a θ cos(θ)]
( unit tangent vector )
T(θ) = [ -a (-cos(θ) + θ sin(θ)) / ( | a | (1 + θ²) ½ ) , a (sin(θ) + θ cos(θ)) / ( | a |(1+θ²)½)]
Next we calculate a vector that is orthogonal to T(θ) :
N(θ) = [-a (sin(θ) + θ cos(θ)) / ( | a |(1+θ²)½), -a (-cos(θ) + θ sin(θ)) / ( | a | (1 + θ²) ½ )]
Taking the magnitude of this orthogonal vector and dividing by the magnitude of the unit
tangent vector gives us the equation for the curvature of the Spiral of Archimedes:
| N(θ) | / | T(θ) | =
(2 + θ²) / ((1 + θ²)3/2|a| )
k(θ) = ( 2 + θ²) / (( 1 + θ²)3/2|a| )
( curvature of the curve )
The radius of the osculating circle is difined as p = 1/k and is therefore :
p = (( 1 + θ²)3/2|a| ) / ( 2 + θ²)
The osculating circle is therefore defined as the set of points
c =[ (θ cos(θ) - sin(θ)- θ²sin(θ))a / (2+ θ²) , a(θsin(θ)+cos(θ)+ θ²*cos(θ))/(2+ θ²)]
History and Background of the Spiral of Archimedes
This spiral curve was introduced and studied by Archimedes in the third century
B.C.. Archimedes used the spiral curve to work on the problem of squaring the circle.
The quadrature of the circle was one of the greatest geometrical problems of antiquity.
The problem was; given a circle, to construct a square with an area equal to that in the
circle using only a compass and a straightedge. Archimedes had shown in his work “On
Spirals” that the area of a circle with a given radius was equal to the area of a right
triangle with the two shortest sides equaling the radius and the circumference of the given
circle.
We can confirm this by using the osculating circle of Archimedes spiral and
constructing a triangle with an equal area and with the two shortest sides equal to the
radius and the circumference of the given circle.
Area
DC = 21.19 cm2
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A D
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C
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G
area of triangle = 21.19 cm2
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The spiral was used for many useful purposes. The Archimedes screw, a spiral
screw turned inside a cylinder, was once commonly used to lift water from canals. The
screw is still used to lift water in the Nile delta in Egypt, and is often used to shift grain in
mills and powders in factories.
Analyzing Archimedes Curve
Who said we had to stop the rotation at 2pi? If we let θ increase from 2pi to 4pi,
the curve would rotate once more around the origin. Since a=1 in our example, the
distance r increases from 2pi to 4pi as well. Basically, the spiral and distance r increase
outward at the same rate.
We can also increase or decrease the distance r from the origin to a*θ, by letting
a equal any real number other than 1. Below, the top three graphs illustrate this point. In
the first graph a=1/2, in the second a=1, and in the third a=2. When the value of a is
changed from 1 to 2, the distance r doubled as well. When the value of a decreased from
1 to ½, the distance decreased by half as well.
So far, we have tackled positive a values and positive θ values, but what about
negative values? Well, it turns out, when θ is negative and a remains positive, the
direction of the spiral is reversed, as displayed below (center graph, second row). The
rotation of the curve, in this case, begins at θ = 0 and rotates as θ values increase in the
negative θ direction. When a is negative and θ remains positive, the position vector r is
negated and the curve is traced out with negative values of r as θ increases (third graph,
second row). Finally, when both a and θ are negative, the curve rotates in the negative θ
direction and the points on the curve are plotted as negative values of r are rotated
through this clockwise direction (first graph, second row).
The Development of a System of Parametric Equations
We take the polar definition of the curve, r = a*θ, and convert it to a parametric
system of equations using the figure below and some algebraic manipulation. We can
use the relationship between polar coordinates and Cartesian coordinates to solve for x
and y.
Looking at the above figure, the connection between polar and Cartesian coordinates can
be seen, in which the pole corresponds to the origin and the polar axis coincides with the
positive x-axis. If the point P has Cartesian coordinates (x,y) and polar coordinates (r, θ),
then from the figure, we have:
cos θ=x/r
sin θ=y/r
so……..
x=rcos θ
y=rsin θ
Since were talking about Archimedes spiral, r = a* θ, so in terms of Cartesian
coordinates…
x=a*θcos θ
y=a*θsin θ
We can also solve for x and y through simple algebraic manipulation
we know: r²=x ²+ y²
lets solve for x first:
r=aθ
r²=a²θ²
square both sides
x²+y²=a²θ²
substitute r²=x ²+ y²
x²=a²θ²-y²
x²=a²θ²-r²sin²θ
y=rsinθ → y²=r²sin²θ
x²=a²θ²- a²θ²sin²θ
x²=a²θ²(1- sin²θ)
x² =a²θ² cos²θ
x=| aθcosθ |
x=aθcosθ
Finally, solve for y:
r=aθ
r²=a²θ²
square both sides
x²+y²=a²θ²
substitute r²=x ²+ y²
y²=a²θ²-x²
y²=a²θ²-r²cos²t
x=rcosθ → x²=r²cos²θ
y²=a²θ²- a²θ²cos²θ
y²=a²θ²(1- cos²θ)
y²=a²θ² sin²θ
y=|aθsinθ|
y=aθsinθ
Thus, the system of parametric equations are x = aθcosθ and y = aθsinθ, where a is any
real number.
We can prove that these are the correct equations by using the distance formula. The
result looks very much like the polar equation defined at the beginning.
r=√(x²+y²)
r=√(a²θ²cos²θ+a²θ²sin²θ)
r=√(a²θ²(cos²θ+sin²θ))
r=√(a²θ²)
r=|aθ|
r=aθ
x²=a²θ²cos²θ and y²=a²θ²sin²θ.
Here is the graph of x = θcosθ and y = θsinθ. As expected, it is just the graph of r = a* θ,
where a=1.
That’s all folks!
Bibliography
http://www.xahlee.org/SpecialPlaneCurves_dir/ArchimedeanSpiral_dir/archimedeanSpir
al.html
Stewart, James. Early Transcendentals: Calculus. Belmont: Brooks/Cole-Thomson
Learning, 2003.
http://www.solarnavigator.net/inventors/archimedes.htm
http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Squaring_the_circle.html