Editing Archimedean spiral (section)

== Derivation of general equation of spiral ==
{{See also|Circular motion}}
A [[Physics|physical approach]] is used below to understand the notion of Archimedean spirals.

Suppose a point object moves in the [[Cartesian plane|Cartesian system]] with a constant [[velocity]] {{mvar|v}} directed parallel to the {{mvar|x}}-axis, with respect to the {{mvar|xy}}-plane. Let at time {{math|''t'' {{=}} 0}}, the object was at an arbitrary point {{math|(''c'', 0, 0)}}. If the {{mvar|xy}} plane rotates with a constant [[angular velocity]] {{mvar|ω}} about the {{mvar|z}}-axis, then the velocity of the point with respect to {{mvar|z}}-axis may be written as:

[[File:Spiral derivation...png|thumb|right|400px|The {{mvar|xy}} plane rotates to an angle {{mvar|ωt}} (anticlockwise) about the origin in time {{mvar|t}}. {{math|(''c'', 0)}} is the position of the object at {{math|''t'' {{=}} 0}}. {{mvar|P}} is the position of the object at time {{mvar|t}}, at a distance of {{math|''R'' {{=}} ''vt'' + ''c''}}.]]
<math display=block>\begin{align}
|v_0|&=\sqrt{v^2+\omega^2(vt+c)^2} \\
v_x&=v \cos \omega t - \omega (vt+c) \sin \omega t \\
v_y&=v \sin \omega t + \omega (vt+c) \cos \omega t
\end{align}</math>

As shown in the figure alongside, we have {{math|''vt'' + ''c''}} representing the modulus of the [[Position (vector)|position vector]] of the particle at any time {{mvar|t}}, with {{mvar|v<sub>x</sub>}} and {{mvar|v<sub>y</sub>}} as the velocity components along the x and y axes, respectively.

<math display="block">\begin{align}
\int v_x \,dt &=x \\
\int v_y \,dt &=y
\end{align}</math>

The above equations can be integrated by applying [[integration by parts]], leading to the following parametric equations:

<math display=block>\begin{align}
x&=(vt + c) \cos \omega t \\
y&=(vt+c) \sin \omega t
\end{align}</math>

Squaring the two equations and then adding (and some small alterations) results in the Cartesian equation
<math display=block>\sqrt{x^2+y^2}=\frac{v}{\omega}\cdot \arctan \frac{y}{x} +c</math>
(using the fact that {{math|''ωt'' {{=}} ''θ''}} and {{math|''θ'' {{=}} arctan {{sfrac|''y''|''x''}}}}) or
<math display=block>\tan \left(\left(\sqrt{x^2+y^2}-c\right)\cdot\frac{\omega}{v}\right) = \frac{y}{x}</math>

Its polar form is
<math display=block>r= \frac{v}{\omega}\cdot \theta +c.</math>