Consider the graph of r = theta for theta > 0 given in radians, also known as the Archimedean spiral. Prove that the consecutive intersection points of this graph with any ray whose initial point is the origin are equidistant from each other. What is that distance?

1 answer

To analyze the intersections of the Archimedean spiral \( r = \theta \) with a ray emanating from the origin, we start by parameterizing the ray. A ray in polar coordinates can be expressed with an angle \( \phi \): thus, any point on this ray can be represented in polar coordinates as \( (r, \phi) \).

The point on the ray corresponding to a distance \( r \) from the origin is given by:

\[ (r, \phi) = (r, \phi) \]

To find the intersection points of the Archimedean spiral \( r = \theta \) with this ray, we set the polar equations equal:

\[ r = \theta = k \phi \]

where \( k \) is some constant that represents a specific distance along the ray and varies as \(\theta\) increases. We can denote the points of intersection according to \( n \) as follows:

\[ \theta_n = k \phi_n \]

This relationship gives us the intersection points of the spiral with the ray. To find the specific points where these intersections occur, we can denote the \( n \)-th intersection point occurring at:

\[ \theta_n = n \cdot k \]

Then \( r \) at this intersection will be:

\[ r_n = \theta_n = n \cdot k \]

The positions along the ray are given by the polar coordinates:

\[ (r_n, \phi) = (n \cdot k, \phi) \]

Now let us consider the distance between two consecutive intersection points, namely, \( (r_n, \phi) \) and \( (r_{n+1}, \phi) \):

  • The \( n \)-th intersection point is: \((n \cdot k, \phi)\)
  • The \((n+1)\)-th intersection point is: \(((n+1) \cdot k, \phi)\)

To find the distance \( d \) between these points in Cartesian coordinates, we can convert from polar to Cartesian:

\[ (r, \theta) \rightarrow (r \cos \theta, r \sin \theta) \]

Calculating the Cartesian coordinates for both points gives:

  • For \( (r_n, \phi) = (n \cdot k, \phi) \):

\[ (x_n, y_n) = (n \cdot k \cos \phi, n \cdot k \sin \phi) \]

  • For \( (r_{n+1}, \phi) = ((n+1) \cdot k, \phi) \):

\[ (x_{n+1}, y_{n+1}) = ((n + 1) \cdot k \cos \phi, (n + 1) \cdot k \sin \phi) \]

The distance \( d \) between these two points is given by:

\[ d = \sqrt{(x_{n+1} - x_n)^2 + (y_{n+1} - y_n)^2} \]

Substituting the Cartesian coordinates, we get:

\[ d = \sqrt{(((n + 1) \cdot k \cos \phi) - (n \cdot k \cos \phi))^2 + (((n + 1) \cdot k \sin \phi) - (n \cdot k \sin \phi))^2} \]

This simplifies to:

\[ d = \sqrt{(k \cos \phi)^2 + (k \sin \phi)^2} = \sqrt{k^2 (\cos^2 \phi + \sin^2 \phi)} = \sqrt{k^2} = k \]

Thus, the distance \( d \) between two consecutive intersection points of the Archimedean spiral with the ray is constant and equal to \( k \). Therefore, we conclude:

The consecutive intersection points of the Archimedean spiral with any ray emanating from the origin are equidistant from each other, and that distance is \( k \).