Archimedean Spiral PLEASE HELP

Proving Equidistance of Intersection Points on an Archimedean Spiral

 

Understanding the Problem

 

We're considering the Archimedean spiral defined by the polar equation r = θ for θ > 0. The task is to prove that consecutive intersection points of this spiral with any ray originating from the origin are equidistant.

 

Solution

 

1. Parametric Representation of the Ray:

 

Let the ray be defined by the angle φ.

 

Any point on this ray can be represented in polar coordinates as (r, φ), where r is the distance from the origin.

 

2. Intersection Points:

 

To find the intersection points of the spiral and the ray, we equate their polar equations:

 

θ = r

 

Substituting r with r = ρcos(φ - θ) (polar coordinate conversion to Cartesian), we get:

 

θ = ρcos(φ - θ)

 

3. Finding Consecutive Intersection Points:

 

Let θ₁ and θ₂ be the angles corresponding to two consecutive intersection points.

 

Then, we have:

 

θ₁ = ρcos(φ - θ₁)

 

θ₂ = ρcos(φ - θ₂)

 

4. Calculating the Distance Between Intersection Points:

 

The distance between two points in polar coordinates is given by:

 

d = sqrt(r₁² + r₂² - 2r₁r₂cos(θ₂ - θ₁))

 

Substituting r₁ = θ₁ and r₂ = θ₂, we get:

 

d = sqrt(θ₁² + θ₂² - 2θ₁θ₂cos(θ₂ - θ₁))

 

5. Simplifying the Distance:

 

Using the trigonometric identity cos(A - B) = cosAcosB + sinAsinB, we can simplify the expression for d:

 

d = sqrt((θ₂ - θ₁)²)

 

d = |θ₂ - θ₁|

 

6. Conclusion:

 

Since θ₂ and θ₁ represent consecutive intersection points, |θ₂ - θ₁| is a constant (equal to 2π for a full rotation).

 

Therefore, d = 2π is also a constant.

 

The Distance

 

The distance between consecutive intersection points of the Archimedean spiral with any ray from the origin is 2π.

 

This proves that the consecutive intersection points are equidistant.