Consider the graph of for 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?

Here's the picture of a few of these intersection points for a particular ray:

Note that we aren't including the origin as one of the intersections in the picture, since we aren't including the origin in the graph!

I know the distance is 2pi, however I dont know how to prove that the points of intersection are equidistant. Could someone please explain why the points are equidistant from each other? Also, I am not completely clear on why the distance is 2pi so if someone could explain that, it would be great.

Thanks!

JRFPLSHelp Aug 3, 2024

#1**0 **

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.

Rangcr897 Aug 4, 2024