Problem 37 - Calculus

Question

Show that the spirals \(r=\theta\) and \(r=1 / \theta\) are perpendicular when they meet at \(\theta=1\)

Step-by-Step Solution

Answer

The spirals are perpendicular at \(\theta = 1\) because their slopes are negative reciprocals.

Step 1: Find the derivative of the polar equation

First, let's find the derivative of the polar equation for each spiral. For the first spiral, the equation is given by \(r = \theta\). In polar coordinates, the derivative of \(r\) with respect to \(\theta\) is simply 1, as \(\frac{dr}{d\theta} = \frac{d}{d\theta}(\theta) = 1\).

Step 2: Convert polar to Cartesian coordinates

To find the angle of intersection, convert each spiral into Cartesian coordinates. For \(r = \theta\):- \(x = r \cos(\theta) = \theta \cos(\theta)\)- \(y = r \sin(\theta) = \theta \sin(\theta)\).For \(r = \frac{1}{\theta}\):- \(x = \frac{1}{\theta} \cos(\theta)\)- \(y = \frac{1}{\theta} \sin(\theta)\).

Step 3: Find the tangents of the spirals

The slope of the tangent in Cartesian coordinates is \(\frac{dy}{dx}\). For each spiral:1. \(r = \theta\): - \(\frac{dy}{d\theta} = \cos(\theta) - \theta \sin(\theta)\) - \(\frac{dx}{d\theta} = \cos(\theta) + \theta \sin(\theta)\) - \(\frac{dy}{dx} = \frac{\cos(\theta) - \theta \sin(\theta)}{{\cos(\theta) + \theta \sin(\theta)}}\).2. \(r = \frac{1}{\theta}\): - \(\frac{dy}{d\theta} = -\frac{\sin(\theta)}{\theta^2} - \frac{\cos(\theta)}{\theta}\) - \(\frac{dx}{d\theta} = -\frac{\cos(\theta)}{\theta^2} + \frac{\sin(\theta)}{\theta}\) - \(\frac{dy}{dx} = \frac{-\theta\sin(\theta)-\cos(\theta)}{-\theta\cos(\theta)+\sin(\theta)}\).

Step 4: Evaluate the derivatives at the intersection

For both spirals, when \(\theta = 1\), compute the slopes:1. For \(r = \theta\), \(\theta = 1\): - \(\frac{dy}{dx} = \frac{\cos(1) - \sin(1)}{\cos(1) + \sin(1)}\).2. For \(r = \frac{1}{\theta}\), \(\theta = 1\): - \(\frac{dy}{dx} = \frac{-\sin(1) - \cos(1)}{-\cos(1) + \sin(1)}\).Note, the slopes are negative reciprocals.

Step 5: Conclude the spirals are perpendicular

If the slopes of two curves at their point of intersection are negative reciprocals of each other, then the curves are perpendicular. Here, since \(\frac{dy}{dx}\) for the first spiral is the negative reciprocal of \(\frac{dy}{dx}\) for the second spiral when \(\theta = 1\), the spirals are perpendicular at that point.

Key Concepts

Derivatives

To begin solving the problem, it is important to understand the concept of derivatives, which are used to determine the rate at which one quantity changes with respect to another. In this exercise, derivatives are applied to polar coordinates. We focus on the derivative of the polar coordinate function. The derivative in polar coordinates is expressed in terms of the variable \( \theta \), showcasing how \( r \), which is the radial distance from the origin, changes as \( \theta \) changes.

For the spiral curve equation \( r=\theta \), the derivative \( \frac{dr}{d\theta} \) is 1, indicating a direct and straightforward change in the radial distance with increasing \( \theta \). This simplicity makes it easy to analyze the curve. Meanwhile, for the spiral \( r=\frac{1}{\theta} \), the process involves finding the derivative of \( r \) as \( \theta \) changes, which results in a more complex relationship.

Spiral Curves

Spiral curves are fascinating in mathematics due to their unique properties. The two spiral equations discussed here are \( r = \theta \) and \( r = \frac{1}{\theta} \).

• The first spiral, given by \( r = \theta \), is known as the Archimedean spiral. It describes a spiral that moves away from the origin at a constant rate as \( \theta \) increases.
• The second spiral, represented by \( r = \frac{1}{\theta} \), is a reciprocal relationship. As \( \theta \) becomes larger or smaller, the radius changes inversely.

When exploring their intersections, we find that these curves provide a rich ground for examining angles and derivatives as their pathways offer dynamic intersections.

Cartesian Coordinates

The conversion from polar to Cartesian coordinates is critical in this exercise because analyzing in the Cartesian system can simplify understanding of a curve. Polar coordinates describe a point based on its distance from a central point (the pole) and the angle from a fixed direction.

In converting the polar equations \( r = \theta \) and \( r = \frac{1}{\theta} \) to Cartesian coordinates:

• For the equation \( r = \theta \), the Cartesian forms become \( x = \theta \cos(\theta) \) and \( y = \theta \sin(\theta) \).
• For \( r=\frac{1}{\theta} \), the coordinates are represented by \( x = \frac{1}{\theta} \cos(\theta) \) and \( y = \frac{1}{\theta} \sin(\theta) \).

This transformation allows us to work with derivatives in a format more familiar to those accustomed to Cartesian systems.

Tangents

Tangents to a curve describe the direction it is heading at any given point. In the Cartesian system, the slope of the tangent to a curve at a given point can be found using the derivative \( \frac{dy}{dx} \). For the polar curves under discussion:

• For \( r = \theta \), transform to Cartesian, then calculate \( \frac{dy}{dx} \) using \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \).
• Similarly, for \( r = \frac{1}{\theta} \), derive \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) to determine the tangent slope \( \frac{dy}{dx} \).

The calculation of tangents at \( \theta = 1 \) reveals how the spiral reacts as \( theta \) changes, offering a glimpse into the nature of its intersection with other curves.

Angle of Intersection

The angle of intersection between curves is one of the elegant parts of calculus. It's fascinating to find how curves interact with one another. Specifically, if two curves meet at a point with tangents whose derivatives (slopes) are negative reciprocals, they intersect perpendicularly.

In this problem, when examining the spirals \( r = \theta \) and \( r = \frac{1}{\theta} \) at \( \theta = 1 \):

• Compute \( \frac{dy}{dx} \) for each spiral at this point.
• The slope of \( r = \theta \) is \( \frac{\cos(1) - \sin(1)}{\cos(1) + \sin(1)} \).
• The slope of \( r = \frac{1}{\theta} \) is the negative reciprocal.

This negative reciprocal relationship confirms they are perpendicular at \( \theta = 1 \), visually and analytically demonstrating the spirals' unique meeting.