Problem 244 - Calculus Volume 3

Question

Find the points on the interval \(-\pi \leq \theta \leq \pi\) at which the cardioid \(r=1-\cos \theta\) has a vertical or horizontal tangent line.

Step-by-Step Solution

Answer

The cardioid has horizontal tangents at \((0, 2)\) and \((0, -2)\); vertical tangents at \((0, 0)\) and \((-2, 0)\).

Step 1: Understanding the Problem

We are given a cardioid equation in polar coordinates, \( r = 1 - \cos \theta \), and an interval \( -\pi \leq \theta \leq \pi \). We need to find the points where the cardioid has vertical or horizontal tangent lines.

Step 2: Convert to Parametric Form

Convert the given polar equation to parametric form using \( x = r \cos \theta \) and \( y = r \sin \theta \). This gives us: \( x = (1 - \cos \theta) \cos \theta \) and \( y = (1 - \cos \theta) \sin \theta \).

Step 3: Compute the Derivatives

Calculate the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).\[ \frac{dx}{d\theta} = \frac{d}{d\theta} \left((1 - \cos \theta)\cos \theta\right) \]\[ \frac{dy}{d\theta} = \frac{d}{d\theta} \left((1 - \cos \theta)\sin \theta\right) \]

Step 4: Differentiate \(x\) and \(y\) with Respect to \(\theta\)

Apply the product rule to find:\[ \frac{dx}{d\theta} = -\sin \theta \cos \theta - (1 - \cos \theta)\sin \theta \]\[ \frac{dy}{d\theta} = \cos \theta \sin \theta + (1 - \cos \theta)\cos \theta \]

Step 5: Horizontal Tangents

A horizontal tangent occurs when \( \frac{dy}{d\theta} = 0 \). Solve \[ (1-\cos\theta)\cos\theta + \cos\theta\sin\theta = 0 \]Factor out \( \cos\theta \):\[ \cos\theta((1-\cos\theta)+\sin\theta) = 0 \]So \( \cos\theta = 0 \), which gives \( \theta = \frac{\pi}{2}, -\frac{\pi}{2} \).

Step 6: Vertical Tangents

A vertical tangent occurs when \( \frac{dx}{d\theta} = 0 \). Solve \[ -(1-\cos\theta)\sin\theta - \sin\theta \cos\theta = 0 \]Factor out \( \sin\theta \):\[ \sin\theta((1-\cos\theta)+\cos\theta) = 0 \]So \( \sin\theta = 0 \), which gives \( \theta = 0, \pi, -\pi \).

Step 7: Identify Points on the Cardioid

Substitute \( \theta = \frac{\pi}{2}, -\frac{\pi}{2}, 0, \pi, -\pi \) in the cardioid equation \( r = 1 - \cos \theta \) to get the respective points:- For \( \theta = \pm \frac{\pi}{2} \), \( r = 2, (0,2), (0,-2) \), which correspond to horizontal tangents.- For \( \theta = 0 \), \( r = 0 \), origin point, corresponding to a vertical tangent.- For \( \theta = \pm \pi \), \( r = 2 \), point \((-2,0)\) for vertical tangents.

Key Concepts

Polar Coordinates

In mathematics, polar coordinates provide a different but extremely useful way to describe the position of a point. Instead of using the typical rectangular coordinates (x, y) in a plane, polar coordinates use a distance and an angle: (r, \(\theta\)).

• \( r \) represents the distance from the origin to the point.
• \( \theta \) is the angle, measured in radians, from the positive x-axis to the line from the origin to the point.

In polar coordinates, many curves, such as the cardioid, take simple forms. While working with these coordinates, especially for curves, you'll often have equations involving \( r \) and \( \theta \). Switching between polar and rectangular (Cartesian) coordinates involves using the relationships:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

This conversion is crucial in certain problems, like finding tangent lines, where you need to work with a Cartesian form to apply calculus easily.

Tangent Lines

Tangent lines are straight lines that "just touch" a curve at a given point. They are used to represent the instantaneous direction of the curve at that point. You'll come across two types of tangent lines:

• **Horizontal Tangents:** These occur when the slope of the tangent line is zero. For polar curves, you find horizontal tangents by identifying when \( \frac{dy}{d\theta} = 0 \).
• **Vertical Tangents:** These occur when the slope is undefined. In this case, \( \frac{dx}{d\theta} = 0 \) is the criterion.

The slopes are expressible in terms of derivatives, \[\text{Slope} = \frac{dy/d\theta}{dx/d\theta}.\]To find these derivatives, you will typically rewrite a polar equation in its Cartesian parametric form and differentiate.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding a derivative of a function. The derivative measures how a function's output value changes as its input changes. For a function \( y = f(x) \), the derivative \(\frac{dy}{dx}\) is the slope of the tangent line at any point \(x\).

For a parametric form, involving separate functions for \( x \) and \( y \) with respect to another variable such as \( \theta \) (common in polar coordinates), you use:

• \( \frac{dx}{d\theta} \)
• \( \frac{dy}{d\theta} \)

These derivatives are combined to find the slope of the curve in a Cartesian frame.
Differentiation can also be used to identify the nature of tangent lines to curves. Applying the chain rule, product rule, and other differentiation techniques will help solve many calculus problems effectively.

Cardioid

A cardioid is a heart-shaped curve that can be defined using polar coordinates. The general equation for a cardioid is \( r = a(1 - \cos \theta) \) or \( r = a(1 + \cos \theta) \). In problems like these, the coefficient \(a\) is typically set to 1 for simplicity. Here, the curve takes the form \( r = 1 - \cos \theta \).

• The cardioid has symmetrical properties around specific axes, making it interesting and sometimes challenging for analysis.
• In practical problems, like finding tangent lines, you will work to identify certain points where properties such as symmetry play a role in simplifications or recognizing behavior.

With cardioids, converting to parametric or Cartesian coordinates is necessary to apply calculus methods for analyzing tangent lines or other curve properties. Understanding how to handle these conversions and calculations will provide deeper insights into their geometric properties and applications.