Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all points on the cardioid \(r=a(1+\cos \theta)\) where the tangent line is (a) horizontal, and (b) vertical.

Step-by-Step Solution

Answer

Horizontal tangents: \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \pi\). Vertical tangents: \(\theta = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3}\).

Step 1: Understanding the Cardioid Equation

The cardioid is given by the polar equation \( r = a(1+\cos \theta) \). Our task is to find points where the tangent line is horizontal and vertical. In polar coordinates, a point is described in terms of \( (r, \theta) \), where \( r \) is the radius and \( \theta \) is the angle.

Step 2: Convert Polar to Parametric Equations

To analyze the tangent's slope, first express the cardioid in parametric form: \( x = r \cos \theta = a(1+\cos \theta) \cos \theta \) and \( y = r \sin \theta = a(1+\cos \theta) \sin \theta \).

Step 3: Finding Expressions for \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)

Differentiate \( x \) and \( y \) with respect to \( \theta \):\[\frac{dx}{d\theta} = \frac{d}{d\theta}[a(1+\cos \theta) \cos \theta]and\frac{dy}{d\theta} = \frac{d}{d\theta}[a(1+\cos \theta) \sin \theta]\]

Step 4: Calculate \(\frac{dx}{d\theta}\)

Use the product rule to differentiate:\[\frac{dx}{d\theta} = a[-\sin \theta \cos \theta + (1+\cos \theta)(-\sin \theta)] = a[-\sin \theta \cos \theta - \sin \theta - \cos \theta \sin \theta]\]Simplify:\[\frac{dx}{d\theta} = -a \sin \theta (1 + 2\cos \theta)\]

Step 5: Calculate \(\frac{dy}{d\theta}\)

Differentiate using the product rule:\[\frac{dy}{d\theta} = a[\cos \theta \sin \theta + (1+\cos \theta) \cos \theta] = a[ \cos \theta \sin \theta + \cos \theta + \cos^2 \theta]\]Simplify:\[\frac{dy}{d\theta} = a \cos \theta (1 + \cos \theta)\]

Step 6: Finding Points with Horizontal Tangents

Horizontal tangents occur when \(\frac{dy}{d\theta} = 0\). Set the expression to zero:\[a \cos \theta (1 + \cos \theta) = 0\]Solutions are \( \cos \theta = 0 \) or \( \cos \theta = -1 \). This leads to \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \pi\).

Step 7: Finding Points with Vertical Tangents

Vertical tangents occur when \(\frac{dx}{d\theta} = 0\). Set the expression to zero and solve:\[-a \sin \theta (1 + 2\cos \theta) = 0\]Solutions are \( \sin \theta = 0 \) or \(1 + 2\cos \theta = 0\). This leads to \(\theta = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3}\).

Step 8: Identify the Points from \(\theta\) Values

Calculate the points \((r, \theta)\) for each solution:- For \( \theta = 0 \), \( r = 2a \).- For \( \theta = \frac{\pi}{2} \), \( r = a \).- For \( \theta = \pi \), \( r = 0 \).- For \( \theta = \frac{3\pi}{2} \), \( r = a \).- For \( \theta = \frac{2\pi}{3}, \frac{4\pi}{3} \), \( r = \frac{a}{2} \).

Key Concepts

Polar Coordinates

Polar coordinates are an alternative to Cartesian coordinates for describing the position of points in a plane. Instead of using x and y coordinates, polar coordinates use a radius and an angle.
This means that any point in the plane is represented as

• \((r, \theta)\)
• where \(r\) is the distance from the point to the origin,
• and \(\theta\) is the angle measured counterclockwise from the positive x-axis.

The relationship between polar and Cartesian coordinates is given by:

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

These formulas are useful for converting polar equations to Cartesian equations and vice versa. Understanding polar coordinates is essential in solving various calculus problems that involve curves such as cardioids.

Parametric Equations

Parametric equations express the coordinates of the points that make up a geometric object as functions of a variable, commonly denoted as \(t\). These equations are especially useful for describing curves in the plane or three-dimensional space.
For example, the parametric equations for a circle of radius \(a\) can be written as:

• \(x = a \cos t\)
• \(y = a \sin t\)

In the given problem, the cardioid's polar equation, \(r = a(1 + \cos \theta)\), is converted to parametric equations for easier analysis of the tangent lines:

• \(x = r \cos \theta = a(1+\cos \theta) \cos \theta\)
• \(y = r \sin \theta = a(1+\cos \theta) \sin \theta\)

This conversion helps in differentiating both \(x\) and \(y\) with respect to \(\theta\) to find the slopes of tangent lines.

Tangent Lines

Tangent lines are straight lines that touch a curve at a single point without crossing it at that point. Finding the equations of tangent lines involves calculating the slope of the curve at a particular point.
For parametric equations, you can find the slope of the tangent to a curve by differentiating the parametric equations and calculating:

• The slope \(m\) is given by \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\).

Horizontal tangent lines occur when the numerator equals zero, which gives \(\frac{dy}{d\theta} = 0\). This is because a horizontal line has a slope of zero.
Vertical tangent lines occur when the denominator equals zero, or \(\frac{dx}{d\theta} = 0\). This is because a vertical line has an undefined slope.
In the cardioid problem, identifying where \(\frac{dy}{d\theta} = 0\) and \(\frac{dx}{d\theta} = 0\) allows us to determine the specific points on the cardioid where the tangent lines are horizontal or vertical, respectively.