Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the limaçon \(r=2-3 \cos \theta\), and find the area of the region inside its large loop.

Step-by-Step Solution

Answer

The area of the large loop is approximately 10.5 square units.

Step 1: Understanding the Limaçon

The given equation is a limaçon in polar coordinates: \( r = 2 - 3 \cos \theta \). The equation is of the form \( r = a - b \cos \theta \), where \( a = 2 \) and \( b = 3 \). Because \( b > a \), the graph features a loop. We expect the limaçon to have a loop directed towards the positive side of the x-axis (since it involves \( \cos \theta \)).

Step 2: Plot Critical Points

To sketch the limaçon, consider when \( \cos \theta = 1, 0, -1 \): - \( \cos \theta = 1: r = 2 - 3 \, (1) = -1 \), a point inside the loop.- \( \cos \theta = 0: r = 2 \), these are points on the main circle.- \( \cos \theta = -1: r = 2 + 3 \, (1) = 5 \), the farthest point on the loop.

Step 3: Graph the Limaçon

Using the critical points and understanding the symmetry about the polar axis (\( \theta = 0 \)), sketch the limaçon. It should display an inner loop crossing the origin and extending outwards to a maximum of \( r = 5 \) on the opposite side.

Step 4: Calculating the Area Inside the Large Loop

The area enclosed by a polar curve \( r = f(\theta) \) from \( \theta = \alpha \) to \( \theta = \beta \) is \( \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \). For the large loop of the limaçon, split the area into complete loops around the limaçon. Symmetric from \( \theta = -\pi \) to \( \theta = \pi \), calculate only half, multiply by two. \[ A = 2 \times \frac{1}{2} \int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}} (2 - 3 \cos \theta)^2 \, d\theta \]

Step 5: Perform the Integration

First, expand the integrand: \[ (2 - 3 \cos \theta)^2 = 4 - 12 \cos \theta + 9 \cos^2 \theta \]Use the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \) to rewrite \( \cos^2 \theta \). Substitute into the integral and solve.Finally, calculate the definite integral to find:\[ A = \int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}} (4 - 15 \cos \theta + \frac{9}{2} + \frac{9}{2} \cos(2\theta)) \, d\theta \] = \[ \frac{9\pi}{2} + 4 \]

Step 6: Result of the Area Calculation

Evaluate the integral to conclude the total area:\[ A = \text{The area of the region inside the large loop is } \frac{9\pi}{2} + 4 \approx 10.5 \text{ square units.}\]

Key Concepts

Polar Coordinates

The polar coordinate system is an alternative to the Cartesian coordinate system. Unlike Cartesian coordinates, which use a grid of squares, polar coordinates use circles centered around a point (the pole).
Each point in this system is represented by a distance from the pole, denoted as \( r \), and an angle \( \theta \) measured from a reference direction, usually the positive x-axis. Polar coordinates are particularly useful when dealing with problems involving circular symmetry, such as the graphing of limaçons, like the one given here: \( r = 2 - 3 \cos \theta \).

• \( r \) represents the radial distance from the pole.
• \( \theta \) indicates the direction of \( r \).

This system can sometimes make solving and visualizing certain types of problems much simpler, especially when it involves rotations or round shapes.
In the case of our limaçon, understanding the role of \( \cos \theta \) helps predict the direction and features of the curve.

Area Calculation

To find the area enclosed by a polar curve, we use the formula: \[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]where \( r = f(\theta) \) is the equation for the curve.
This works because we essentially "sweep" an infinitesimally small sector and integrate over the interval from \( \alpha \) to \( \beta \).For our specific limaçon, the goal is to calculate the area inside the large loop. Since the curve is symmetric, evaluate the area for one loop of the curve and multiply it by two. This simplifies the calculations significantly. It is also essential to keep in mind the function of the polar radius \( r \), as it may become negative, representing points that "loop back" on themselves. You'd consider these characteristics to calculate only the desired loop's area.Perform the integration by expanding and simplifying the expression for \( r^2 \), applying trigonometric identities to manage terms like \( \cos^2 \theta \), and then solving the resulting integral.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values in their domains. They can simplify complex trigonometric expressions and are particularly valuable in integrals involving these functions.One useful identity is \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \). This helps transform powers of cosine into more manageable forms, often reducing them to expressions more suitable for calculus operations.

• Use identities like \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \) to simplify terms.
• Reduce complex expressions into simpler trigonometric forms.

In solving the area of the limaçon, trigonometric identities are crucial for simplifying the squared cosine term. This makes the integration feasible, as it reduces the problem into solving basic integrals. Knowledge of these identities allows you to solve such integrals faster and with greater ease.

Graphing Polar Equations

Graphing polar equations involves plotting points based on their polar coordinates. The plot of \( r = 2 - 3 \cos \theta \) is a limaçon, a type of polar curve with interesting symmetry properties.To start graphing, it's helpful to calculate critical points and understand the symmetry of the function. For example, because the graph involves \( \cos \theta \), it's symmetric about the polar axis.
When graphing, consider these steps:

• Identify critical points by setting \( \cos \theta \) to 1, 0, and -1.
• Calculate \( r \) for these values to find maximum, minimum, and notable intercepts.

The graph should display a loop, as \( b > a \), which means it features an inner loop crossing the pole. Plotting goes hand-in-hand with understanding what these functions represent in relation to changes in \( \theta \). As you plot, adjust for the wrap around the origin, and focus on key points where the curve reaches its extremes in distance from the pole.