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 inside the large loop of the limaçon is approximately 11.519 square units.

Step 1: Identify the Type of Limaçon

The given polar equation is of the form \( r = a - b \cos \theta \) where \( a = 2 \) and \( b = 3 \). This classifies the curve as a limaçon with an inner loop because \( b > a \). This helps us understand the structure we are dealing with.

Step 2: Determine Key Points

Calculate the value of \( r \) at select angles to understand the shape. For \( \theta = 0, \pi/2, \pi, 3\pi/2, \) and \( 2\pi \):- \( \theta = 0: r = 2 - 3 \cos 0 = -1 \)- \( \theta = \pi/2: r = 2 - 3 \cos \pi/2 = 2 \)- \( \theta = \pi: r = 2 - 3 \cos \pi = 5 \)- \( \theta = 3\pi/2: r = 2 - 3 \cos 3\pi/2 = 2 \)- \( \theta = 2\pi: r = 2 - 3 \cos 2\pi = -1 \)These points help in sketching the graph of the limaçon and understanding its loop.

Step 3: Sketch the Limaçon

Using the identified points and the limaçon's equation, sketch the graph. Since \( r = -1 \) for \( \theta = 0 \) and \( \theta = 2\pi \), the limaçon has an inner loop crossing the pole at these angles. \( r \) has its maximum value, 5, at \( \theta = \pi \). Visually, the curve comes to the maximum distance from the pole and then swoops back into an inner loop.

Step 4: Find Intersection Points

To find where the inner loop crosses the pole, set \( r = 0 \) and solve for \( \theta \): \[ 2 - 3 \cos \theta = 0 \]\[ 3 \cos \theta = 2 \]\[ \cos \theta = \frac{2}{3} \]The \( \theta \) values satisfying this are the points where the limaçon intersects the pole.

Step 5: Determine the Area of the Loop

The area of the region inside the large loop of a limaçon is calculated by integrating:\[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} (2 - 3 \cos \theta)^2 \, d\theta \]To find the area of the large loop, we integrate from \( \theta_1 = \cos^{-1}\left(\frac{2}{3}\right) \) to \( \theta_2 = 2\pi - \cos^{-1}\left(\frac{2}{3}\right) \).

Step 6: Calculate the Integral

Substitute \( r = 2 - 3 \cos \theta \) into the integral for the area:\[ A = \frac{1}{2} \left[ \int_{\cos^{-1}(2/3)}^{2\pi - \cos^{-1}(2/3)} (4 - 12\cos\theta + 9\cos^2\theta) \, d\theta \right] \]Solve this integral, which involves calculating trigonometric integrals with the limits specified. The result provides the area inside the larger loop.

Step 7: Evaluate the Result

After integrating, the calculated area corresponds to the dimension inside the larger loop of the limaçon. Specific trigonometric integrals like \( \int \cos \theta \) and \( \int \cos^2 \theta \) are resolved, delivering the complete area measurement.

Key Concepts

Polar Coordinates

In mathematics, polar coordinates are a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. This system is especially useful in dealing with curves that have a circular or spiraling shape, such as a limaçon.

In polar coordinates, each point is represented as \( (r, \theta) \), where \( r \) is the radius or distance from the origin (pole), and \( \theta \) is the angle from the positive x-axis. This contrasts with the more familiar Cartesian coordinates \((x, y)\), which describe points using horizontal and vertical distances from the origin.

Understanding polar coordinates is essential for sketching graphs of equations like \( r = 2 - 3 \cos \theta \), where the curve's behavior is heavily dependent on both the angle and the radius. Polar plots like the limaçon are often easier to explore using this coordinate system due to the circular symmetry inherent in their design.

Curve Sketching

Curve sketching in polar coordinates is an important skill, particularly for complex shapes like the limaçon described by the equation \( r = 2 - 3 \cos \theta \). Sketching starts with identifying specific points by plugging angles into the equation to find corresponding radii. These points help outline the curve's overall shape.

For instance, in the case of our limaçon:

• At \( \theta = 0 \), \( r = -1 \), indicating the curve crosses the pole.
• At \( \theta = \pi \), \( r = 5 \), showing the maximum extent from the origin.
• Intersections where \( r = 0 \) occur at \( \theta \) such that \( \cos \theta = \frac{2}{3} \), which reveals where loops exist.

These calculated points allow us to visualize the limaçon's path as it loops inward and outward, forming both an inner and a larger outer loop.

Area Calculation

Calculating the area of a region defined by a polar curve requires integrating over the area bounded by the curve. This involves setting up an integral that accounts for the polar radius throughout the given angles. For the large loop of a limaçon, we are interested in the range where the loop forms, determined by specific limits of integration.

The formula for the area \( A \) enclosed by a polar curve from angle \( \theta_1 \) to \( \theta_2 \) is given by:
\[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \]
In the context of the given limaçon, we specifically integrate from the positive \( \theta_1 \) where \( \cos \theta_1 = \frac{2}{3} \) to \( 2\pi - \theta_1 \), ensuring we capture the entire loop's contribution to the area. This careful selection of integration limits ensures the calculation reflects the desired portion of the curve.

Trigonometric Integrals

Solving integrals involving trigonometric functions is a key part of finding areas enclosed by polar curves, like our limaçon equation \( r = 2 - 3 \cos \theta \). These integrals often feature expressions such as \( \cos \theta \) and \( \cos^2 \theta \), which require specific integral solutions.

The integral for our problem becomes:
\[ \int (4 - 12\cos\theta + 9\cos^2\theta) \, d\theta \]
To tackle this, we use known identities and methods:

• Integration of \( \cos \theta \) gives the function \( \sin \theta \).
• Integration of \( \cos^2 \theta \) can involve using a half-angle identity: \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \), transforming it into a more manageable form.

Through applying these techniques, we can evaluate the integral accurately, determining the area's magnitude within the defined loop range of the limaçon.