Problem 26 - Thomas Calculus

Question

Sketch the region defined by the inequalities \(0 \leq r \leq 2 \sec \theta\) and \(-\pi / 4 \leq \theta \leq \pi / 4.\)

Step-by-Step Solution

Answer

The region is bounded by lines at \( \pm \pi/4 \) and the curve \( r = 2 \sec \theta \), within these angles.

Step 1: Understand the Polar Coordinate System

In polar coordinates, we use two values: the radial distance \( r \) from the origin and the angle \( \theta \) measured from the positive x-axis. The inequalities given are in terms of \( r \) and \( \theta \). Our range for \( \theta \) is from \(-\pi / 4\) to \(\pi / 4\), creating an arc around the origin.

Step 2: Analyze the radial inequality

The inequality \( 0 \leq r \leq 2 \sec \theta \) governs the radial distance. Here, \( \sec \theta = \frac{1}{\cos \theta} \). Thus, \( r \leq \frac{2}{\cos \theta} \) implies that as \( \theta \) approaches the bounds of its interval, \( r \) will increase towards infinity as \( \cos \theta \to 0 \). However, \( r \) should remain bounded.

Step 3: Determine boundary curves

For the boundary of the region, consider the curve defined by \( r = 2 \sec \theta = \frac{2}{\cos \theta} \). This curve behaves like a vertical line at the angle \( \theta = \pm \pi/4 \), as \( \cos(\pm \pi/4) = \frac{\sqrt{2}}{2} \), leading to \( r = 2\sqrt{2} \).

Step 4: Sketch the region

Begin by drawing polar axes with an origin. From the origin, draw lines at \( \theta = -\pi/4 \) and \( \theta = \pi/4 \) which provides horizontal bounding angles. Next, plot the boundary curve \( r = 2 \sec \theta \). Between these angles, shade the region from \( r = 0 \) up to the curve. This shaded region is the solution area for given inequalities.

Key Concepts

Inequalities

Inequalities play a crucial role in defining regions in polar coordinates. They establish the range of possible values for the radial distance \( r \) and the angle \( \theta \). In the given exercise, the inequalities are \( 0 \leq r \leq 2 \sec \theta \) and \(-\pi/4 \leq \theta \leq \pi/4\). Each inequality helps in constructing a different aspect of the defined region:

• The angle inequality \( -\pi/4 \leq \theta \leq \pi/4 \) determines the angular span of the region. This is the arc around the origin.
• The radial inequality \( 0 \leq r \leq 2 \sec \theta \) specifies the extent of the region from the origin outwards.

Understanding these inequalities helps in finding the boundaries and the shape of the region. The relationship \( \sec \theta = \frac{1}{\cos \theta} \) means that \( r \) will be small near the center and potentially larger near the bounds, as long as \( \cos \theta \) isn't zero. This guides how the region will stretch as \( \theta \) changes.

Sketching Regions

Sketching regions in polar coordinates involves creating a visual representation of the area defined by the inequalities. This is a hands-on way to better understand and interpret the mathematical relationships involved. Let's break down the steps for sketching the given area:

• Begin by plotting the polar coordinate plane. The polar axes typically include a central origin and concentric circles representing different values of \( r \).
• Next, draw the boundary angles, \( \theta = -\pi/4 \) and \( \theta = \pi/4 \). These form the sector bounds of the sketch. Think of this as slicing the polar plane into a wedge from \( -\pi/4 \) to \( \pi/4 \).
• The radial boundary, described by \( r = 2 \sec \theta \), also needs to be drawn. This will form an outer curve, stretching as far as \( \theta \) allows. When \( \theta = \pm \pi/4 \), the radial distance \( r \) becomes \( 2\sqrt{2} \).

Once these boundaries are in place, you can shade the region between \( r = 0 \) to \( r = 2 \sec \theta \) within the specified angles. This shaded area represents the region solution for the inequalities.

Graphing Polar Equations

Graphing polar equations is an essential skill for visualizing problems stated in polar coordinates. The equation \( r = 2 \sec \theta \) in the task functions as a boundary. Here's how to graph it effectively:

• Polar equations describe how \( r \) varies with \( \theta \). For \( r = 2 \sec \theta \), \'sec\' implies the reciprocal of cosine, and knowing \( \sec \theta = \frac{1}{\cos \theta} \) helps in plotting.
• Identify key points by finding values of \( \theta \) where the function has notable behavior, such as at \( \theta = \pm \pi/4 \), where \( \cos(\theta) = \pm \frac{\sqrt{2}}{2} \). This results in a boundary \( r = 2\sqrt{2} \).
• The graph will appear symmetric about the polar axis, as \( \theta \) spans equal angles in either direction from the origin.

Plot these points and connect them smoothely to form the curve. You'll observe that as \( \theta \) nears its boundary, the graph's radial distance increases, sketching a curve that displays the solutions region effectively. Precise graphing aids in accurate representation of regions defined in polar systems.