Problem 33 - College Trigonometry

Question

In Exercises \(31-40\), sketch the region in the \(x y\) -plane described by the given set. $$ \left\\{(r, \theta) \mid 0 \leq r \leq 3 \cos (\theta),-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\right\\} $$

Step-by-Step Solution

Answer

The region is a semicircle centered at \\(x = 1.5, y = 0\\) with radius 1.5 in the upper half-plane.

Step 1: Understand Polar Coordinates

The set is described using polar coordinates, \(r\) and \(\theta\). In polar coordinates, a point is defined by its distance \(r\) from the origin and an angle \(\theta\) from the positive x-axis.

Step 2: Describe the Bounds

The given set specifies that \(0 \leq r \leq 3 \cos(\theta)\) and \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). The angle \(\theta\) varies from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), which represents a semi-circle over the first and fourth quadrants.

Step 3: Convert to Cartesian Coordinates

The equation \(r = 3 \cos(\theta)\) can be converted using the relationships \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Substituting, we obtain \(x = 3 \cos^2(\theta)\) and \(y = 3 \cos(\theta) \sin(\theta)\). Alternatively, set \(r = 3 \cos(\theta)\) to find \(x = 3 - x\) giving \(((x - 1.5)^2 + y^2 = 2.25)\). This describes a circle.

Step 4: Sketch the Region

The equation \((x-1.5)^2 + y^2 = 2.25\) describes a circle centered at \(x = 1.5, y = 0\) with radius 1.5. However, since \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), we only take the upper half of the circle, where \(y\geq0\), giving a semicircle.

Key Concepts

Conversion to Cartesian Coordinates

To transition from polar to Cartesian coordinates, we utilize the fundamental trigonometric relationships:

• For the x-coordinate: \(x = r \cos(\theta)\)
• For the y-coordinate: \(y = r \sin(\theta)\)

These equations allow us to express polar equations in a format that can be plotted on the familiar Cartesian \((x, y)\) plane. For the exercise provided, we use \(r = 3 \cos(\theta)\). Plugging this into our conversions results in \(x = 3 \cos^2(\theta)\) and \(y = 3 \cos(\theta) \sin(\theta)\). By squaring and adding these expressions and utilizing the identity \(\cos^2(\theta) + \sin^2(\theta) = 1\), we can find a more useful form: For the given polar equation, transforming it fully yields a circle equation \((x - 1.5)^2 + y^2 = 2.25\). This describes a circle centered at \((1.5, 0)\) with radius 1.5. This transformation simplifies understanding and sketching of the region on the Cartesian plane.

Sketching Polar Regions

Sketching a polar region first involves understanding the shape defined by the polar equations. In this exercise, the given equation \(r = 3 \cos(\theta)\) represents a circle because it describes all the points equidistant from a point within a plane.The bounds \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) specify angles that span a semi-circle that opens along the x-axis. This is crucial because although the algebraic equation describes an entire circle, the angle interval confines us to sketch a semi-circle only. Visualizing polar regions helps:

• Identify symmetries and bounds in the angles and radii
• Clarify how the shapes look in contrast to their algebraic descriptions

In this scenario, the exercise requires sketching the upper semicircle (where \(y \geq 0\)) since the angles range between the negative and positive halves of the x-axis.

Bounds in Polar Equations

Understanding bounds in polar coordinates is key to accurately interpreting the region involved. The given exercise provides bounds as \(0 \leq r \leq 3 \cos(\theta)\) and \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). These boundaries define the range within which the values of \(r\) and \(\theta\) become valid:

• Radius \(r\) begins from 0 (origin) up to a maximum defined by the curve \(3 \cos(\theta)\).
• As \(\theta\) varies from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), the result is a semi-circle manifested in the upper half of the plane.

These constraints delineate a semicircular area starting at the origin, curving through the positive x-direction and looping back, supplying a visual boundary. Hence, the completion of the polar-to-Cartesian conversion reveals a segment of a circle in the Cartesian plane.