Problem 60 - Calculus Early Transcendentals

Question

Use a graphing utility to graph the following equations. In each case, give the smallest interval \([0, P]\) that generates the entire curve. \(r=1-2 \sin 5 \theta\)

Step-by-Step Solution

Answer

Answer: The smallest interval [0, P] that generates the entire curve of the polar equation \(r = 1 - 2\sin(5\theta)\) is \([0, \frac{2\pi}{5}]\).

Step 1: Graph the polar equation r = 1 - 2sin(5θ)

To graph the given polar equation, we can begin by using a graphing calculator or an online graphing tool that allows for inputting polar equations. Simply input the polar equation \(r = 1 - 2\sin(5\theta)\) and generate the graph. Through this, we can better understand the properties of the polar curve, and we will need this in order to find the smallest interval \([0, P]\) that generates the entire curve.

Step 2: Identify the properties of the sine function

To understand the polar curve, we need to recognize properties of the sine function. In general, the sine function has a periodic property and we can determine the period by looking at the coefficient of \(\theta\). In this case, the coefficient of \(\theta\) is \(5\). This information is useful for figuring out P.

Step 3: Determine the period of the given polar equation

In order to find the smallest interval \([0, P]\) that generates the entire curve, we need to determine the period of the polar equation. Keep in mind that the sine function has a period of \(2\pi\). Since the coefficient of \(\theta\) in \(1 - 2\sin(5\theta)\) is \(5\), we divide the period of the sine function by the coefficient to get the period of the given polar equation: $$P = \frac{2\pi}{5}$$

Step 4: Write the final answer

Based on our calculations, the smallest interval \([0, P]\) that generates the entire curve of the polar equation \(r = 1 - 2\sin(5\theta)\) is: $$[0, \frac{2\pi}{5}]$$

Key Concepts

Graphing Utilities

Graphing utilities are powerful tools that help us visualize equations and explore their properties. When dealing with polar equations like \(r = 1 - 2\sin(5\theta)\), these utilities can make the process straightforward.

Here’s how they work:

• Graphing tools can plot equations in different formats, including Cartesian and polar coordinates.
• For polar equations, you simply input \(r\) and \(\theta\) values.
• Once the data is entered, the utility generates a graph showing the curve on a polar plane.

Using graphing utilities saves time and allows for exploration of curve behavior and characteristics, aiding in comprehending complex equations.

Periodicity of Sine Function

The sine function is key in polar equations and has distinctive periodic properties. Understanding its periodicity aids in interpreting curves like \(r = 1 - 2\sin(5\theta)\).

The periodicity of the sine function involves its recurring pattern, usually completing a cycle every \(2\pi\). However, when modified by a coefficient, this period changes. For example:

• The standard period of \(\sin(\theta)\) is \(2\pi\).
• When you have \(\sin(k\theta)\), the period becomes \(\frac{2\pi}{k}\).

In our equation, the coefficient is 5, thus the period is altered to \(\frac{2\pi}{5}\). This knowledge is crucial for finding the interval that fully represents the graph.

Polar Equations

Polar equations represent points on a plane using radius and angle, \((r, \theta)\). Unlike Cartesian coordinates, this system utilizes circles and angles to express locations and shapes.

Here are some aspects to consider:

• Polar coordinates are especially useful for curves and shapes that naturally align with circular geometry.
• In polar equations like \(r = 1 - 2\sin(5\theta)\), \(r\) determines the distance from the origin, and \(\theta\) sets the angle.
• By manipulating parameters in polar equations, interesting patterns and symmetries can emerge.

Polar equations provide a different perspective and are instrumental in understanding wave functions, spirals, and intricate patterns.