Problem 53

Question

Identify and graph each polar equation. $$ r=4 \sin (5 \theta) $$

Step-by-Step Solution

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Answer
Graph the rose curve with 5 petals. Petals are evenly spaced around the origin.
1Step 1: Identify the type of polar equation
The given equation is in the form of \( r = a \, \text{sin}(k\theta) \), which is known as the rose curve equation. Here, \( a = 4 \) and \( k = 5 \).
2Step 2: Determine the petals number and symmetry
To determine the number of petals, use the formula: \( \text{Number of petals} = k \). Since \( k = 5 \), there will be 5 petals. The rose curve \( r = a \, \text{sin}(k\theta) \) is symmetry with respect to the origin.
3Step 3: Plot key points
Calculate key points by substituting values for \( \theta \) and solving for \( r \). For example: when \( \theta = 0 \), \( r = 4 \, \text{sin}(0) = 0 \); when \( \theta = \frac{\pi}{10} \), \( r = 4 \, \text{sin}(0.5\pi) = 4 \), and so forth. Repeat for several values of \( \theta \).
4Step 4: Draw the polar graph
Using the calculated points, plot the curve on polar graph paper. Due to the symmetry and periodic nature of sin functions, the petals will be evenly spaced around the origin. Draw the five evenly-spaced petals extending from the origin.

Key Concepts

Rose CurvePolar GraphSine Function
Rose Curve
The rose curve is a fascinating and visually appealing type of polar curve. It gets its name because it resembles a flower with petals. The general form of a rose curve equation is \(r = a \sin(k\theta)\), where \(a\) and \(k\) are constants.
In the equation given in the exercise, \(r = 4 \sin(5\theta)\), the values \(a = 4\) and \(k = 5\). This tells us that:
  • The length of each petal is 4, since \(a\) determines the peak radius.
  • There are 5 petals, since \(k = 5\).
The rose curve has interesting properties such as symmetry and periodic patterns. For \(\sin(k\theta)\), the curve is symmetric with respect to the origin and possibly other axes depending on \(k\). When \(k\) is odd, like here with \(k = 5\), the number of petals is exactly \(k\). This simple identification helps in easily predicting the shape of a rose curve before even graphing it.
Polar Graph
A polar graph uses a two-dimensional coordinate system where each point on the plane is determined by an angle and a distance. In simple terms, instead of using x and y coordinates like in Cartesian systems, we use radius \(r\) and angle \(\theta\).

To plot points on a polar graph:
  • Measure the angle from a fixed direction (usually the positive x-axis).
  • Move outwards by the given radius from the origin.
In the case of our rose curve \(r = 4 \sin(5\theta)\), you plot several points by varying \(\theta\) and calculating the corresponding \(r\). For instance, when \(\theta\) is 0, \(r\) is 0 because \(\sin(0) = 0\). As \(\theta\) increases, \(r\) changes following the sine function, forming the petals symmetrically around the origin.
Polar graphs are particularly useful for plotting equations that describe circles, spirals, and other complex curves that are simpler to describe in terms of angles and radii.
Sine Function
The sine function, denoted as \(\sin\), is fundamental in trigonometry and describes a smoothly oscillating wave. It is periodic, meaning it repeats its values in regular intervals.

Key points about the sine function \(\sin(x)\):
  • The function ranges from -1 to 1.
  • It has a period of \(2\pi\), which means \(\sin(x + 2\pi) = \sin(x)\) for any \(x\).
  • The sine curve crosses the x-axis at integer multiples of \(\pi\): \(\sin(0) = 0, \sin(\pi) = 0\).
In polar equations like \(r = 4 \sin(5\theta)\), the sine function dictates the radial distance \(r\) from the origin. Because sine oscillates between 4 and -4 (after being multiplied by 4), the petals of the rose curve are formed by these oscillations as \(\theta\) varies.
Understanding the sine function’s behavior is crucial for analyzing and graphing polar equations, especially those that describe curves like the rose curve. It helps predict how the radius \(r\) will change as the angle \(\theta\) sweeps around the circle.