Problem 94 - Algebra and Trigonometry

Question

Graph \(r_{1}\) and \(r_{2}\) in the same polar coordinate system. What is the relationship between the two graphs? $$r_{1}=2 \sin 3 \theta, r_{2}=2 \sin 3\left(\theta+\frac{\pi}{6}\right)$$

Step-by-Step Solution

Answer

The graph of \(r_{2}\) is a rotation of the graph of \(r_{1}\) counterclockwise by \(\frac{\pi}{6}\) radians.

Step 1: Graphing \(r_{1}=2 \sin 3 \theta\)

To graph this equation, the student should recognize that the coefficient 2 will affect the size of the shape, while the 3 inside the sine function will determine the number of leaves in a rose curve. Due to the number being odd, there will be 3 leaves. Plotting several points by varying \(\theta\) from 0 to \(2\pi\) will help in creating an accurate graph.

Step 2: Graphing \(r_{2}=2 \sin 3(\theta+ \frac{\pi}{6})\)

This equation, like the first, will yield a rose curve with 3 leaves due to the 3 inside the sine function. However, the \(\frac{\pi}{6}\) added to \(\theta\) represents a phase shift. It means the entire graph of \(r_{1}\) will be rotated counterclockwise by \(\frac{\pi}{6}\) radians. Again, plotting points for \(\theta\) ranging from 0 to \(2\pi\) will give an accurate graph.

Step 3: Relationship between the two graphs

After graphing both equations, it can be seen that the graph for \(r_{2}\) is the same shape as \(r_{1}\), but it is rotated counterclockwise by \(\frac{\pi}{6}\) radians. This demonstrates the effect a phase shift has on the graph of a polar equation.