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Chapter 7: Problem 94

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

Short Answer

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

Step by step solution

01

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.
02

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.
03

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.

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Most popular questions from this chapter

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