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Chapter 6: Problem 12

Test for symmetry with respect to a. the polar axis. b. the line \(\theta=\frac{\pi}{2}\) c. the pole. $$r^{2}=16 \sin 2 \theta$$

Short Answer

Expert verified
The graph of the given equation is only symmetric about the pole.

Step by step solution

01

Test for Symmetry about the Polar Axis

Replace \(\theta\) with \(-\theta\). In our equation, we get: \( r^{2} = 16 \sin 2(-\theta) = 16 \sin -2\theta = -16 \sin 2\theta \). These results show that the graph isn't symmetric with respect to the polar axis since the equation was altered.
02

Test for Symmetry about the line \(\theta = \frac{\pi}{2}\)

Replace \(\theta\) with \(\pi - \theta\). In our equation, we get: \( r^{2} = 16 \sin 2(\pi - \theta) = 16 \sin 2\pi - 16 \sin 2\theta = -16 \sin 2\theta \). This shows that the graph isn't symmetric with respect to the line \(\theta = \frac{\pi}{2}\) since the equation was altered.
03

Test for Symmetry about the Pole

Replace \(\theta\) with \(\theta + \pi\). In our equation, we get: \( r^{2} = 16 \sin 2(\theta + \pi) = 16 \sin (2\theta + 2\pi) = 16 \sin 2\theta \). The equation remains unchanged. Therefore, the graph is symmetric with respect to the pole.

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