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Chapter 6: Problem 11

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

Short Answer

Expert verified
The given equation \(r^{2}=16 \cos 2 \theta\) is found to be symmetric with respect to the polar axis, the line \(\theta=\frac{\pi}{2}\), and the pole.

Step by step solution

01

Test for symmetry with respect to the polar axis

A polar graph is symmetric with respect to the polar axis if the equation remains the same when \(\theta\) is replaced by \(-\theta\). Let's replace \(\theta\) in the given equation with \(-\theta\). We have \[r^{2}=16 \cos 2 (-\theta)\] Simplifying this, we get \[r^{2}=16 \cos (-2\theta)\] And from the properties of cosine function, we have \(\cos (-2\theta)=\cos 2\theta\), which means the given equation is unchanged, so there is symmetry with respect to the polar axis.
02

Test for symmetry with respect to the line \(\theta=\frac{\pi}{2}\)

A polar graph is symmetric with respect to the line \(\theta=\frac{\pi}{2}\) if the equation remains the same when \(\theta\) is replaced by \(\pi-\theta\). Let鈥檚 replace \(\theta\) in the given equation with \(\pi-\theta\): This results in \[r^{2}=16 \cos 2 (\pi-\theta)\] Simplifying, we get \[r^{2}=16 \cos (2\pi-2\theta)\]. But since the cosine function has a period of \(2\pi\), \(\cos (2\pi-2\theta) = \cos (-2\theta)\) which is equal to \(\cos 2\theta\). Hence, the given equation is unchanged and there is symmetry with respect to the line \(\theta=\frac{\pi}{2}\).
03

Test for symmetry with respect to the pole

A polar graph is symmetric with respect to the point of the pole if the equation remains the same when \(r\) is replaced by \(-r\). In the equation, there is no explicit \(r\) but there is \(r^{2}\). Since \((-r)^{2}\) is same as \(r^{2}\), for this case, there is symmetry with respect to the pole.

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