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Chapter 6: Problem 36

Test for symmetry and then graph each polar equation. $$r=\sin \frac{\theta}{2}$$

Short Answer

Expert verified
The given polar equation, \(r=\sin \frac{\theta}{2}\), does not show symmetry about the origin, the line \(\theta=\pi/2\), or the polar axis. The graph of this equation is a spiral starting from the origin, with distance from the origin (r) depending on angle (\(\theta\)).

Step by step solution

01

Symmetry Test for the Origin

The equation is symmetric with respect to the origin if replacing \(\theta\) with \(-\theta\) gives the same equation. Substitute \(-\theta\) for \(\theta\) in \(r=\sin \frac{\theta}{2}\). Doing so gives \(r=\sin \frac{-\theta}{2}=-\sin \frac{\theta}{2}=-r\). The equation is not unchanged, hence it does not show symmetry about the origin.
02

Symmetry Test for the line \(\theta=\pi/2\)

The equation is symmetric about the line \(\theta=\pi/2\) if replacing \(\theta\) with \(\pi - \theta\) gives the same equation. Thus substitute \(\pi-\theta\) for \(\theta\) in \(r =\sin \frac{\theta}{2}\), which yields \(r =\sin \frac{\pi-\theta}{2}\). This does not give the original equation hence it does not show symmetry about the line \(\theta=\pi/2\).
03

Symmetry Test for the polar axis

The equation is symmetric about the polar axis if replacing \(\theta\) with \(-\theta\) gives the same equation. Thus substitute \(-\theta\) for \(\theta\) in \(r =\sin \frac{\theta}{2}\), which yields \(r =\sin \frac{-\theta}{2}=-\sin \frac{\theta}{2} = -r\). This does not yield the unchanged equation, so it does not have symmetry around the polar axis.
04

Graph the Polar Equation

Since the given equation is not symmetric about the origin, line \(\theta=\pi/2\), or polar axis, we will proceed to draw it directly. The graph will look like a spiral starting from the origin, with distance from the origin \((r)\) depending on angle \((\theta)\).

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