Problem 36 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The polar equation \(r=\sin \frac{\theta}{2}\) does not possess any form of symmetry (origin, polar axis or the line \(\theta=\pi/2\)). Its graph is a spiral that originates from the origin and gets progressively farther away from it as \(\theta\) increases.

Step 1: Test for Symmetry

To test for symmetry with respect to the origin, replace \(r\) with \(-r\) and \(\theta\) with \(\theta + \pi\). If the equation remains the same, it possesses symmetry around the origin. Here, that gives \(-r = \sin \frac{(\theta + \pi)}{2}\). As the equation has changed, it is not symmetric about the origin.\n\nNext, to test for symmetry with respect to the polar axis, replace \(\theta\) with \(-\theta\). If the equation remains unchanged, it has symmetry about the polar axis. The alteration yields \(r= \sin \frac{-\theta}{2}\), which can be simplified to \(r= \sin|-\frac{\theta}{2}|\) due to the absolute nature of sine function. The resulting form is not equivalent to the initial equation, so there's no symmetry about the polar axis. \n\nLastly, to verify symmetry about the line \(\theta=\pi/2\), replace \(r\) with \(-r\) and \(\theta\) with \(-\theta\). If the equation remains the same, it is symmetric about \(\theta=\pi/2\). Substituting gives us \(-r= \sin \frac{-\theta}{2}\), which simplifies to -r= \sin|\frac{-\theta}{2}|\). Since the modified equation differs from the original one, the graph has no symmetry about the line \(\theta=\pi/2\)

Step 2: Graph the Equation

First, create a coordinate system for polar equations with a radial line (for the variable \(r\)) and an angle line (for variable \(\theta\)). Since our equation does not show any symmetry, we need to plot points for different values of \(\theta\) to see where they land in our polar grid. This may require trial and error to determine the right range and density of plotted points to reveal the shape. \n\nAfter plotting numerous points (for instance, points at \( \theta = 0, \pi/4, \pi/2, 3\pi/4, \pi, 5\pi/4, 3\pi/2, 7\pi/4 \), and \(\ 2\pi \)), the graph of the equation will start to appear. The shape of this function will be a spiral starting at the origin, becoming farther away from the origin as \(\theta\) increases.