Problem 13 - Algebra and Trigonometry

Question

Test for symmetry and then graph each polar equation. $$r=2 \cos \theta$$

Step-by-Step Solution

Answer

The polar equation \(r=2 \cos \theta\) is symmetric with respect to the polar axis. However, it is not symmetric with respect to the pole or the line \( \theta=\pi/2\). The graph of the equation represents a circle centered at (1, 0) with a radius of 1.

Step 1: Symmetry Test

There are three types of symmetry to test for in polar coordinates: symmetry about the polar axis, symmetry about the pole and symmetry about the line \( \theta=\pi/2\). For symmetry about the polar axis, replace \( \theta \) with \( -\theta \) in the equation. If the equation remains unchanged, then it's symmetric about the polar axis. In our equation, \(r=2 \cos (-\theta) = 2 \cos \theta\), the equation does not change, so it's symmetric about the polar axis. For symmetry about the pole, replace \( r \) with \( -r \) in the equation. If the equation remains the same, then it's symmetric about the pole. However, if we replace \( r \) with \( -r \), the equation changes to \( -r=2 \cos \theta\), which is not symmetric about the pole. For symmetry about \( \theta=\pi/2 \), replace \( \theta \) with \( \pi-\theta \) in the equation. If this results in the same equation, then it's symmetric about \( \theta=\pi/2 \). In our case, \(r = 2 \cos (\pi - \theta) = - 2 \cos \theta\), which is different from the original equation, it's not symmetric about \( \theta=\pi/2 \).

Step 2: Graphing

With symmetry details in hand and considering the equation is in the form \( r = a \cos \theta \), it represents a circle with centre at \( (a/2, 0) \), and radius \( a/2 \), where a > 0. here a = 2. Hence it's a circle with center at (1, 0) and radius 1. The plot points are (2, 0), (1, π/2), (0, π), and (1, 3π/2) in polar coordinates.