Problem 18 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The polar equation \(r=2-2\cos\theta\) demonstrates symmetry about the x-axis but not about the y-axis or the origin. The graph represents a circle known as a cardioid with a minimum radius of 0 and a maximum radius of 2.

Step 1: Test for symmetry

First, test the given equation \(r=2-2\cos\theta\) for symmetry. Replace \(\theta\) with \(-\theta\), \(\theta\) with \((\pi-\theta)\), and \(r\) with \(-r\) respectively. If any of these substitutions gives the original equation, then there is symmetry according to the associated axis or origin.

Step 2: Substitute \(\theta\) with \(-\theta\)

Substitute \(\theta\) with \(-\theta\) for testing symmetry about the x-axis. The equation becomes \(r=2-2\cos(-\theta)\). Because the cosine function is even, we know that \(\cos(-\theta)=\cos\theta\), so the equation simplifies to \(r=2-2\cos\theta\), which is the original equation. This implies symmetry about the x-axis.

Step 3: Substitute \(\theta\) with \((\pi-\theta)\)

Now, substitute \(\theta\) with \((\pi-\theta)\) in the original equation to test for symmetry about the y-axis. The given equation becomes \(r=2-2\cos(\pi-\theta)\). But we know that \(\cos(\pi-\theta)=-\cos\theta\), so the equation simplifies to \(r=2+2\cos\theta\), which is not the original equation, therefore there is no symmetry about the y-axis.

Step 4: Substitute \(r\) with \(-r\)

Substitute \(r\) with \(-r\) in the original equation to test for symmetry about the origin, The equation becomes \(-r=2-2\cos\theta\), which can also be written as \(r=-2+2\cos\theta\). This is not the original equation, therefore there is no symmetry about the origin.

Step 5: Graph the polar equation

The graph of the polar equation \(r=2-2\cos\theta\) is a circle known as a cardioid. The circle is symmetrical about the x-axis (horizontal line). It has a minimum radius of 0 and a maximum radius of 2, with the maximum occurring at values \(\theta = 0\) and \(\theta = 2\pi\), and a minimum radius occurring at \(\theta = \pi\). Due to symmetry about x-axis, we only need to graph for values from \(\theta = 0\) to \(\theta = \pi\) and the rest of the graph till \(\theta = 2\pi\) will just be a reflection about x-axis.