91Ó°ÊÓ

The given equation is \(r=\frac{1}{1-2 \cos\theta}\). First, we need to determine the domain for \(\theta\) and \(r\). Since the problem asks for the curve as \(\theta\) increases from 0 to \(2 \pi\), we can use that as our domain for \(\theta\). The only potential restriction to this domain could come from the denominator of the fraction, which can't be equal to zero. This would occur when \(1-2 \cos\theta = 0\). Solving for \(\theta\), we find that \(\cos\theta = \frac{1}{2}\), which corresponds to \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{5 \pi}{3}\). These points create discontinuity; thus, we will split our domain into intervals: \([0, \frac{\pi}{3}), (\frac{\pi}{3}, \frac{5\pi}{3})\) and \((\frac{5\pi}{3}, 2\pi]\). There are no restrictions on the variable \(r\), except that it will be defined within the domain for \(\theta\).

Now we'll determine the behavior of the curve. To do so, we might want to analyze how cosine behaves and how it affects our equation: • \(\cos \theta > 0\) when \(0 < \theta < \pi\). The denominator in our equation will be less than 1, making \(r\) positive. • \(\cos \theta < 0\) when \(\pi < \theta < 2 \pi\). The denominator in our equation is greater than 1, making \(r\) negative. The cases we are most interested in happen near the discontinuity points at \(\theta=\frac{\pi}{3}\) and \(\theta=\frac{5\pi}{3}\). As \(\theta\) approach the discontinuity points: • \(r \rightarrow \infty\) as \(\theta \rightarrow \frac{\pi}{3}\) from the left. • \(r \rightarrow \infty\) as \(\theta \rightarrow \frac{5\pi}{3}\) from the right.

Using the information from steps 1 and 2, we can sketch the given polar equation. Since we're dealing with a discontinuity, we need to graph the curve in segments. 1. Start by drawing the curve for \(0 \leq \theta < \frac{\pi}{3}\). As the angle approaches \(\frac{\pi}{3}\), \(r\) becomes extremely large. 2. Skip the angle \(\theta = \frac{\pi}{3}\). 3. Begin graphing again for \(\frac{\pi}{3} < \theta \leq \frac{5\pi}{3}\). For these angles, the denominator will be greater than 1, making \(r\) negative and placing the points in the opposite direction. 4. As we approach the discontinuity at \(\theta = \frac{5\pi}{3}\), we will again observe \(r \rightarrow \infty\). 5. Finally, draw the curve for \(\frac{5\pi}{3} < \theta \leq 2\pi\). As the angle moves towards \(2\pi\), \(r\) becomes positive and the curve joins the starting point. Be sure to indicate the direction of the curve as \(\theta\) increases with arrows, and label points demonstrating the behavior of the curve near the discontinuity points and at the starting and ending points.