Plot each polar curve to find the first range of positive -values for which \( r \) is negative.

Assume we wish to determine, for a given polar curve \( r = f(\theta) \), the first range of positive \(\theta\) values for which the radius is negative (that is, where \( f(\theta) < 0 \)). In many textbook examples the procedure is similar. We illustrate with two common examples. --- ### Example 1. \( r = 1 + 2\sin\theta \) 1. **Find when \( r \) is negative.** We require \[ 1 + 2\sin\theta < 0. \] Subtract 1 from both sides: \[ 2\sin\theta < -1. \] Divide by 2: \[ \sin\theta < -\frac{1}{2}. \] 2. **Determine the corresponding \(\theta\) values.** The sine function is less than \(-\frac{1}{2}\) for angles in the third and fourth quadrants. In one cycle, we have \[ \theta \in \left(\frac{7\pi}{6}, \frac{11\pi}{6}\right). \] Since we seek the first range for positive \(\theta,\) the interval \(\left(\frac{7\pi}{6}, \frac{11\pi}{6}\right)\) is the answer. 3. **Plotting insight.** When you plot \( r = 1+2\sin\theta \), the curve (a limacon) crosses the pole (origin) at \(\theta\) where \( r=0 \). As you progress through increasing \(\theta\) from 0, the value of \( r \) eventually becomes negative as soon as \(\theta\) exceeds \(\frac{7\pi}{6}\) and remains negative up to \(\theta = \frac{11\pi}{6}\). --- ### Example 2. \( r = 2\cos\theta \) 1. **Find when \( r \) is negative.** Set \[ 2\cos\theta < 0. \] Dividing by 2 (and noting that 2 is positive so the inequality direction is preserved) gives: \[ \cos\theta < 0. \] 2. **Determine the corresponding \(\theta\) values.** The cosine function is negative when \(\theta\) is in the second and third quadrants: \[ \theta \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right). \] The **first** range (the smallest positive angles for which \( r \) becomes negative) is obtained by taking the part until the graph “repeats” its behavior. In many texts, one shows that as \(\theta\) increases from 0, the graph first has \( r>0 \) until \(\theta = \frac{\pi}{2}\) and then \( r \) is negative for \(\theta\) just beyond \(\frac{\pi}{2}\). For many purposes you might denote the answer as \[ \theta \in \left(\frac{\pi}{2}, \pi\right), \] although the complete interval of \(\theta\) giving negative \(r\) is \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\). (Often students choose the smallest interval that “traces” the negative part of the graph uniquely.) 3. **Plotting insight.** When you graph \( r = 2\cos\theta \) you see that the right half of the circle is traced with \( r>0 \) and the left half (which corresponds to angles between \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\)) is drawn with negative values of \( r \). If you wish to “start” your graph at a positive \(\theta,\) the very first negative \( r \) values occur beginning just after \(\theta=\frac{\pi}{2}\). --- ### General Strategy For any polar curve \( r = f(\theta) \): 1. **Set up the inequality:** Solve \( f(\theta) < 0 \). 2. **Solve for \(\theta\):** Use algebra and knowledge of the trigonometric functions to find the intervals. 3. **Interpret the graph:** Sketch (or use a graphing utility) to check your results. Notice where the curve crosses the pole (where \( r=0 \)); these crossing points typically mark the boundaries of the intervals where the sign of \( r \) changes. --- ### Final Answers for the Examples - For \( r = 1+2\sin\theta \), the first range of positive \(\theta\) for which \( r \) is negative is \[ \theta \in \left(\frac{7\pi}{6}, \frac{11\pi}{6}\right). \] - For \( r = 2\cos\theta \) (if considering the graph’s initial negative region), one acceptable answer is \[ \theta \in \left(\frac{\pi}{2}, \pi\right), \] though the full interval is \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\). Use these methods to analyze any other polar curve by finding when \( r < 0 \).