91Ó°ÊÓ

To identify the key points of the curve, we need to evaluate the function \(r(\theta) = \cos 3\theta\) at the domain limits (i.e., at \(0\) and \(\frac{1}{2}\pi\)), and other critical points if there are any. Let's find r at these values: 1. When \(\theta=0\), \(r(\theta) = \cos 3\cdot0 = \cos 0 = 1\). As a result, point A is \((1,0)\). 2. When \(\theta=\frac{1}{2}\pi\), \(r(\theta) = \cos 3\cdot\frac{1}{2}\pi = \cos\frac{3}{2}\pi = 1\). As a result, point B is \((1,\frac{1}{2}\pi)\). Since we only have two key points, we can proceed to the next step directly.

We will now sketch the curve by connecting point A to point B using the function \(r(\theta) = \cos 3\theta\). To do this, let's divide the domain [0, \(\frac{1}{2}\pi\)] into smaller intervals, let's say \(0, \frac{1}{6}\pi, \frac{1}{3}\pi, \frac{1}{2}\pi\). For each of these values, let's compute the corresponding r value and represent the resulting points in the polar plane. 1. When \(\theta=\frac{1}{6}\pi\), \(r(\theta) = \cos 3\cdot\frac{1}{6}\pi = \cos\frac{1}{2}\pi = 0\). As a result, point C is \((0,\frac{1}{6}\pi)\). 2. When \(\theta=\frac{1}{3}\pi\), \(r(\theta) = \cos 3\cdot\frac{1}{3}\pi = \cos\pi = -1\). As a result, point D is \((-1,\frac{1}{3}\pi)\). 3. When \(\theta=\frac{1}{2}\pi\), we have already found the point B earlier, that is \((1,\frac{1}{2}\pi)\). So the key points on the curve are point A, C, D, and B. Now connect these points in the polar plane to form the curve. Begin from point A, move down to point C, then out to point D, and finally back to point B. This completes the sketch of the polar curve \(r=\cos 3\theta\) for \(0 \leq \theta \leq \frac{1}{2}\pi\).