91Ó°ÊÓ

First, let's find the key points by analyzing the equation \(r = \sin 2\theta\). We will start by finding the points where the curve intersects the polar axis (i.e. when \(r = 0\)). Setting \(r = 0\): \[0 = \sin 2\theta \] To solve for \(\theta\), remember that the sine function equals to zero at the angles which are integer multiples of \(\pi\). Therefore: \[2\theta = k\pi\] \[\theta = \frac{k\pi}{2}\] where \(k\) is an integer. In order to find one full period of the curve, we take the values of \(k = 0, 1, 2, ..., 7\), which gives us the polar angles: \[\theta = \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, 3\pi, \frac{7\pi}{2}\}\] Now, let's find the maximum radius value for each interval. To do this, find when the derivative of the function equals to zero: \[\frac{dr}{d\theta} = 2\cos 2\theta = 0\] \[\cos 2\theta = 0\] This cosine function is equal to zero at angles \(\frac{(2n + 1)\pi}{2}\), where \(n\) is an integer. Therefore, for each interval, we take the midpoint angle as the maximum radius: \[\theta_{max} = \{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\}\] Since the sine function oscillates between -1 and 1, the maximum radius values will be: \[r_{max} = \{\sin\frac{\pi}{2}, \sin\frac{3\pi}{2}, \sin\frac{5\pi}{2}, \sin\frac{7\pi}{2}\} = \{1, -1, 1, -1\}\] Now, we have the key points on the curve: \[\theta: \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, 3\pi, \frac{7\pi}{2}\}\] \[r: \{0, 1, 0, -1, 0, 1, 0, -1\}\]

Using the key points found in the previous step, we will now sketch the curve. Begin at the origin (0,0) and move in the direction of increasing \(\theta\) while keeping track of the changing radius. 1. We begin at \(\theta = 0\), with a radius of \(r = 0\). This point lies at the origin. 2. At \(\theta = \frac{\pi}{4}\), the radius is at its maximum value, \(r = 1\). Plot this point and notice the curve extends in the direction of increasing \(\theta\). 3. When \(\theta = \frac{\pi}{2}\), the radius becomes \(r = 0\), which means we are back at the origin. This completes one leaf of the rose. 4. As \(\theta\) continues increasing, the next maximum value for the radius occurs at \(\theta = \frac{3\pi}{4}\), with a radius of \(r = -1\). Since the radius is negative, this will plot the point at the same distance but opposite direction in the polar coordinate plane, that is, with angle \(\theta = \frac{7\pi}{4}\) and radius \(r = 1\). 5. Repeat this process for the other maximum values to complete the remaining leaves of the rose. Now we have sketched the curve of the polar equation \(r = \sin2\theta\), which describes a four-leaved rose.