91Ó°ÊÓ

We are given a polar equation, \(r = 2\csc{\theta}\). We know that the cosecant function is the reciprocal of the sine function, meaning \(\csc{\theta} = \frac{1}{\sin{\theta}}\). With this information, we can rewrite the equation as \(r = \frac{2}{\sin{\theta}}\).

As \(\theta\) changes from 0 to \(2\pi\), we must analyze the behavior of the curve for various values of \(\theta\). Important angles to consider are: - \(\theta = 0\) - \(\theta = \frac{\pi}{2}\) - \(\theta = \pi\) - \(\theta = \frac{3\pi}{2}\) At \(\theta = 0\), \(\sin{\theta} = 0\), since the sine function is 0 at 0. Thus, the equation becomes \( r = \frac{2}{0}\), which is undefined. This means that there will be no point on the curve with \(\theta = 0\). At \(\theta = \frac{\pi}{2}\), \(\sin{\theta} = 1\), since the sine function is 1 at \(\frac{\pi}{2}\). Thus, the equation becomes \(r = \frac{2}{1}\), which is 2. There will be a point on the curve at radius 2 with angle \(\frac{\pi}{2}\). Similarly, at \(\theta = \pi\), the curve will again be undefined as \(\sin(\pi) = 0\). At \(\theta = \frac{3\pi}{2}\), the sine function is -1, so the equation becomes \(r = \frac{2}{-1}\), which means there will be a point on the curve at radius -2 with angle \(\frac{3\pi}{2}\).

Now we can plot the points we found in the cartesian coordinate system using the polar to cartesian conversion \((x, y) = (r\cos{\theta}, r\sin{\theta})\). The points are: - \(r = 2, \theta = \frac{\pi}{2}\), which gives us \((x, y) = (0, 2)\) - \(r = -2, \theta = \frac{3\pi}{2}\), which gives us \((x, y) = (0, -2)\) Now, plot these points on the cartesian coordinate system and connect them with a smooth curve to create the final sketch. You will find that the curve resembles a vertical line at \(x = 0\), except it does not include the points where the equation is undefined, which are \((0,0)\), \((0,-2\pi)\), etc.