Chapter 6.5, Problem 43E

Solution

To analyze: The graph given by the polar equation $r^2=\sin 2\theta$ such that $0\le \theta \le 2\pi$ .

The equation represents a lemniscate. Domain of the graph is $\left[0,\frac{\pi }{2}\right]\cup \left[\pi ,\frac{3\pi }{2}\right]$ . The range is $\left[0,1\right]$ . The graph is continuous and symmetric about the origin. Maximum $r$ value is $1$ . There are no asymptotes to the graph.

Given information:

The given polar equation is $r^2=\sin 2\theta$ .

Calculation:

Rewrite the polar equation as, $r=\pm \sqrt{\sin 2\theta }$ .

Plot the graph for the polar equation.

  

Figure (1)

From the figure (1), it can be observed that the equation represents a lemniscate.

The equation is defined only when $\sin 2\theta \ge 0$ . Therefore, the domain of the graph is $\left[0,\frac{\pi }{2}\right]\cup \left[\pi ,\frac{3\pi }{2}\right]$ .

The range of $\sin 2\theta$ , is $0\le \sin 2\theta \le 1$ . Therefore, the range of $r^2=\sin 2\theta$ is $\left[0,1\right]$ by taking maximum value of $\sin 2\theta$ as $1$ and minimum value as $0$ .

The graph is continuous on its domain. The graph is symmetric about the origin.

Since $r$ values are bounded by $\left[0,1\right]$ , so the graph is bounded.

The maximum value of $r$ is $1$ .

There are no asymptotes to the graph.

Therefore, the polar equation $r^2=\sin 2\theta$ is analyzed with the help of figure (1).