Chapter 6.5, Problem 56E

Solution

(a)

To describe: The graph of the polar equation $r=1+3\sin 3\theta$ .

The graph of the polar equation $r=1+3\sin 3\theta$ shows graph of rose cure have three long petals of each length $4\text{units}$ and three short petal each of length $2\text{units}$ .

Given information:

The given polar equation is $r=1+3\sin 3\theta$ .

Calculation:

Graph for the polar equation $r=1+3\sin 3\theta$ .

  

Figure (1)

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

The range of $\sin 3\theta$ , $-1\le \sin 3\theta \le 1$ .

Here, the range of the polar equation $r=1+3\sin 3\theta$ is $\left[2,4\right]$ by taking maximum value of $\sin 3\theta$ as $1$ and minimum value of $\sin 3\theta$ as $-1$ .

Therefore, the maximum value of the petal is $4$ and minimum value of the petal is $\left|-2\right|=2$ .

(b)

To find: If there exist any symmetry in the graph.

The graph is symmetric about the $y-$ axis.

Given information:

The given polar equation is $r=1+3\sin 3\theta$ .

Calculation:

From part (a), the graph represents a rose curve symmetric about the $y-$ axis.

c)

To find: The maximum value of $\left|r\right|$ if it exists.

The maximum value of $r$ is $4$ .

Given information:

The given polar equation is $r=1+3\sin 3\theta$ .

Calculation:

From part (a), the maximum value of $\left|r\right|$ exists.

The maximum value of $r$ is $\left|r\right|=4$ .