Chapter 6.5, Problem 68E

Solution

a)

To prove: The graph of the polar equation $r=a\sin n\theta$ is not symmetric about the $y$ -axis.

It is proved that $r=a\sin n\theta$ is symmetric about the $y$ -axis.

Given information:

The given equation is $r=a\sin n\theta$ .

Formula used:

Odd-even identities: $\sin \left(-\theta \right)=-\sin \theta$ .

Sine of a difference identity: $\sin \left(A-B\right)=\sin A\cos B-\cos A\sin B$ .

Sine of a sum identity: $\sin (A+B)=\sin A\cos B+\cos A\sin B$ .

Calculation:

Symmetry about the $y$ -axis.

Case1: Replace $\left(r,\theta \right)$ by $\left(-r,-\theta \right)$ .

  $\begin{array}{c}-r=a\sin \left(-n\theta \right)\\ r=a\sin \left(n\theta \right)\end{array}$

Case2: Replace $\left(r,\theta \right)$ by $\left(r,\pi -\theta \right)$ .

  $\begin{array}{c}r=a\sin \left[n\left(\pi -\theta \right)\right]\\ -r=a\left(\sin n\pi \cos n\theta -\cos n\pi \sin n\theta \right)\\ r=a\sin n\theta \end{array}$

Both replacements give the same original polar equation.

Thus, it is proved that $r=a\sin n\theta$ is symmetric about the $y$ -axis.

b)

To find: The graph of the polar equation $r=a\sin n\theta$ is not symmetric about $x$ -axis.

It is proved that $r=a\sin n\theta$ is not symmetric about the $x$ -axis.

Given information:

The given equation is $r=a\sin n\theta$ .

Formula used:

Odd-even identities: $\sin \left(-\theta \right)=-\sin \theta$ .

Sine of a difference identity: $\sin \left(A-B\right)=\sin A\cos B-\cos A\sin B$ .

Sine of a sum identity: $\sin (A+B)=\sin A\cos B+\cos A\sin B$ .

Calculation:

Symmetry about the $x$

-axis:

Case1: Replace $\left(r,\theta \right)$ by $\left(r,-\theta \right)$ .

  $\begin{array}{c}r=a\sin \left(-n\theta \right)\\ r=-a\sin \left(n\theta \right)\end{array}$

Case 2: Replace $\left(r,\theta \right)$ by $\left(-r,\pi -\theta \right)$ .

  $\begin{array}{c}-r=a\sin \left[n\left(\pi -\theta \right)\right]\\ -r=a\left(\sin n\pi \cos n\theta -\cos n\pi \sin n\theta \right)\\ r=-a\sin n\theta \end{array}$

Both replacements do not give the same original polar equation.

Thus, it is proved that $r=a\sin n\theta$ is not symmetric about the $x$ -axis.

c)

To find: The graph of the polar equation $r=a\sin n\theta$ is not symmetric about the origin.

It is proved that $r=a\sin n\theta$ is not symmetric about the origin.

Given information:

The given equation is $r=a\sin n\theta$ .

Formula used:

Odd-even identities: $\sin \left(-\theta \right)=-\sin \theta$ .

Sine of a difference identity: $\sin \left(A-B\right)=\sin A\cos B-\cos A\sin B$ .

Sine of a sum identity: $\sin (A+B)=\sin A\cos B+\cos A\sin B$ .

Calculation:

Symmetry about the origin:

Case1: Replace $\left(r,\theta \right)$ by $\left(-r,\theta \right)$ .

  $\begin{array}{c}-r=a\sin \left(n\theta \right)\\ r=-a\sin \left(n\theta \right)\end{array}$

Case2: Replace $\left(r,\theta \right)$ by $\left(r,\pi +\theta \right)$ .

  $\begin{array}{c}r=a\sin \left[n\left(\pi +\theta \right)\right]\\ -r=a\left(\sin n\pi \cos n\theta +\cos n\pi \sin n\theta \right)\\ r=-a\sin n\theta \end{array}$

Both replacements do not give the same original polar equation.

Thus, it is proved that $r=a\sin n\theta$ is not symmetric about the origin.

d)

To prove: The maximum $\left|r\right|$ value is $\left|a\right|$ .

It is proved that the maximum $\left|r\right|$ value is $\left|a\right|$ .

Given information:

The given equation is $r=a\sin n\theta$ .

Formula used:

The range of $\sin n\theta$ is $-1\le \sin n\theta \le 1$ .

Calculation:

The maximum value of $\sin n\theta$ is $1$ , then maximum $\left|r\right|$ value is $\left|a\right|$ .

The minimum value of $\sin n\theta$ is $-1$ , then minimum $\left|r\right|$ value is $-\left|a\right|$ .

Therefore, it is proved that the maximum $\left|r\right|$ value is $\left|a\right|$ .

e)

To analyze: The graph of the curve given by $r=a\sin n\theta$ .

The polar equation represents a rose curve with $n$ petals.

The domain of the graph is $\left(\infty ,-\infty \right)$ .

The range of the polar equation is $\left[-\left|a\right|,\left|a\right|\right]$ .

The graph is symmetric about both the axes and the origin.

The graph is continuous and bounded between the minimum and maximum $r-$ values.

The maximum value of $r=-\left|a\right|$ and minimum value of $r=\left|a\right|$ .

Given information:

The given equation is $r=a\sin n\theta$ .

Calculation:

Plot the graph for $r=a\sin n\theta$ .

Consider the value of $a$ as $4$ and value of $n=1,3,5$ .

From the equation it can be deduced, it can be deduced that the polar equation represents a rose curve with $n$ petals.

The domain of the graph is $\left(\infty ,-\infty \right)$ since it is valid for all real values of $\theta$ .

The range of $\cos n\theta$ is, $-1\le \cos n\theta \le 1$ .

So, the range of the polar equation is $\left[-\left|a\right|,\left|a\right|\right]$ taking $-1$ as minimum value of $\cos n\theta$ and $1$ as maximum value of $\cos n\theta$ .

The graph is symmetric about both the axes and the origin.

The graph is continuous and bounded between the minimum and maximum $r-$ values.

The maximum value of $r=-\left|a\right|$ and minimum value of $r=\left|a\right|$ .