Chapter 9.6, Problem 32E

To determine

To calculate: To identify and graph the polar equation. Also to identify any symmetry and zeros of $r$

Answer to Problem 32E

Type of polar equation is rose with five petals.

Polar equation is symmetric with respect to line $\theta =\frac{\pi }{2}$

Zeros of polar equations are $\theta =0,\frac{\pi }{5},\frac{2\pi }{5},\frac{3\pi }{5}$

Explanation of Solution

Given information: Polar equation is $r=\sin 5\theta$

Formula Used:

Cardioid: Heart shape curve and general form is $r=a+a\cos \theta$

Limacons: General form of cardioid and general form is $r=a+b\cos \theta$

When $\frac{a}{b}<1$ , polar equation represents limacon with inner loop

When $\frac{a}{b}=1$ , polar equation represents Cardiod

When $1<\frac{a}{b}<2$ , polar equation represents Dimpled limacon

When $\frac{a}{b}=2$ , polar equation represents convex limacon

Rose Curves: Sinusoidal curve that have a flower shape and general equation is $r=a\cos \left(k\theta \right)$

Archimedian Spirals: Curve that extends indefinitely outward from a pole and general form is $r=a+b\theta$

Lemniscate: Eight shaped curve and general form is $r^2=a^2\cos \left(2\theta \right)$

Test for Symmetry:

1. To test symmetry with respect to the line $\theta =\frac{\pi }{2}$ , replace $\left(r,\theta \right)$ by $\left(r,\pi -\theta \right)$

2. To test symmetry with respect to the polar axis, replace $\left(r,\theta \right)$ by $\left(r,-\theta \right)$

3. To test symmetry with respect to the pole, replace $\left(r,\theta \right)$ by $\left(r,\pi +\theta \right)$

Calculation:

Polar equation is given as follows:

  $r=\sin 5\theta$

Type of Polar Equation:

Polar equation is of form $r=a\sin \left(k\theta \right)$

Thus, graph of polar equation is a rose with five petals

Test Symmetric of Polar Equation:

To test Symmetry respect with respect to the line $\theta =\frac{\pi }{2}$ :

Replacing $\left(r,\theta \right)$ by $\left(r,\pi -\theta \right)$ , polar equation is

  $\begin{array}{l}r=\sin 5\left(\pi -\theta \right)\\ r=\sin 5\theta \end{array}$

Since above equation is same as the given polar equation

Thus, polar equation is symmetric with respect to the line $\theta =\frac{\pi }{2}$

To test Symmetry with respect to the polar axis

Replacing $\left(r,\theta \right)$ by $\left(r,-\theta \right)$ , polar equation is

  $\begin{array}{l}r=\sin 5\left(-\theta \right)\\ r=-\sin 5\theta \end{array}$

Since above equation is NOT same as the given polar equation

Thus, polar equation is NOT symmetric with respect to the polar axis

To test Symmetry with respect to the pole

Replacing $\left(r,\theta \right)$ by $\left(r,\pi +\theta \right)$ , polar equation is

  $\begin{array}{l}r=\sin 5\left(\pi +\theta \right)\\ r=-\sin 5\theta \end{array}$

Since above equation is NOT same as the given polar equation

Thus, polar equation is NOT symmetric with respect to the pole

Calculating Zeros of Polar Equation:

In order to calculate zeros of polar equation, substitute $r=0$

Thus,

  $\begin{array}{l}\sin 5\theta =0\\ 5\theta ={\sin }^{-1}0\\ 5\theta =0,\pi ,2\pi ,3\pi \\ \theta =0,\frac{\pi }{5},\frac{2\pi }{5},\frac{3\pi }{5}\end{array}$

Graph of Polar Equation:

Graph of polar equation is as follows:

  

Conclusion:

Hence, type of polar equation is rose with five petals.

Polar equation is symmetric with respect to line $\theta =\frac{\pi }{2}$

Zeros of polar equations are $\theta =0,\frac{\pi }{5},\frac{2\pi }{5},\frac{3\pi }{5}$