Chapter 9.6, Problem 31E

To determine

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

Answer to Problem 31E

Type of polar equation is rose with three petals.

Polar equation is symmetric with respect to polar axis

Zeros of polar equations are $\theta =\frac{\pi }{6},\frac{3\pi }{6},\frac{5\pi }{6}$

Explanation of Solution

Given information: Polar equation is $r=5\cos 3\theta$

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:

  $r=5\cos 3\theta$

Type of Polar Equation:

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

Thus, graph of polar equation is a rose with three 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

  $\begin{array}{l}r=5\cos 3\left(\pi -\theta \right)\\ r=-5\cos 3\theta \end{array}$

Since above equation is NOT same as the given polar equation

Thus, polar equation is NOT 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

  $\begin{array}{l}r=5\cos 3\left(-\theta \right)\\ r=5\cos 3\theta \end{array}$

Since above equation is same as the given polar equation

Thus, polar equation is 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

  $\begin{array}{l}r=5\cos 3\left(\pi +\theta \right)\\ r=-5\cos 3\theta \end{array}$

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,

  $\begin{array}{l}5\cos 3\theta =0\\ \cos 3\theta =0\\ 3\theta ={\cos }^{-1}0\\ 3\theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\\ \theta =\frac{\pi }{6},\frac{3\pi }{6},\frac{5\pi }{6}\end{array}$

Graph of Polar Equation:

Graph of polar equation is as follows:

  

Conclusion:

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

Polar equation is symmetric with respect to polar axis

Zeros of polar equations are $\theta =\frac{\pi }{6},\frac{3\pi }{6},\frac{5\pi }{6}$