Chapter 9.6, Problem 16E

To determine

To calculate: To test for symmetry with respect to the line $\theta =\frac{\pi }{2}$ , the polar axis and the pole

Answer to Problem 16E

The polar equation is NOT symmetric with respect to the line $\theta =\frac{\pi }{2}$ and pole, while equation is symmetric with respect to polar axis

Explanation of Solution

Given information: Polar equation is $r=5+4\cos \theta$

Formula Used:

Test for Symmetry:

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

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

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:

  $r=5+4\cos \theta$

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

  $\begin{array}{l}r=5+4\cos \left(\pi -\theta \right)\\ r=5+4\left(-\cos \theta \right)\\ r=5-4\cos \theta \end{array}$

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

  $\begin{array}{l}r=5+4\cos \left(-\theta \right)\\ r=5+4\left(\cos \theta \right)\\ r=5+4\cos \theta \end{array}$

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

  $\begin{array}{l}r=5+4\cos \left(\pi +\theta \right)\\ r=5+4\left(-\cos \theta \right)\\ r=5-4\cos \theta \end{array}$

Since above equation is not same as the given polar equation

Thus, polar equation is NOT with respect to the pole

Conclusion:

Hence, polar equation is NOT symmetric with respect to the line $\theta =\frac{\pi }{2}$ and pole, while equation is symmetric with respect to polar axis