Chapter 9.6, Problem 15E

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 15E

The polar equation is symmetric with respect to the line $\theta =\frac{\pi }{2}$ , polar axis and pole

Explanation of Solution

Given information: Polar equation is $r=3$

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:

  $r=3$

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

  $r=3$

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

  $r=3$

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

  $r=3$

Thus, polar equation is symmetric with respect to the pole

Conclusion:

Hence, polar equation is symmetric with respect to the line $\theta =\frac{\pi }{2}$ , polar axis and pole