Chapter 9.6, Problem 21E

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 21E

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

Explanation of Solution

Given information: Polar equation is $r^2=16\sin 2\theta$

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:

  $r^2=16\sin 2\theta$

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

  $\begin{array}{l}{r}^2=16\sin 2\left(\pi -\theta \right)\\ r^2=16\sin \left(2\pi -2\theta \right)\\ r^2=16\left(-\sin 2\theta \right)\\ r^2=-16\sin 2\theta \end{array}$

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

  $\begin{array}{l}{r}^2=16\sin 2\left(-\theta \right)\\ r^2=16\sin \left(-2\theta \right)\\ r^2=16\left(-\sin 2\theta \right)\\ r^2=-16\sin 2\theta \end{array}$

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

  $\begin{array}{l}{r}^2=16\sin 2\left(\pi +\theta \right)\\ r^2=16\sin \left(2\pi +2\theta \right)\\ r^2=16\left(\sin 2\theta \right)\\ r^2=16\sin 2\theta \end{array}$

Since above equation is same as the given polar equation

Thus, polar equation is with respect to the pole

Conclusion:

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