Chapter 9.2, Problem 87AYU

Explain why the following test for symmetry is valid: Replace $r$ by $-r$ and $\theta$ by $-\theta$ in a polar equation. If an equivalent equation results, the graph is symmetric with respect to the line $\theta \text{ = }\frac{\pi }{2}(y\text{-axis})$.

(a) Show that the test on page 591 fails for $r^2\text{ = cos}\theta$, yet this new test works.

(b) Show that the test on page 591 works for $r^2\text{ = }\sin \theta$, yet this new test fails.

To determine

To show: The test rule is given as: Replace $r$ by $- r$ and $$ \theta $$ by $- \theta$ in a polar equation. If an equivalent equation results, the graph is symmetric with respect to the line $\theta = \frac{\pi }{2}$ ($y$-axis).

a) Symmetry test fails for $r^2 = \text{cos}\theta$, yet this new test works.

b) Symmetry test works for $r^2 = \text{sin}\theta$, yet this new test fails.

Explanation of Solution

Given:

It is asked to show that the graph of the polar equations $r^2 = \text{cos}\theta \text{and} r^2 = \text{sin}\theta$ are fails in symmetry tests but this works for the new rule given.

Check for symmetry test:

Polar axis:

Replace $\theta$ by $- \theta$.

For a), the result is $r^2 = \text{cos}\left(- \theta \right) = \text{cos}\theta .$

The test satisfied. So the graph is symmetric with respect to the polar axis.

For b), the result is $r^2 = \text{sin}\left(- \theta \right) = - \text{sin}\theta .$

The test fails. So the graph may or may not be symmetric with respect to the polar axis.

The line $\theta = \frac{\pi }{2}$:

Replace $\theta$ by $\pi - \theta$.

For a), the result is

$r^2 = \text{cos}\left(\pi - \theta \right) = \text{cos}\pi \text{cos}\theta + \text{sin}\pi \text{sin}\theta = \left(- 1\right)\text{cos}\theta + \left(0\right)\text{sin}\theta = - \text{cos}\theta .$

The test fails, so the graph may or may not symmetric with respect to the line $\theta = \frac{\pi }{2}$.

For b), the result is

$r^2 = \text{sin}\left(\pi - \theta \right) = \text{sin}\pi \text{cos}\theta - \text{cos}\pi \text{sin}\theta = \left(0\right) \text{cos}\theta - \left(- 1\right) \text{sin}\theta = \text{sin}\theta .$

The test satisfied. So the graph is symmetric with respect to the line $\theta = \frac{\pi }{2}$.

The Pole:

Replace $r$ by $- r$.

For a), the result is ${\left(- r\right)}^2 = \text{cos}\theta$, so $r^2 = \text{cos}\theta$. This test satisfied.

For b), the result is ${\left(- r\right)}^2 = \text{sin}\theta$, so $r^2 = \text{sin}\theta$. This test satisfied.

Replace $\theta$ by $\theta + \pi$.

For a), the result is $r^2 = \text{cos}\left(\theta + \pi \right) = \text{cos}\theta \text{cos}\pi - \text{sin}\theta \text{sin}\pi = \left(- 1\right)\text{cos}\theta = - \text{cos}\theta .$

For a), the result is $r^2 = \text{sin}\left(\theta + \pi \right) = \text{sin}\theta \text{cos}\pi + \text{cos}\theta \text{sin}\pi = - \text{sin}\theta .$

This test also fails, so the graph may or may not be symmetric with respect to the pole.

Let’s do the new test:

Rule: Replace $r$ by $- r$ and $\theta$ by $- \theta$ in a polar equation. If an equivalent equation results, the graph is symmetric with respect to the line $\theta = \frac{\pi }{2}$ ($y$-axis).

For a), the result is ${\left(- r\right)}^2 = \text{cos}\left(- \theta \right)$

$r^2 = \text{cos}\theta$

Therefore, the symmetry test fails for $r^2 = \text{cos}\theta$, yet the new test works.

For b), the result is ${\left(- r\right)}^2 = \text{sin}\left(- \theta \right) = - \text{sin}\theta$

$r^2 = - \text{sin}\theta$

Therefore, the symmetry test works for $r^2 = \text{sin}\theta$, yet the new test fails.