Question

For each of the following, give an example of a polar equation whose graph has the type(s) of symmetry listed, if possible. If such an equation doesn’t exist, explain why.

(a) symmetry about the x-axis

(b) symmetry about the y-axis

(c) symmetry about the origin

(d) symmetry about the origin but not about the x-axis

(e) symmetry about the x-axis, y-axis, and origin

(f) symmetry about the x-axis and y-axis but not about the origin

Short answer

Part (a)$r=3\cos \theta$

Part (b)$r=3\sin \theta$

Part (c)$r=\sin 2\theta$

Part (d)$r^2=\sin 2\theta$

Part (e) $r^2=\cos \theta$

Part (f) The answer is not possible.

Step-by-step solution

Part (a) Step 1: Given information

A polar equation with the type(s) of symmetry specified on its graph

Part (a) Step 1: Explanation

Symmetry about $x-$axis.

If a curve is symmetric with respect to an axis, then every point $(r,\theta )$on the graph is symmetrical about the $x-$axis if$(r,-\theta )$is also on the graph.

Example : $r=3\cos \theta$

Replace $(r,\theta )$by $(r,-\theta )$

Then,

$r=3\cos (-\theta )$

$r=3\cos \theta$

Thus $r=3\cos \theta$is symmetrical about $x-$axis.

Therefore, the answer is $r=3\cos \theta$

Part (b) Step 1: Explanation

Symmetry about $y-$axis.

If a curve is symmetric with regard to the $y$axis, then every point $(r,\theta )$on the graph is symmetrical about the $y$axis if$(-r,-\theta )$is also on the graph.

Example: $r=3\sin \theta$

Replace $(r,\theta )$by $(-r,-\theta )$

Then,

$$\begin{gathered} -r=3\sin (-\theta ) \\ -r=-3\sin \theta \\ r=3\sin \theta \end{gathered}$$

Thus, $r=3\sin \theta$is symmetrical about $y-$axis.

Therefore, the answer is $r=3\sin \theta$

Part (c) Step 1: Explanation

Symmetry about the origin.

A curve is symmetric with respect to the origin if the point $(r,\theta +\pi )$is also on the graph for every point $(r,\theta )$

Example: $r=\sin 2\theta$

Replace $(r,\theta )$by $(r,\pi +\theta )$

Then,

$$\begin{gathered} r=\sin 2(\pi +\theta ) \\ r=\sin 2\theta \end{gathered}$$

Thus, $r=\sin 2\theta$is symmetrical in origin.

Therefore, the answer is $r=\sin 2\theta$

Part (d) Step 1: Explanation

There is symmetry around the origin, but not around the $x-$axis.

A curve is symmetric with respect to the origin if the point $(r,\theta +\pi )$is also on the graph for every point $(r,\theta )$

Example: $r^2=\sin 2\theta$

Replace $(r,\theta )$by $(r,\pi +\theta )$

Then,

$$\begin{gathered} r^2=\sin 2(\pi +\theta ) \\ {r}^2=\sin 2\theta \end{gathered}$$

Thus, $r^2=\sin 2\theta$is symmetrical in origin.

If a curve is symmetric with regard to the $x-$axis, then every point $(r,\theta )$ on the graph is symmetrical about the $x-$axis if $(r,-\theta )$ is also on the graph. $(r,\theta )$ should be replaced with$(r,-\theta )$

Then,

$$\begin{gathered} r^2=\sin 2(-\theta ) \\ {r}^2=-\sin 2\theta \end{gathered}$$

Not symmetric about $x-$axis.

Therefore, the answer is $r^2=\sin 2\theta$

Part (e) Step 1: Explanation

Symmetry about $x-$axis, $y-$axis and origin.

The equation symmetric about $x-$axis, $y-$axis and origin is $r^2=\cos \theta$

Therefore the answer is $r^2=\cos \theta$

Part (f) Step 1: Explanation

Symmetry is about $x-$axis, $y-$axis but not about the origin.

Symmetry about the $x-$and $y-$axes but not about the origin is not achievable for an equation. Therefore, the answer is Not possible.