**10.7 Graphs of Polar Equations**

**A. Graphing a Polar Equation by Point Plotting by**:

1

**. Convert to Rectangular Form. Multiply both sides of the Polar Equation by "r" and convert the result to rectangular form**.

Example 1:

r = 3 cos q

r

^{2}= r 3 cos q

x

^{2}+ y

^{2}= 3x

x

^{2}-3x + y

^{2}= 0

Using complete the square:

x

^{2}- 3x + (-3/2)

^{2}+ y

^{2}= 0

(x - 3/2)

^{2}+ y

^{2}= 9/4

This is the graph of a circle with center point (3/2, 0) and r = 3/2

**2. Use a Polar Coordinate Mode**

Put your calculator in polar mode and place the equation in the r =

r = 3 cos q

Use 0 £ q £ 2p , -6 £ x £ 6, and -4 £ y £ 4

This produces the same graph.

**3. Use the Parametric Mode**:

The graph of the Polar Equation

*r*=

*f*(q ) can be written in parametric form, using

*t*as a parameter, as follows:

*x*=

*f*(

*t*) cos

*t*and

*y*=

*f*(

*t*) sin

*t*

*r*= 3 cos q

*x*= 3 cos

*t*cos

*t*

*y*= 3 cos

*t*sin

*t*

Again, this will produce the same graph.

**B. Tests for Symmetry**:

The graph of a polar equation is symmetric with respect to the following if the given substitution yields an equivalent equation.

**1.**

**The line q = p/2**

Replace (r, q ) by (r, p-q ) or (-r, -q )

**2. The polar axis:**

Replace (r, q) by (r, -q) or (-r, p-q)

**3. The Pole:**

Replace (r, q) by (r, p+ q) or (-r, q)

**Example 2:**Use symmetry to sketch the graph r = 16 cos 3q

1. The line q = p /2 replace (r, q) by (r, p - q) or (-r, -q)

-r = 16 cos (3(-q))

-r = 16 cos (-3q )

-r = 16 cos 3q

Þthis is not the same equation so it is not equivalent

2. The polar axis replace (r, q) by (r, -q)

r = 16 cos 3(-q)

r = 16 cos 3q

Þthis is the same equation so it is equivalent

3. The pole replace (r, q) by (r, p + q) or (-r, q)

-r = 16 cos 3q

Þthis is not equivalent

Therefore r = 16 cos 3q is symmetric with respect to polar axis.

Plotting the points in the table and using the polar axis symmetry, you obtain the graph which is called limacon. (the c has a ' attached to the bottom of it)

The points are (r, q) = (0, 16), (p/6, 0), (p/3, -16), (p/2, 0), (2p/3, 16), (5p/6, 0), (p, -16)

Example 3: r = 4 - sin q

1. line q = p/2

r = 4 - sin (p- q )

r = 4 - (sin p cos q - cosp sin q)

r = 4 - ((0) cos q - (-1) sinq)

r = 4 - sin q

Þsame equation

2. Polar axis

r = 4 - sin (-q)

r = 4 + sin q

Þnot same equation

or -r = 4 - sin ( p- q)

-r = 4 - (sinp cos q - cos p sin q)

-4 = 4 - sin q

Þnot equivalent

3. Pole

-r = 4 - sin q not equivalent

or

r = 4 - sin ( p+ q)

r = 4 - (sinp cos q + cosp sin q)

r = 4 - (-sin q)

r = 4 + sin q

Þnot equivalent

Therefore r = 4 - sin q is symmetric with respect to q = p/2

Example 4: r = 2 csc q cos q

r = 2(1/sinq)(cos q) = 2 cot q

1. q = p /2

-4 = 2 cot (-q)

-r = -2 cot q

r = 2 cot q

Þtherefore it is the same equation or equivalent

2. Polar axis:

-r = 2 cot ( p- q)

-r = 2 cot (-q)

-r = -2 cot q

r = 2 cot q

ÞEquivalent

3. Pole:

r = 2 cot ( p+ q)

r = 2 cot q

Þ Equivalent

ÞTherefore r = 2 csc q cos q is symmetric with respect to q = /2, polar axis and pole.

**B. Zeros and Maximum**

**r - values**

½ r ½ is the maximum value for r and knowing the q - values for which r = 0 are two helpful ways to sketch the graph.

Example 4: Find the maximum value of r = 6 + 12 cos q

½r ½ = ½6 + 12 cosq ½

½ r ½ = ½ 6 + 12 cos q ½ £ ½6 ½ + ½ 12 cos q½

½ r ½ = 6 + 12½ cos q½£ 18 (from the thought that 6 + 12 = 18)

so if cos q = 1 then q = 0

So when q = 0, ½ r ½ = 18

Zeros:

Let r = 0

0 = 6 + 12 cos q

-6 = 12 cos q

-1/2 = cos q

Therefore q = 2p/3, 4p/3

Now for the table:

(r , q) goes with (x, y)

(18, 0°) and (18, 0)

(6, 90°) and (0, 6)

(0 , 120°) and (0 , 0)

(-6, 180 °) and (6, 0)

(0, 240°) and (0, 0)

(6, 270°) and (0, -6)

(18, 360°) and (18, 0)

Put your calculator in polar mode and you can get the polar coordinate. Graph the equation and go to 2nd trace, value and place the angle measure in, and the x - values and y - values will be displayed.

Example 5: r = 5 sin (2q)

½ r ½ = ½ 5 sin 2q ½

½ r ½ = 5

If we let sin b = 1 so b = 90° and b = 2q so therefore q = 45° or p/4

so when q = p/4, 3p/4 , 5p /4, 7p/4

So again to get the table of values:

(r, q) goes with (x, y)

(0, 0°) and (0, 0)

(5, 45°) and (3.5, 3.5)

(0, 90°) and (0, 0)

(-5, 135°) and (3.5, 3.5)

(0, 180°) and (0, 0)

(5, 225°) and (-3.5, -3.5)

(0, 270°) and (0, 0)

(-5, 315 °) and (-3.5, 3.5)

(0, 360°) and (0, 0)

This is a rose curve.

**check out this video at**

http://sam.ntpi.spcollege.edu/spjc/view/eventListing.jhtml?eventid=5323&c=13169