8. Curves in Polar Coordinates

Using a Table of Values to Sketch Polar Coordinate Curves

If we don't have a computer and we need to sketch the function on paper, we need to set up a table of values as follows:

θ (degrees)	`0°`	`30°`	`60°`	`90°`	`120°`	`150°`	`180°`
θ (radians)	`0`	`π/6`	`π/3`	`π/2`	`(2π)/3`	`(5π)/6`	`π`
r = 2 + 3 sin θ	`2`	`3.5`	`4.60`	`5`	`4.6`	`3.5`	`2`

θ (degrees)	`180°`	`210°`	`240°`	`270°`	`300°`	`330°`	`360°`
θ (radians)	`π`	`(7π)/6`	`(4π)/3`	`(3π)/2`	`(5π)/3`	`(11π)/6`	`2π`
r = 2 + 3 sin θ	`2`	`0.5`	`-0.60`	`-1`	`-0.60`	`0.5`	`2`

The first 7 points from this table are (2, 0^○), (3.5, 30^○), (4.60, 60^○), (5, 90^○), (4.6, 120^○), (3.5, 150^○), and (2, 180^○).

We plot these points (they are numbered) on the polar graph. I have also indicated with arrows the direction you need to go when joining the points.

Recall: A negative "r" means we need to be on the opposite side of the origin.

Here's the complete graph.

Conversion to Rectangular Form

Once again we convert our polar function into rectangular form so we can see how much simpler polar form is for certain functions.

`r = 2 + 3\ sin\ θ`

In rectangular form, this is:

`sqrt(x^2+y^2)=2+3y/sqrt(x^2+y^2)`

`x^2+y^2=2sqrt(x^2+y^2)+3y`

`x^2+y^2-3y=2sqrt(x^2+y^2)`

`(x^2+y^2-3y)^2=4(x^2+y^2)`

`x^4+2x^2y^2+y^4-6y(x^2+y^2)+` `9y^2` `=4(x^2+y^2)`

`x^4+2x^2y^2+y^4-6y(x^2+y^2)-` `4x^2+` `5y^2` `=0`

Notice how much simpler the polar form is compared to the rectangular form.

Using a Table of Values to Sketch this Curve

What if you can't use a computer to draw the graph?

You'll need to set up a table of values, as follows. I've put degrees and the radian equivalents.

θ (degrees)	0○	30○	60○	90○	120○	150○	180○
θ (radians)	`0`	`π/6`	`π/3`	`π/2`	`(2π)/3`	`(5π)/6`	`π`
`r = 3\ cos\ 2θ`	`3`	`1.5`	`-1.5`	`-3`	`-1.5`	`1.5`	`3`

θ (degrees)	210○	240○	270○	300○	330○	360○
θ (radians)	`(7π)/6`	`(4π)/3`	`(3π)/2`	`(5π)/3`	`(11π)/6`	`2π`
`r = 3\ cos\ 2θ`	`1.5`	`-1.5`	`-3`	`-1.5`	`1.5`	`3`

The first 7 points from this table are (3, 0°), (1.5, 30°), (-1.5, 60°), (-3, 90°), (-1.5, 120°), (1.5, 150°), and (3, 180°).

Placing those first 7 points on a polar coordinate grid gives us the following:

We start at Point 1, (3, 0°), and move around the graph by increasing the angle and changing the distance from the origin (determined by substituting the angle into r = 3 cos 2θ. I have drawn arrows to indicate the basic direction we have to head in to get to the next point.

Recall: A negative "r" means we need to be on the opposite side of the origin.

I have only plotted the first 7 points above to keep the graph simple. Clearly, we would need to calculate more than this number of points to get a good sketch. (You would need at least twice as many points as I have in the table above - every 15° would be adequate.)

Here's the complete graph.

[In the above graph, angles are in radians, where π radians = 180°. To learn more, see: Radians.]

Notice the curve is fully drawn once θ takes all values between 0 and 2π.

Conversion from Polar to Rectangular Coordinates

Next, here's the answer for the conversion to rectangular coordinates.

Why? We convert the function given in this question to rectangular coordinates to see how much simpler it is when written in polar coordinates.

To convert `r = 3\ cos\ 2θ` into rectangular coordinates, we use the fact that

cos 2θ = cos² θ - sin² θ.

So r = 3 cos 2θ = 3(cos² θ - sin² θ).

Now since `cos\ theta=x/r`, `sin\ theta=y/r` and `r^2=x^2+y^2`, we have

`cos^2theta=(x^2)/(r^2)=(x^2)/(x^2+y^2)`

and

`sin^2theta=(y^2)/(r^2)=(y^2)/(x^2+y^2)`

Taking the positive square root of `r^2=x^2+y^2` gives us:

`r=sqrt(x^2+y^2)`

So `r = 3 cos\ 2θ` in polar coordinates is equivalent to

`r=3(cos^2theta-sin^2theta)`

`sqrt(x^2+y^2)=3(x^2/(x^2+y^2)-y^2/(x^2+y^2))`

`sqrt(x^2+y^2)=3/(x^2+y^2)(x^2-y^2)`

`(x^2+y^2)^(3"/"2)=3(x^2-y^2)`

in rectangular coordinates.

We see that our equation in polar coordinates, r = 3 cos 2θ, is much simpler than the rectangular equivalent.

We need to sketch `r=sin theta-1`.

Using the same process for the earlier examples, we obtain:

For the curve above, when θ = 0, `r = −1`, so the curve starts on the left side of the origin.

Conversion to Rectangular Form

To convert to rectangular form, we use r² = x² + y² and

`sin^2theta=(y^2)/(r^2)=(y^2)/(x^2+y^2)`

In rectangular form, r = sin θ − 1 is:

`sqrt(x^2+y^2)=y/sqrt(x^2+y^2)-1`

`x^2+y^2=y-sqrt(x^2+y^2)`

`x^2+y^2-y=-sqrt(x^2+y^2)`

`(x^2+y^2-y)^2=x^2+y^2`

`x^4+2x^2y^2+y^4-2y(x^2+y^2)+y^2` `=x^2+y^2`

`x^4+2x^2y^2+y^4-2y(x^2+y^2)-x^2` `=0`

We need to sketch `r = 2.5`

In this example, we cannot see "θ" in the function we are given. This means the radius, r is constant, no matter what value angle θ takes.

Here's the graph:

What is the Equivalent in Rectangular Coordinates?

We convert the function given in this question to rectangular coordinates to see how much simpler it is when written in polar coordinates.

To convert `r = 2.5` into rectangular coordinates, we use

r² = x² + y²

In this example, `r = 2.5`, so `r^2= 6.25`.

So this gives us: x² + y² = 6.25

Not surprisingly, this is the similar to the equation for a circle we obtained in The Circle section earlier.

This time we are graphing r = 0.2 θ in polar coordinates.

Note the value of r is always `0.2` times the value of θ (of course, we are in radians).