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# Examples using Polar Coordinates Help (page 2)

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## Three-leafed Rose

The general form of the equation of a three-leafed rose centered at the origin in the polar plane is given by either of the following two formulas:

r = a cos 3 θ

r = a sin 3 θ

where a is a real-number constant greater than 0. The cosine version of the curve is illustrated in Fig. 10-8A. The sine version is illustrated in Fig. 10-8B.

Fig. 10-8 . (A) Polar graph of a three-leafed rose with equation r = a cos 3 θ . (B) Polar graph of a three-leafed rose with equation r = a sin 3 θ

## Four-leafed Rose

The general form of the equation of a four-leafed rose centered at the origin in the polar plane is given by either of the following two formulas:

r = a cos 2 θ

r = a sin 2 θ

where a is a real-number constant greater than 0. The cosine version is illustrated in Fig. 10-9A. The sine version is illustrated in Fig. 10-9B.

Fig. 10-9. (A) Polar graph of a four-leafed rose with equation r = a cos 2 θ. (B) Polar graph of a four-leafed rose with equation r = a sin 2 θ.

## Spiral

The general form of the equation of a spiral centered at the origin in the polar plane is given by the following formula:

r =

where a is a real-number constant greater than 0. An example of this type of spiral, called the spiral of Archimedes because of the uniform manner in which its radius increases as the angle increases, is illustrated in Fig. 10-10.

Fig. 10-10.

## Cardioid

The general form of the equation of a cardioid centered at the origin in the polar plane is given by the following formula:

r = 2 a (1 + cos θ )

where a is a real-number constant greater than 0. An example of this type of curve is illustrated in Fig. 10-11.

Fig. 10-11.

## Examples Using Polar Coordinates Practice Problems

#### PROBLEM 1

What is the value of the constant, a, in the spiral shown in Fig. 10-10? What is the equation of this spiral? Assume that each radial division represents 1 unit.

Fig. 10-10 . Polar graph of a spiral; illustration for Problem 1.

#### SOLUTION 1

Note that if θ = π, then r = 2. Therefore, we can solve for a by substituting this number pair in the general equation for the spiral. We know that ( θ , r ) = (π,2), and that is all we need. Proceed like this:

r =

2 =

2/π = a

Therefore, a = 2/π , and the equation of the spiral is r = (2/π) θ or, in a somewhat simpler form without parentheses, r = 2 θ /π.

#### PROBLEM 2

What is the value of the constant, a , in the cardioid shown in Fig. 10-11? What is the equation of this cardioid? Assume that each radial division represents 1 unit.

Fig. 10-11 . Polar graph of a cardioid; illustration for Problem 2.

#### SOLUTION 2

Note that if θ = 0, then r = 4. We can solve for a by substituting this number pair in the general equation for the cardioid. We know that ( θ , r ) = (0,4), and that is all we need. Proceed like this:

r = 2 a (1 + cos θ )

4 = 2 a (1 + cos 0)

4 = 2 a (1 + 1)

4 = 4 a

a = 1

This means that the equation of the cardioid is r = 2(1 + cos θ ) or, in a simpler form without parentheses, r = 2 + 2 cos θ.

Practice problems for these concepts can be found at: Polar Coordinates Practice Test.

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