Polar Equations

Project by Brenna Nelson,

Stewart Foster, Kathy Huynh

Converting From Polar to Rectangular Coordinates A point P in a polar coordinate system is represented

by an ordered pair of numbers (r, θ) (r, θ): polar coordinates r: radius θ: angle

A point with the polar coordinates (r, θ) can also be represented by either of the following: (r, measure

(in radians) θ ± 2kπ) or (-r, θ + π + 2kπ) where k is any integer

Polar coordinates of the pole are (0, θ) where θ can be any angle

Polar to Rectangular Coordinates

If P is a point with polar coordinates (x,y) of P are given by x = rcosθ cosθ = x/r y = rsinθ sinθ = y/r

tanθ = y/x

r² = x² + y² where r is the hypotenuse and x and y are the corresponding sides to the triangle

Plug the values of r and θ into the x and y equations to find the values of the rectangular coordinates

Converting from Polar to Rectangular

Polar Coordinates (r, θ) Given: (6, π/6)

r = 6 θ = π/6 Use the equations x=rcosθ and y=rsinθ to find the

values for x and y by plugging in the given values of r and θ

x=rcosθ y=rsinθ x=(6)cos(π/6) y=(6)sin(π/6)

x=(6) · ( ) y=(6) · (1/2)x=3 y=3

(x, y) = (3 , 3)

3 / 23

3

Insert the values for r and θ into the equations

Find the numerical values from solving the found equations

The found values for x and y are the rectangular coordinates

Rectangular to Polar Coordinates

r = tanθ = y/x so θ =

Plug the values of the x and y coordinates into the equations to find the values of the polar coordinates

Steps for conversion: Step 1) Always plot the point (x,y) first Step 2) If x=0 or y=0, use your illustration to find

(r, θ) polar coordinates Step 3) If x does not equal zero and y does not

equal zero, then r= Step 4) To find θ, first determine the quadrant

that the point lies in

2 2x y1tan ( / )y x

2 2x y

Converting from Rectangular to Polar

Rectangular Coordinates (x, y) Given (2, -2)

x = 2 y = -2 By plugging the values of x and y into the polar

coordinate equations r = and , you can thus find the values of r and θ.

r =

θ =

Polar coordinates (r, θ) = ( , -π/4)

2 2x y 1tan

2 2(2) ( 2) 4 4 8 2 2

1 1tan ( 2 / 2) tan ( 1) / 4

2 2

Try these on your own:

Convert r = 4sinθ from the polar equation the rectangular equation.

Convert 4xy=9 from the rectangular equation to the polar equation.

Solutions: Example 1

Convert r = 4sinθ from the polar equation the rectangular equation.

r = 4sinθ Given equationr² = 4rsinθ Multiply each side by rr² = 4y y = rsinθx² + y² = 4y r² = x² + y² Equation of a circlex² + (y² - 4y) = 0 Subtract 4y from each sidex² + (y² - 4y + 4) = 4 Complete the square in yx² + (y – 2)² = 4 Factor y

This is the standard form of the equation of a circle with center (0,2) and radius 2.

Solutions: Example 2

Convert 4xy = 9 from the rectangular equation to the polar equation. Use x =rcosθ and y = rsinθ to substitute into the

equation

4(rcosθ)(rsinθ) = 9 x = rcosθ, y =rsinθ

4r²cosθsinθ = 9

2r²(2sinθcosθ) = 9

2r²sin(2θ) = 9

This is the standard polar equation for the rectangular equation 4xy = 9

Double Angle Formula 2sinθcosθ = sin(2θ)

Polar Equations

Limaçons: Gen. equation: r = a ± bcosθ

(0 < a, 0 < b) r = a ± bsinθ

Rose Curves: Gen. equation: r = a ± acos(nθ)

r = a ± asin(nθ)

(n petals if n is odd,

2n petals if n is even)

Polar Equations

Circles: Gen. equation: r = a

r = cos(θ)

Lemniscates: Gen. equation: r2 = a ± a2cos(2θ)

r2 = a ± a2sin(2θ)

How to Sketch Polar Equations

Sketch the graph of the polar equation:

r = 2 + 3cosθ The function is a graph of a limaçon because it

matches the general formula: r = a ± bcosθ

Method 1

r = 2 + 3cosθ

1. Convert the equation from polar to rectangular

~ x = rcosθ ~ y = rsinθ

x = (2 + 3cosθ)cosθ

y = (2 + 3cosθ)sinθ Substitute different values

for θ to find the remaining

coordinates

60

56

3

2

23

x5

3.982

1.75

0

-.25

.518

1

y0

2.299

3.031

2

.433

-.299

0

Method 2

r = 2 + 3cosθ

2. Substitute values of θ and use radial lines to plot points

Use a number of radial lines to ensure that the entire graph of the polar function is sketched

Radial line: the lines that extends from the origin, forming an angle equivalent to the radian value Ex. Because = 90 , the radial line for is…

REMEMBER: draw arrows to show in

which direction the polar function is

being sketched

2

2

Method 2 (con’t.)

r = 2 + 3cosθ Method 2 is used to sketch the polar equation The work is shown below:

60

56

3

2

23

r5

4.598

3.5

2

.5

-.598

-1

Because you know that the equation is a limaçon, you can roughly sketch the rest of the graph.

NOTE: this method is only an approximation; it should not be used for calculations.

Method 3

r = 2 + 3cosθ

3. Using a calculator The easiest way to graph a polar equation is

to just put the equation into the calculator

The method for graphing the polar equations with the calculator are explained in a later slide.

Try these on your own:

Graph the polar equation, r = 3cosθ,

using Method 1

Graph the polar equation, r = 2, using

Method 2

Solutions:

Graph the polar equation, r = 3cosθ,using Method 1 x = 3cosθ(cosθ) y = 3cosθ(sinθ)

Graph the polar equation, r = 2, using

Method 2

60

56

3

2

23

r 2 2 2 2 2 2 2

x 3 2.25 .75 0 .75 2.25 3

0 1.299 1.299 0 -1.299 -1.299 0

0 6 5

6

3

2 2

3

y

Finding Polar Intersection Points

Method 1:

Set equations equal to each other.

Solve for θ.

Method 2, for θ values not on unit circle:

Set calculator mode to polar.

Graph equations.

Find approximate intersection points using TRACE and

then find exact intersection points using method 1.

Use Method 1 to find the intersection points for the two polar equations. r = cos(θ) r = sin(θ)

tanθ = 1

θ = 45º , 225º

θ = and

sin( )

cos( )

cos( )

cos( )

=

4

5

8

Try these on your own:

Find the intersection points of the equations using Method 1:

r = 3 + 3sin(θ)

r = 2 – cos(2 θ)

Solutions:

Find the intersection points of the equations using Method 1:

r = 3 + 3sin(θ)

r = 2 – cos(2θ)

3 + 3sinθ = 2 – cos(2θ)

1 + 3sinθ = −cos(2θ)

1 + 3sinθ = −2cos2θ +1

3sinθ + 2(1 – sin2θ) = 0

3sinθ + 2 – 2sin2θ = 0 Factor

Double Angle Formula

cos(2θ) = 2cos2θ + 1

Trig Propertycos2θ = 1 – sin2θ

Solutions:

3sinθ + 2 – 2sin2θ = 0

2sin2θ – 3sinθ – 2 = 0

(2sinθ + 1)(sinθ – 2) = 0

2sinθ = −1 sinθ = 2

sinθ = −1/2

θ = and

5

3

6

Use unit circle

to solve for θ

Doesn’t exist

Factor

Method 2r = 1 + 3cosθ

r = 2

Bibliography

Sullivan, Michael. Precalculus. Upper Saddle River: Pearson Education, 2006.

Foerster, Paul. Calculus: Concepts and Applications. Emeryville: Key Curriculum Press, 2005.

http://curvebank.calstatela.edu/index/lemniscate.gif http://curvebank.calstatela.edu/index/limacon.gif http://curvebank.calstatela.edu/index/rose.gif http://www.libraryofmath.com/pages/graphing-polar-equations/Images/

graphing-polar-equations_gr_3.gif