Polar Equations

Report
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 measure
(in radians)
 A point with the polar coordinates (r, θ) can also be
represented by either of the following:
 (r, 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
 y = rsinθ

cosθ = x/r
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
 Insert the values for r and
 x=rcosθ
y=rsinθ
 x=(6)cos(π/6)
y=(6)sin(π/6) θ into the equations
x=(6) · ( 3 / 2 )
y=(6) · (1/2)  Find the numerical values
from solving the found
x=3 3
y=3
equations

(x, y) = (3 3, 3)  The found values for x and y
are the rectangular coordinates
Rectangular to Polar Coordinates
 r = x2  y2
 tanθ = y/x so θ =


1
tan ( y / x )
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= x 2  y 2
Step 4) To find θ, first determine the quadrant
that the point lies in
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 = x 2  y 2 and tan  1  , you can
thus find the values of r and θ.
2
2
 r = (2)  (  2)  4  4  8  2 2
1
1
(  2 / 2)  tan (  1)    / 4

θ = tan

Polar coordinates (r, θ) = ( 2 2 , -π/4)
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
r² = 4rsinθ
r² = 4y
x² + y² = 4y
x² + (y² - 4y) = 0
x² + (y² - 4y + 4) = 4
x² + (y – 2)² = 4
Given equation
Multiply each side by r
y = rsinθ
r² = x² + y² Equation of a circle
Subtract 4y from each side
Complete the square in y
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  Double Angle 2sinθcosθ = sin(2θ)
Formula
2r²sin(2θ) = 9

This is the standard polar equation for the
rectangular equation 4xy = 9
Polar Equations
 Limaçons:

Gen. equation:
(0 < a, 0 < b)
r = a ± bcosθ
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
x

y = (2 + 3cosθ)sinθ
0
5

 Substitute different values
6
3.982

for θ to find the remaining
1.75
3

coordinates
0
2
2
0
2.299
3.031
2
3
-.25
.433
6
.518
-.299
1
0
5

y
Method 2
2.



r = 2 + 3cosθ
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 2 = 90 , the radial line for 2 is…
REMEMBER: draw arrows to show in
which direction the polar function is
being sketched
Method 2 (con’t.)
r = 2 + 3cosθ
 Method 2 is used to sketch the polar equation
 The work is shown below:

r
0
5

6
4.598
3
3.5
2
2
3
.5
6
-.598


2
5

-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)

x
y
0
3


6

3
2.25
2
.75
0
0 1.299 1.299
6
3
.75
0

5
2
2.25
3
-1.299 -1.299
0
 Graph the polar equation, r = 2, using
Method 2

r
0
2



6
3
2
2
2
2
5
2
6
3
2
2

2
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)
sin( )
cos( )
=
cos( )
cos( )
tanθ = 1
θ = 45º , 225º

5
θ=
and
4
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
 Double Angle cos(2θ) = 2cos2θ + 1
Formula
 Trig Property
cos2θ = 1 – sin2θ
 Factor
Solutions:
3sinθ + 2 – 2sin2θ = 0
 Factor
2sin2θ – 3sinθ – 2 = 0
(2sinθ + 1)(sinθ – 2) = 0
2sinθ = −1
sinθ = 2
 Doesn’t exist
sinθ = −1/2
θ=
5
3
and

6
 Use unit circle
to solve for θ
Method 2
r = 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-polarequations/Images/graphing-polar-equations_gr_3.gif

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