5.2 INDENTIFYING AND GRAPHING POLAR EQUATIONS
Example 1 Identify and graph r = 3.
There is no θ in this equation, which means θ can be anything. This is
the graph of all the points at a distance r = 3 from the pole.
The set of all points equidistant from a center is a circle. Recall
r2 = x2 + y2, which is a circle if r is constant. In this case, x2 + y2=32=9
π/2 = -3π/2
90°=-270°
2π/3 =- 4π/3
120°=-240°
3π/4= -5π/4
135°=- 225°
π/3 =- 5π/3
60°= -300°
π/4 =- 7π/4
45°=-315°
5π/6= - 7π/6
150°=-210 °
π= - π
5
180°=- 180°
π/6 = -11π/6
30°=-330°
4
3
2
1
1
2
4
3
5 0 =2π = -2π
0° = 360°
7π/6=- 5π/6
210°=- 150°
11π/6= - π/6
330°= - 30°
7π/4= - π/4
315°= - 45°
5π/4= -3π/4
225°=- 135° 4π/3=- 2π/3
240°= - 120°
5π/3 = -π/3
300°=- 60°
3π/2=- π/2
270°=- 90°
By the way, if you graph a circle using your calculator in
rectangular points, you have to break it into 2 functions:
Y1 = r 2 − x 2 and Y2 = − r 2 − x 2
(Upper semicircle)
(Lower semicircle)
since x2 + y2 = r2 is not a function, but in polar coordinates, it r=3 is
a function, because for each choice of θ there is only one
corresponding value or r, namely, r =3, so your calculator will have
no problem.
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Example 2
Identify and graph θ = π/4
This is the graph of all points for any value of r along the ray at
θ = π/4, with positive r values along the terminal side and
negative r values point in the opposite direction. Converting to
rectangular coordinates we get: y/x = tan-1 π/4 = 1
So multiplying both sides by x gives y = x.
π/2 = -3π/2
90°=-270°
2π/3 =- 4π/3
120°=-240°
3π/4= -5π/4
135°=- 225°
π/3 =- 5π/3
60°= -300°
π/4 =- 7π/4
45°=-315°
5π/6= - 7π/6
150°=-210 °
π= - π
5
180°=- 180°
π/6 = -11π/6
30°=-330°
4
3
2
1
1
2
4
3
5 0 =2π = -2π
0° = 360°
7π/6=- 5π/6
210°=- 150°
11π/6= - π/6
330°= - 30°
5π/4= -3π/4
225°=- 135° 4π/3=- 2π/3
240°= - 120°
7π/4= - π/4
315°= - 45°
5π/3 = -π/3
300°=- 60°
3π/2=- π/2
270°=- 90°
Now you do #13 and 15 on p.361
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Example 3/ Example 5: Identify and graph the equations rsin θ = 2 and rcos θ = -3
This one is not as easily identifiable when only looking at polar coordinates. But if
we convert to rectangular coordinates it is simple. x = rcos θ and y = rsin θ, so
substituting in y for rsin θ gives y = 2 and substituting x for rcos θ gives x = -3.
On a polar grid, rcos θ = -3 is 3 units on the opposite side of the polar axis.
π/2 = -3π/2
90°=-270°
2π/3 =- 4π/3
120°=-240°
3π/4= -5π/4
135°=- 225°
π/3 =- 5π/3
60°= -300°
π/4 =- 7π/4
45°=-315°
rsin θ = 2
5π/6= - 7π/6
150°=-210 °
7π/6=- 5π/6
210°=- 150°
4
3
2
1
1
2
4
3
rcos θ = -3
π= - π
5
180°=- 180°
π/6 = -11π/6
30°=-330°
5 0 =2π = -2π
0° = 360°
11π/6= - π/6
330°= - 30°
5π/4= -3π/4
225°=- 135° 4π/3=- 2π/3
240°= - 120°
7π/4= - π/4
315°= - 45°
5π/3 = -π/3
300°=- 60°
3π/2=- π/2
270°=- 90°
For future reference:
Any polar equation in the form rsin θ = a is a horizontal line going
through the rectangular coordinate point (0,a).
And rcos θ = a is a vertical line going through the rectangular
coordinate point (a,0).
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GRAPING POLAR EQUATIONS USING A GRAPHING
CALCULATOR
See p. 166 -175 in
http://education.ti.com/guidebooks/graphing/83p/83m$book-eng.pdf
Step 1: Solve the equation for r in terms of θ
Step 2: Change MODE to Radian (third line down, first
entry, though Degree mode will be easier when making
tables) and also change to Pol (fourth line down, third entry
from the left). Notice that the viewing window is different
now. In addition to setting Xmin, Xmax, Xscl and so forth,
the viewing window in polar mode requires setting
minimum and maximum values for θ and an increment
setting for θ (θstep). Set the ZOOM to 6:ZStandard. This
automatically sets θmin to 0, θmax to 2π and θstep to π/24.
Step 3: Press Y= (notice it now shows r1 = now instead of
Y1) and enter the expression involving θ that you found in
Step 1 and press GRAPH.
Step 4: If you cannot see the graph well, select ZOOM
0:ZFit and then ZOOM 5:ZSquare so that the graph is not
distorted.
POLAR CIRCLES
Equation
Description
r = 2asin θ
Circle; radius a; center at
rectangular coordinate (0,a)
r = -2asin θ
Circle; radius a; center at
rectangular coordinate (0,-a)
r = 2acos θ
Circle; radius a; center at
rectangular coordinate (a,0)
r = -2acos θ
Circle; radius a; center at
rectangular coordinate (-a,0)
(0,a)
If you have a polar equation in one of these forms, you don’t have
to convert it to rectangular coordinates to graph it.
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SYMMETRY IN POLAR COORDINATES
Tests for Symmetry
Symmetry with Respect to the Polar Axis (x-axis)
In a polar equation, replace θ by –θ and if an equivalent equation results,
the graph is symmetric with respect to the polar axis (positive x-axis).
[Recall that cos(θ) = cos(-θ)].
Symmetry with Respect to the Line θ = π/2 (y-axis)
In a polar equation, replace θ by π – θ. If an equivalent equation results,
the graph is symmetric with respect to the line θ = π/2 [Recall that
sin(θ) = sin(π-θ)]
Symmetry with Respect to the Pole (origin)
In a polar equation, replace r by -r. If an equivalent equation results, the
graph is symmetric with respect to the pole.
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Cardioid
A cardioid is a curve given by one of the following polar
equations:
r = a(1 + cos θ)
r = a(1 - cos θ)
(a, π/2)
(a, π/2)
(0, π)
(2a, 0)
(0, 0)
(2a, π)
(a, 3π/2)
(a, 3π/2)
r = a(1 + sin θ)
r = a(1 - sin θ)
(2a, π/2)
(0, π/2)
(a, π)
(a,0)
(a, π)
(a, 0)
(0, 3π/2)
(2a, 3π/2)
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Rose
A curve which has the shape of a petalled flower. This curve was named rhodonea by the
Italian mathematician Guido Grandi between 1723 and 1728 because it resembles a rose
(MacTutor Archive). The polar equation of the rose is
r = a sin (nθ)
or
r = a cos (nθ)
If n is odd, the rose is n-petalled. If n is even, the rose is 2n-petalled.
If n is a rational number = r/s then the curve closes* at a polar angle of πsp , where
p = 1 if rs is odd and p=2 if rs is even. The following graphs assume a=1.
r=1, s=2,
rs=(1)(2)=2 (even)
So p=2
Closes at πsp
π(2)(2)=4π
Let θmax=4π or
4*180=720°
r=1, s=3,
rs=(1)(3)=3 (odd)
So p=1
Closes at πsp
π(3)(1) =3π
Let θmax =3π
or 3*180=540°
r=3, s=4,
rs=(3)(4)=12 (even)
So p=2
Closes at πsp
π(4)(2) =8π
Let θmax=8π
or 8*180=1440°
DEGREE MODE
RADIAN MODE
Eric W. Weisstein. "Rose."
From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/Rose.html
* If you want to reproduce these graphs on your graphing calculator, you have to increase the
θmax in the WINDOW screen until the equation has repeated its cycle enough to complete its
pattern, which is the angle at which the curve “closes.” (See above).
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If n
is irrational, then there are an infinite number of petals.
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Limaçon
r = 1 + 2sinθ
r = 1 - 2sinθ
r = 1 – 1sinθ
r = 3 + 2cosθ
(-1,90°)
(3,270°)
Notice that sin(90°)=1, so r = 1-2(1)=-1 at θ= 90°
Also, sin(270°)=-1, so r = 1-2(-1)=3 at θ= 270°
The limaçon is a polar curve of the form
r = a ± bcosθ
(2) r = a ± bsinθ
also called the limaçon of Pascal. It was first investigated by Dürer, who gave a
method for drawing it in Underweysung der Messung (1525). It was rediscovered by
Étienne Pascal, father of Blaise Pascal , and named by Gilles-Personne Roberval in
1650 (MacTutor Archive). The word "limaçon" comes from the Latin limax,
meaning "snail."
If a ≥ 2b, the limaçon is convex.
If 2b > a > b, the limaçon is dimpled.
If b=a, the limaçon degenerates to a cardioid.
If a > 0, and a < b, the limaçon has an inner loop.
Eric W. Weisstein. "Limaçon." From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/Limacon.html
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Logarithmic Spiral
The logarithmic spiral is a spiral whose polar equation is given by
r = aebθ
where is the distance from the origin, is the angle from the xaxis, and a and b are arbitrary constants.
Eric W. Weisstein. "Logarithmic Spiral." From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/LogarithmicSpiral.html
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POLAR EQUATIONS AND GRAPHS
Cardioid
r = a ± acos θ, a > 0
r = a ± acos θ, a > 0
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HOMEWORK
p. 361
19, 21, 29-36,43,49,51,55,59
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