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Section 9.2 – Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify the type of equations we have and then sketch the graph. Objective 1: Identify Polar Equations and Graph Convert the polar equation to rectangular and sketch the graph: Ex. 1 r = 5 Solution: r = 5 (square both sides) r2 = 25 (replace r2 by x2 + y2) x2 + y2 = 25 This is a circle of radius 5 centered at the Pole.
Ex.2 θ =
π
4
Solution:
θ =
π
4 (take the tangent of both sides)
tan(θ) = tan(
π
4) = 1 (tan(θ) =
y
x)
y
x = 1
y = x This is a line with slope of 1 passing through the Pole.
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
181
Ex. 3 rsin(θ) = 4 Solution:
rsin(θ) = 4 (y = rsin(θ)) y = 4 This is a horizontal line passing through (0, 4).
Ex. 4 rcos(θ) = – 2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
182
Solution:
rcos(θ) = – 2 (x = rcos(θ)) x = – 2 This is a vertical line passing through (– 2, 0).
Theorem If a ≠ 0, then
1) rsin(θ) = a is a horizontal line that passes through (0, a).
2) rcos(θ) = a is a vertical line that passes through (a, 0). Convert the polar equation to rectangular and sketch the graph:
Ex. 5 r = 2sin(θ) Solution:
r = 2sin(θ) (multiply both sides by r)
r2 = 2rsin(θ) (r2 = x2 + y2 and y = rsin(θ)) x2 + y2 = 2y (subtract 2y from both sides) x2 + y2 – 2y = 0 (complete the square; add 1 to both sides) x2 + y2 – 2y + 1 = 1 (write y2 – 2y + 1 as (y – 1)2) x2 + (y – 1)2 = 1 This is a circle with radius of 1 centered at (0, 1).
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
183
Ex. 6 r = – 4cos(θ) Solution:
r = – 4cos(θ) (multiply both sides by r)
r2 = – 4rcos(θ) (r2 = x2 + y2 and x = rcos(θ)) x2 + y2 = – 4x (add 4x from both sides) x2 + 4x + y2 = 0 (complete the square; add 4 to both sides) x2 + 4x + 4 + y2 = 4 (write x2 + 4x + 4 as (x + 2)2) (x + 2)2 + y2 = 4 This is a circle with radius of 2 centered at (– 2, 0).
π
2
3π
4
π
4
π 0 1 2
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
184
Circles in Polar Coordinates: 1) r = a This a circle of radius a centered at the Pole.
2) r = 2asin(θ) This a circle of radius a centered at (0, a).
3) r = – 2asin(θ) This a circle of radius a centered at (0, – a).
4) r = 2acos(θ) This a circle of radius a centered at (a, 0).
5) r = – 2acos(θ) This a circle of radius a centered at (– a, 0). Objective 2: Test for Symmetry in Polar Coordinates. When graphing functions, we sometimes found it helpful to use symmetry to help construct the graph. We checked to see if the function was : 1) Symmetric with respect to the x-axis, 2) Symmetric with respect to the y-axis, or 3) Symmetric with respect to the origin. We can do the same thing in polar coordinates, but instead of using the x-
axis, y-axis, and the origin, we will be using the polar axis, the θ = π/2 line, and the pole. With this tests listed below, it is possible for a test to fail, but the function to still exhibit that particular symmetry, Tests for Symmetry in Polar Coordinates:
1) Polar Axis Symmetry (x-axis) (r, θ)
Replace θ by – θ. If the resulting equation is equivalent to the original equation, then the graph is symmetric
with respect to the polar axis. (r, – θ)
2) Line θ =
π
2 Symmetry (y-axis) (r, π – θ) (r, θ)
Replace θ by π – θ. If the resulting equation is equivalent to the original equation, then the graph is
symmetric with respect to the line θ =
π
2.
(r, θ) 3) Pole Symmetry (origin)
Replace r by – r. If the resulting equation is equivalent to the original equation, then the graph is
symmetric with respect to the pole. (– r, θ)
185
Objective 3: Graph Polar Equations by Plotting Points. Graph the following:
Ex. 7 r = 2 – 2cos(θ) Solution: First, we apply the tests for symmetry:
Polar Axis (replace θ by – θ)
r = 2 – 2cos(– θ) (cosine is an even function)
r = 2 – 2cos(θ) This is an equivalent equation so the graph is symmetric with respect to the polar axis.
Line θ = π/2 (replace θ by π – θ)
r = 2 – 2cos(π – θ) (cosine of a difference of angles)
r = 2 – 2[cos(π)cos(θ) + sin(π)sin(θ)]
r = 2 – 2[(– 1)cos(θ) + (0)sin(θ)]
r = 2 + 2cos(θ) This test fails. Pole (replace r by – r)
– r = 2 – 2cos(θ)
r = – 2 + 2cos(θ) This test fails. Since the function is symmetric with respect the Polar Axis, we need
to just plot the function for values of θ between 0 and π inclusively and then reflect it across the polar axis.
θθθθ r = 2 – 2cos(θθθθ)
0 r = 2 – 2cos(0) = 2 – 2 = 0
π
6 r = 2 – 2cos(
π
6) = 2 – 3 ≈ 0.27
π
4 r = 2 – 2cos(
π
4) = 2 – 2 ≈ 0.59
π
3 r = 2 – 2cos(
π
3) = 2 – 1 = 1
π
2 r = 2 – 2cos(
π
2) = 2 – 0 = 2
2π
3 r = 2 – 2cos(
2π
3) = 2 + 1 = 3
3π
4 r = 2 – 2cos(
3π
4) = 2 + 2 ≈ 3.41
5π
6 r = 2 – 2cos(
5π
6) = 2 + 3 ≈ 3.73
π r = 2 – 2cos(π) = 2 + 2 = 4
186
Now, we will plot the points and then draw a smooth curve through the points:
Finally, use symmetry to get the rest of the graph:
This curve is called a Cardioid (heart-shaped curve). These types of curves are in the form of:
r = a ± acos(θ) or r = a ± asin(θ) where 0 < a
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
187
Ex. 8 r = 3 + 2sin(θ) Solution: First, we apply the tests for symmetry:
Polar Axis (replace θ by – θ)
r = 3 + 2sin(– θ) (sine is an odd function)
r = 3 – 2sin(θ) This test fails..
Line θ = π/2 (replace θ by π – θ)
r = 3 + 2sin(π – θ) (sine of a difference of angles)
r = 3 + 2[sin(π)cos(θ) – cos(π)sin(θ)]
r = 3 + 2[(0)cos(θ) – (– 1)sin(θ)]
r = 3 + 2sin(θ) This is an equivalent equation so the graph is symmetric with
respect to the line θ = π/2. Pole (replace r by – r)
– r = 3 + 2sin(θ)
r = – 3 – 2sin(θ) This test fails.
Since the function is symmetric with respect the line θ = π/2, we need to just plot the function for values between – π/2 and π/2 inclusively
and then reflect it across the line θ = π/2.
θθθθ r = 3 + 2sin(θθθθ)
–
π
2 r = 3 + 2sin(–
π
2) = 3 – 2 = 1
–
π
3 r = 3 + 2sin(–
π
3) = 3 – 3 ≈ 1.27
–
π
4 r = 3 + 2sin(–
π
4) = 3 – 2 ≈ 1.59
–
π
6 r = 3 + 2sin(–
π
6) = 3 – 1 = 2
0 r = 3 + 2sin(0) = 3 + 0 = 3
π
6 r = 3 + 2sin(
π
6) = 3 + 1 = 4
π
4 r = 3 + 2sin(
π
4) = 3 + 2 ≈ 4.41
π
3 r = 3 + 2sin(
π
3) = 3 + 3 ≈ 4.73
π
2 r = 3 + 2sin(
π
2) = 3 + 2 = 5
188
Now, we will plot the points and then draw a smooth curve through the points:
Finally, use symmetry to get the rest of the graph:
This curve is called a Limaçon (snail) without an inner loop. These types of curves are in the form of:
r = a ± bcos(θ) or r = a ± bsin(θ) where 0 < b < a
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
189
Ex. 9 r = 1 + 4cos(θ) Solution: First, we apply the tests for symmetry:
Polar Axis (replace θ by – θ)
r = 1 + 4cos(– θ) (cosine is an even function)
r = 1 + 4cos(θ) This is an equivalent equation so the graph is symmetric with respect to the polar axis.
Line θ = π/2 (replace θ by π – θ)
r = 1 + 4cos(π – θ) (cosine of a difference of angles)
r = 1 + 4[cos(π)cos(θ) + sin(π)sin(θ)]
r = 1 + 4[(– 1)cos(θ) + (0)sin(θ)]
r = 1 – 4cos(θ) This test fails. Pole (replace r by – r)
– r = 1 + 4cos(θ)
r = – 1 – 4cos(θ) This test fails. Since the function is symmetric with respect the Polar Axis, we need
to just plot the function for values of θ between 0 and π inclusively and then reflect it across the polar axis.
θθθθ r = 1 + 4cos(θθθθ)
0 r = 1 + 4cos(0) = 1 + 4 = 5
π
6 r = 1 + 4cos(
π
6) = 1 + 2 3 ≈ 4.46
π
4 r = 1 + 4cos(
π
4) = 1 + 2 2 ≈ 3.83
π
3 r = 1 + 4cos(
π
3) = 1 + 2 = 3
π
2 r = 1 + 4cos(
π
2) = 1 – 0 = 1
2π
3 r = 1 + 4cos(
2π
3) = 1 – 2 = – 1
3π
4 r = 1 + 4cos(
3π
4) = 1 – 2 2 ≈ – 1.83
5π
6 r = 1 + 4cos(
5π
6) = 1 – 2 3 ≈ – 2.46
π r = 1 + 4cos(π) = 1 – 4 = – 3
190
Now, we will plot the points and then draw a smooth curve through the points:
Finally, use symmetry to get the rest of the graph:
This curve is called a Limaçon (snail) with an inner loop. These types of curves are in the form of:
r = a ± bcos(θ) or r = a ± bsin(θ) where 0 < a < b
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
191
Ex. 10 r = 3cos(4θ) Solution: First, we apply the tests for symmetry:
Polar Axis (replace θ by – θ)
r = 3cos(4(– θ)) (cosine is an even function)
r = 3cos(4θ) This is an equivalent equation so the graph is symmetric with respect to the polar axis.
Line θ = π/2 (replace θ by π – θ)
r = 3cos(4[π – θ]) (cosine of a difference of angles)
r = 3[cos(4π)cos(4θ) + sin(4π)sin(4θ)]
r = 3[(1)cos(4θ) + (0)sin(4θ)]
r = 3cos(4θ) This is an equivalent equation so the graph is symmetric with
respect to the line θ = π/2. Pole (replace r by – r)
– r = 3cos(4θ)
r = – 3cos(4θ) Even though this test fails, because the function is symmetric to the
lines θ = π/2 and the polar axis, it is symmetric with respect to the pole. Thus, we need to plot the function for values between 0 and π/2
inclusively and then reflect it across the line θ = π/2 and the polar axis.
θθθθ 4θθθθ r = 3cos(4θθθθ)
0 0 r = 3cos(0) = 3
π
16
π
4 r = 3cos(
π
4) ≈ 2.12
π
8
π
2 r = 3cos(
π
2) = 0
3π
16
3π
4 r = 3cos(
3π
4) ≈ – 2.12
π
4 π r = 3cos(π) = – 3
5π
16
5π
4 r = 3cos(
5π
4) ≈ – 2.12
3π
8
3π
2 r = 3cos(
3π
2) = 0
7π
16
7π
4 r = 3cos(
7π
4) ≈ 2.12
π
2 2π r = 3cos(2π) = 3
192
Now, we will plot the points and then draw a smooth curve through the points:
Finally, reflect the graph across the θ = π/2 line and then the polar axis to get the graph:
This curve is called a Rose. The equations are in the form:
r = acos(nθ) or r = asin(nθ) where a ≠ 0 and n is an natural number greater than one. If n is even, the graph will have 2n petals and if n is odd, the graph will have n petals.
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
193
Ex. 11 r2 = 9sin(2θ) Solution: First, we apply the tests for symmetry:
Polar Axis (replace θ by – θ)
r2 = 9sin(2[– θ]) (sine is an odd function)
r2 = – 9sin(2θ) This test fails.
Line θ = π/2 (replace θ by π – θ)
r2 = 9sin(2[π – θ]) (sine of a difference of angles)
r2 = 9[sin(2π)cos(2θ) – cos(2π)sin(2θ)]
r2 = 9[(0)cos(2θ) – (1)sin(2θ)]
r2 = – 9sin(2θ) This test fails.
Pole (replace r by – r)
(– r)2 = 9sin(2θ)
r2 = 9sin(2θ) This is an equivalent equation so the graph is symmetric with respect to the pole. Since the function is symmetric with respect the pole, we need to just plot the function for values between – π/2 and π/2 inclusively and then reflect it across the pole.
θθθθ 2θθθθ r2 = 9sin(2θθθθ) r
–
π
2 – π r2 = 9sin(– π) = 0 0
–
π
3 –
2π
3 r2 = 9sin(–
2π
3) ≈ – 7.79 undefined
–
π
4 –
π
2 r2 = 9sin(–
π
2) = – 9 undefined
–
π
6 –
π
3 r2 = 9sin(–
π
3) ≈ – 7.79 undefined
0 0 r2 = 9sin(0) = 0 0
π
6
π
3 r2 = 9sin(
π
3) ≈ 7.79 ≈ ± 2.79
π
4
π
2 r2 = 9sin(
π
2) = 9 = ± 3
π
3
2π
3 r2 = 9sin(
2π
3) ≈ 7.79 ≈ ± 2.79
π
2 π r2 = 9sin(π) = 0 = 0
194
Now, we will plot the points and then draw a smooth curve through the points:
Using symmetry about the pole does not produce any additional points so our graph is:
This curve is called a Lemniscate (ribbon). These types of curves are in the form of:
r2 = a2cos(2θ) or r2 = a2sin(2θ) where a ≠ 0.
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
195
Ex. 12 r = 4sin(3θ) Solution: First, we apply the tests for symmetry:
Polar Axis (replace θ by – θ)
r = 4sin(3[– θ]) (sine is an odd function)
r = – 4sin(3θ) This test fails.
Line θ = π/2 (replace θ by π – θ)
r = 4sin(3[π – θ]) (sine of a difference of angles)
r = 4[sin(3π)cos(3θ) – cos(3π)sin(3θ)]
r = 4[(0)cos(3θ) – (– 1)sin(3θ)]
r = 4sin(3θ) This is an equivalent equation so the graph is symmetric with
respect to the line θ = π/2. Pole (replace r by – r)
– r = 4sin(3θ)
r = – 4sin(3θ) This test fails.
Since the function is symmetric with respect the line θ = π/2, we need to just plot the function for values between – π/2 and π/2 inclusively and then
reflect it across the line θ = π/2.
θθθθ 3θθθθ r = 4sin(3θθθθ)
–
π
2 –
3π
2 r = 4sin(–
3π
2) = 4
–
π
3 – π r = 4sin(– π) = 0
–
π
4 –
3π
4 r = 4sin(–
3π
4) ≈ – 2.82
–
π
6 –
π
2 r = 4sin(–
π
2) = – 4
0 0 r = 4sin(0) = 0
π
6
π
2 r = 4sin(
π
2) = 4
π
4
3π
4 r = 4sin(
3π
4) = 2.82
π
3 π r = 4sin(π) = 0
π
2
3π
2 r = 4sin(
3π
2) = – 4
196
If we plot the points, we see that we really do not have enough points to draw the curve.
θθθθ 3θθθθ r = 4sin(3θθθθ)
–
π
2 –
3π
2 r = 4sin(–
3π
2) = 4
– 5π
12 –
5π
4 r = 4sin(–
5π
4) ≈ 2.82
–
π
3 – π r = 4sin(– π) = 0
–
π
4 –
3π
4 r = 4sin(–
3π
4) ≈ – 2.82
–
π
6 –
π
2 r = 4sin(–
π
2) = – 4
– π
12 –
π
4 r = 4sin(–
π
4) ≈ – 2.82
0 0 r = 4sin(0) = 0
π
12
π
4 r = 4sin(
π
4) ≈ 2.82
π
6
π
2 r = 4sin(
π
2) = 4
π
4
3π
4 r = 4sin(
3π
4) = 2.82
π
3 π r = 4sin(π) = 0
5π
12
5π
4 r = 4sin(
5π
4) ≈ – 2.82
π
2
3π
2 r = 4sin(
3π
2) = – 4
Since the difference
between
π
4 and
π
6 is
π
12
and 12 is divisible by 3 from
the 3θ, let's try multiples of
π
12 for our angles.
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
197
Now, we will plot the points and then draw a smooth curve through the points:
Using symmetry does not produce any additional points so our graph is:
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
198
Ex. 13 r = (
7
5 )θ
First, we apply the tests for symmetry:
Polar Axis (replace θ by – θ)
r = (
7
5 ) - θ
= (
5
7 )θ
This test fails.
Line θ = π/2 (replace θ by π – θ)
r = (
7
5 )π - θ (division property of exponents)
r = (
7
5 )π
÷ (
7
5 )θ
(invert and multiply)
= (
7
5 )π
•(
5
7 )θ This test fails.
Pole (replace r by – r)
– r = (
7
5 )θ
r = – (
7
5 )θ
This test fails.
There is no symmetry that we can use. Recall that an exponential function is increasing on its domain when the base is greater than 1. Thus, we will make a table of values and plot some points.
θθθθ r = (
7
5 )θθθθ
–
π
2 r = (
7
5 )– π/2
≈ 0.59
–
π
4 r = (
7
5 )– π/4
≈ 0.77
0 r = (
7
5 )0
= 1
π
4 r = (
7
5 )π/4
≈ 1.30
π
2 r = (
7
5 )π/2
≈ 1.70
3π
4 r = (
7
5 )3π/4
≈ 2.21
π r = (
7
5 )π
≈ 2.88
5π
4 r = (
7
5 )5π/4
≈ 3.75
3π
2 r = (
7
5 )3π/2
≈ 4.88
199
Now, we will plot the points and then draw a smooth curve through the points:
This curve is called a Logarithmic Spiral. These types of curves can be rewritten in the form of: r = a ln(r) where a > 0. Classification of Polar Equations
Name Line through the Pole
Vertical Line Horizontal Line
Polar Equation θ = α rcos(θ) = a rsin(θ) = a
Graph
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
π
2
3π
4
π
4
π 0 1 2 3 4 5
5π
4
7π
4
3π
2
200
Name Circle centered at the Pole
Circle centered at (0, a)
Circle centered at (a, 0)
Polar Equation r = a r = 2asin(θ) r = 2acos(θ)
Graph
Name Cardioid Limaçon without an inner loop
Limaçon with an inner loop
Polar Equation r = a ± acos(θ)
r = a ± asin(θ)
r = a ± bcos(θ)
r = a ± bsin(θ) 0 < b < a
r = a ± bcos(θ)
r = a ± bsin(θ) 0 < a < b
Graph
Name Lemniscate Rose with odd number (n) of petals
Rose with even number (2n) of petals
Polar Equations r2 = a2cos(2θ)
r2 = a2sin(2θ)
r = asin(nθ)
r = acos(nθ) n is odd
r = asin(nθ)
r = acos(nθ) n is even
Graph
