# Section 9.2 Polar Equations and Graphs

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1 180 Section 9. 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 = r = (square both sides) r = (replace r by x + y ) x + y = This is a circle of radius centered at the Pole. 0 1 Ex. θ = θ = (take the tangent of both sides) tan(θ) = tan( ) = 1 (tan(θ) = y x ) y x = 1 y = x This is a line with slope of 1 passing through the Pole.

2 Ex. rsin(θ) = rsin(θ) = (y = rsin(θ)) y = This is a horizontal line passing through (0, ). 0 1 Ex. rcos(θ) =

3 18 rcos(θ) = (x = rcos(θ)) x = This is a vertical line passing through (, 0). 0 1 Theorem If a 0, then 1) rsin(θ) = a is a horizontal line that passes through (0, a). ) rcos(θ) = a is a vertical line that passes through (a, 0). Convert the polar equation to rectangular and sketch the graph: Ex. r = sin(θ) r = sin(θ) (multiply both sides by r) r = rsin(θ) (r = x + y and y = rsin(θ)) x + y = y (subtract y from both sides) x + y y = 0 (complete the square; add 1 to both sides) x + y y + 1 = 1 (write y y + 1 as (y 1) ) x + (y 1) = 1 This is a circle with radius of 1 centered at (0, 1).

4 Ex. 6 r = cos(θ) r = cos(θ) (multiply both sides by r) r = rcos(θ) (r = x + y and x = rcos(θ)) x + y = x (add x from both sides) x + x + y = 0 (complete the square; add to both sides) x + x + + y = (write x + x + as (x + ) ) (x + ) + y = This is a circle with radius of centered at (, 0). 0 1

5 Circles in Polar Coordinates: 1) r = a This a circle of radius a centered at the Pole. ) r = asin(θ) This a circle of radius a centered at (0, a). ) r = asin(θ) This a circle of radius a centered at (0, a). ) r = acos(θ) This a circle of radius a centered at (a, 0). ) r = acos(θ) This a circle of radius a centered at ( a, 0). 18 Objective : 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, ) Symmetric with respect to the y-axis, or ) 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 = / 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) ) Line θ = Symmetry (y-axis) (r, θ) (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 line θ =. (r, θ) ) 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)

6 18 Objective : Graph Polar Equations by Plotting Points. Graph the following: Ex. r = cos(θ) First, we apply the tests for symmetry: Polar Axis (replace θ by θ) r = cos( θ) (cosine is an even function) r = cos(θ) This is an equivalent equation so the graph is symmetric with respect to the polar axis. Line θ = / (replace θ by θ) r = cos( θ) (cosine of a difference of angles) r = [cos()cos(θ) + sin()sin(θ)] r = [( 1)cos(θ) + (0)sin(θ)] r = + cos(θ) This test fails. Pole (replace r by r) r = cos(θ) r = + cos(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 inclusively and then reflect it across the polar axis. θ r = cos(θ) 0 r = cos(0) = = r = cos( 6 ) = 0. r = cos( ) = 0.9 r = cos( ) = 1 = 1 r = cos( ) = 0 = r = cos( ) = + 1 = r = cos( ) = +.1 r = cos( 6 ) = +. r = cos() = + =

7 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: 0 1 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

8 Ex. 8 r = + sin(θ) First, we apply the tests for symmetry: Polar Axis (replace θ by θ) r = + sin( θ) (sine is an odd function) r = sin(θ) This test fails.. Line θ = / (replace θ by θ) 18 r = + sin( θ) (sine of a difference of angles) r = + [sin()cos(θ) cos()sin(θ)] r = + [(0)cos(θ) ( 1)sin(θ)] r = + sin(θ) This is an equivalent equation so the graph is symmetric with respect to the line θ = /. Pole (replace r by r) r = + sin(θ) r = sin(θ) This test fails. Since the function is symmetric with respect the line θ = /, we need to just plot the function for values between / and / inclusively and then reflect it across the line θ = /. θ r = + sin(θ) r = + sin( ) = = 1 r = + sin( ) = 1. r = + sin( ) = r = + sin( 6 ) = 1 = 0 r = + sin(0) = + 0 = 6 r = + sin( 6 ) = + 1 = r = + sin( ) = +.1 r = + sin( ) = +. r = + sin( ) = + =

9 188 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: 0 1 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

10 Ex. 9 r = 1 + cos(θ) First, we apply the tests for symmetry: Polar Axis (replace θ by θ) r = 1 + cos( θ) (cosine is an even function) r = 1 + cos(θ) This is an equivalent equation so the graph is symmetric with respect to the polar axis. Line θ = / (replace θ by θ) r = 1 + cos( θ) (cosine of a difference of angles) r = 1 + [cos()cos(θ) + sin()sin(θ)] r = 1 + [( 1)cos(θ) + (0)sin(θ)] r = 1 cos(θ) This test fails. Pole (replace r by r) r = 1 + cos(θ) r = 1 cos(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 inclusively and then reflect it across the polar axis. θ r = 1 + cos(θ) 0 r = 1 + cos(0) = 1 + = 6 6 r = 1 + cos( 6 ) = r = 1 + cos( ) = r = 1 + cos( ) = 1 + = r = 1 + cos( ) = 1 0 = 1 r = 1 + cos( ) = 1 = 1 r = 1 + cos( ) = r = 1 + cos( 6 ) = 1.6 r = 1 + cos() = 1 = 189

11 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: 0 1 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

12 191 Ex. 10 r = cos(θ) First, we apply the tests for symmetry: Polar Axis (replace θ by θ) r = cos(( θ)) (cosine is an even function) r = cos(θ) This is an equivalent equation so the graph is symmetric with respect to the polar axis. Line θ = / (replace θ by θ) r = cos([ θ]) (cosine of a difference of angles) r = [cos()cos(θ) + sin()sin(θ)] r = [(1)cos(θ) + (0)sin(θ)] r = cos(θ) This is an equivalent equation so the graph is symmetric with respect to the line θ = /. Pole (replace r by r) r = cos(θ) r = cos(θ) Even though this test fails, because the function is symmetric to the lines θ = / and the polar axis, it is symmetric with respect to the pole. Thus, we need to plot the function for values between 0 and / inclusively and then reflect it across the line θ = / and the polar axis. θ θ r = cos(θ) 0 0 r = cos(0) = r = cos( ).1 r = cos( ) = 0 r = cos( ).1 r = cos() = r = cos( ).1 r = cos( ) = 0 r = cos( ).1 r = cos() =

13 Now, we will plot the points and then draw a smooth curve through the points: Finally, reflect the graph across the θ = / 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 n petals and if n is odd, the graph will have n petals.

14 19 Ex. 11 r = 9sin(θ) First, we apply the tests for symmetry: Polar Axis (replace θ by θ) r = 9sin([ θ]) (sine is an odd function) r = 9sin(θ) This test fails. Line θ = / (replace θ by θ) r = 9sin([ θ]) (sine of a difference of angles) r = 9[sin()cos(θ) cos()sin(θ)] r = 9[(0)cos(θ) (1)sin(θ)] r = 9sin(θ) This test fails. Pole (replace r by r) ( r) = 9sin(θ) r = 9sin(θ) 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 / and / inclusively and then reflect it across the pole. θ θ r = 9sin(θ) r 6 r = 9sin( ) = 0 0 r = 9sin( r = 9sin( ) = 9 r = 9sin( ).9 ).9 undefined undefined undefined 0 0 r = 9sin(0) = r = 9sin( ).9 ±.9 r = 9sin( ) = 9 = ± r = 9sin( ).9 ±.9 r = 9sin() = 0 = 0

15 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: 0 1 This curve is called a Lemniscate (ribbon). These types of curves are in the form of: r = a cos(θ) or r = a sin(θ) where a 0.

16 19 Ex. 1 r = sin(θ) First, we apply the tests for symmetry: Polar Axis (replace θ by θ) r = sin([ θ]) (sine is an odd function) r = sin(θ) This test fails. Line θ = / (replace θ by θ) r = sin([ θ]) (sine of a difference of angles) r = [sin()cos(θ) cos()sin(θ)] r = [(0)cos(θ) ( 1)sin(θ)] r = sin(θ) This is an equivalent equation so the graph is symmetric with respect to the line θ = /. Pole (replace r by r) r = sin(θ) r = sin(θ) This test fails. Since the function is symmetric with respect the line θ = /, we need to just plot the function for values between / and / inclusively and then reflect it across the line θ = /. θ θ r = sin(θ) 6 r = sin( ) = r = sin( ) = 0 r = sin( ).8 r = sin( ) = 0 0 r = sin(0) = 0 6 r = sin( ) = r = sin( ) =.8 r = sin() = 0 r = sin( ) =

17 196 If we plot the points, we see that we really do not have enough points to draw the curve. 0 1 Since the difference between and 6 is 1 and 1 is divisible by from the θ, let's try multiples of 1 for our angles. θ θ r = sin(θ) r = sin( ) = r = sin( ).8 r = sin( ) = 0 r = sin( ).8 r = sin( ) = r = sin( ) r = sin(0) = r = sin( ).8 r = sin( ) = r = sin( ) =.8 r = sin() = 0 r = sin( ).8 r = sin( ) =

18 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: 0 1

19 Ex. 1 r = ( ) θ First, we apply the tests for symmetry: Polar Axis (replace θ by θ) r = ( ) - θ = ( ) θ This test fails. Line θ = / (replace θ by θ) r = ( ) - θ (division property of exponents) r = ( ) ( ) θ (invert and multiply) = ( ) ( ) θ This test fails. Pole (replace r by r) r = ( ) θ r = ( ) 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. θ r = ( ) θ r = ( ) / 0.9 r = ( ) / 0. 0 r = ( ) 0 = 1 r = ( ) / 1.0 r = ( ) / 1.0 r = ( ) /.1 r = ( ).88 r = ( ) /. r = ( ) /

20 199 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 Vertical Line Horizontal Line Pole Polar Equation θ = α rcos(θ) = a rsin(θ) = a Graph

21 00 Name Circle centered at the Pole Circle centered at (0, a) Circle centered at (a, 0) Polar Equation r = a r = asin(θ) r = acos(θ) Graph Name Cardioid Limaçon without an inner loop Polar Equation r = a ± acos(θ) r = a ± bcos(θ) r = a ± asin(θ) r = a ± bsin(θ) 0 < b < a Graph Limaçon with an inner loop r = a ± bcos(θ) r = a ± bsin(θ) 0 < a < b Name Lemniscate Rose with odd number (n) of petals Polar Equations Graph r = a cos(θ) r = a sin(θ) r = asin(nθ) r = acos(nθ) n is odd Rose with even number (n) of petals r = asin(nθ) r = acos(nθ) n is even

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