8.2 Graphs of Polar Equations

Transcription

1 8. Graphs of Polar Equations Definition: A polar equation is an equation whose variables are polar coordinates. One method used to graph a polar equation is to convert the equation to rectangular form. Example 1: Identify and graph the equation r = 3 on the xy-plane. Solution: Convert the polar equation to a rectangular equation. r = 3 = r = 9 = x + y = 9 The graph of r = 3 on the xy-plane is a circle with center at the pole and radius 3. 1

2 Example : Identify and graph the equation θ = 4. Solution: Convert the polar equation to a rectangular equation. θ = ( ) = tan θ = tan 4 4 = y x = 1 = y = x The graph of θ = 4 on the xy-plane is a line through the origin with slope 1. Tests for Symmetry Let r = f(θ) be a polar equation. Symmetry with Respect to the Polar Axis (x-axis) If f( θ) = f(θ) for all θ, then the graph is symmetric with respect to the polar axis.

3 Symmetry with Respect to the Line θ = (y-axis) If f( θ) = f(θ) for all θ, then the graph is symmetric with respect to the line θ =. 3

4 Symmetry with Respect to the Pole (Origin) If f(θ + ) = f(θ) for all θ or if replacing r with r results in an equivalent equation, then the graph is symmetric with respect to the pole. Note: The three tests for symmetry given here are sufficient conditions for symmetry, but they are not necessary conditions. In other words, an equation may fail any one of these tests and still have a graph that is symmetric with respect to the polar axis, the line θ =, or the pole. For example, the graph of r = sin(θ) is symmetric with respect to the polar axis, the line θ, and the pole, but only the test for symmetry with respect to the pole is satisfied. Example 3: Graph the equation: r = 1 sin θ on the xy-plane. Solution: Here, f(θ) = 1 sin θ. Let s check for symmetry first. Polar Axis: Replace θ with θ. f( θ) = 1 sin( θ) = 1 + sin θ f(θ) The test fails, so the graph may or may not be symmetric with respect to the polar axis. 4

5 The Line θ = : Replace θ with θ. f( θ) = 1 sin( θ) = 1 [sin cos θ cos sin θ] = 1 [0 + sin θ] = 1 sin θ = f(θ) The test is satisfied, so the graph is symmetric with respect to the line θ =. The Pole: Replace r with r. r = 1 sin θ = r = 1 + sin θ Since replacing r with r yields an equation that is not equivalent to the original one, the test fails. Let s try replacing θ with θ +. f(θ + ) = 1 sin(θ + ) = 1 [sin θ cos + cos θ sin ] = 1 [ sin θ + 0] = 1 + sin θ f(θ) This test also fails, so the graph may or may not be symmetric with respect to the pole. Since the graph is symmetric with respect to the line θ =, we only need to assign values of θ from to. Once we trace the part of the graph on θ, we will need to reflect the graph about the line θ = to obtain the complete graph. 5

6 θ r = 1 sin θ 1 ( 1) = ( ) ( 1 ) = = = 1 6 ( 3 ) = 0 This curve is called a cardioid (a heart-shaped curve). Definition: Cardioids are characterized by equations of the form: r = a(1 + cos θ) r = a(1 + sin θ) r = a(1 cos θ) r = a(1 sin θ) where a > 0. The graph of a cardioid passes through the pole. Example 4: Graph the equation: r = 1 + cos θ on the xy-plane. Solution: Here, f(θ) = 1 + cos θ. Let s check for symmetry first. Polar Axis: Replace θ with θ. f( θ) = 1 + cos( θ) = 1 + cos θ = f(θ) The test is satisfied, so the graph is symmetric with respect to the polar axis. 6

7 The Line θ = : Replace θ with θ. f( θ) = 1 + cos( θ) = 1 + [cos cos θ + sin sin θ] = 1 cos θ f(θ) The test fails, so the graph may or may not be symmetric with respect to the line θ =. The Pole: Replace r with r. r = 1 + cos θ = r = 1 cos θ Since replacing r with r yields an equation that is not equivalent to the original one, the test fails. Let s try replacing θ with θ +. f(θ + ) = 1 + cos(θ + ) = 1 + [cos θ cos sin θ sin ] = 1 + [ cos θ 0] = 1 cos θ f(θ) This test also fails, so the graph may or may not be symmetric with respect to the pole. Since the graph is symmetric with respect to the polar axis, we only need to assign values of θ from 0 to. Once we trace the part of the graph on 0 θ, we will need to reflect the graph about the polar axis to obtain the complete graph. 7

8 θ r = 1 + cos θ (1) = 3 ( 3 ) ( 1 3 ) = 1 + (0) = ( 1) = 0 3 ( ) ( 1) = 1 This curve is called a limaçon with an inner loop. Definition: Limaçons are characterized by equations of the form: r = a + b cos θ r = a + b sin θ r = a b cos θ r = a b sin θ where a > 0 and b > 0. If a > b, then the limaçon has no loop and does not pass through the pole. If a < b, then the limaçon has a loop that passes through the pole twice. If a = b, then the limaçon is a cardioid. Example 5: Graph the equation r = cos(θ) on the xy-plane. Solution: Here f(θ) = cos(θ). As usual, let s check for symmetry first. Polar Axis: Replace θ with θ. f( θ) = cos[( θ)] = cos(θ) = f(θ) The test is satisfied, so the graph is symmetric with respect to the polar axis. 8

9 The Line θ = : Replace θ with θ. f( θ) = cos[( θ)] = cos( θ) = cos( θ) = cos(θ) = f(θ) The test is satisfied, so the graph is symmetric with respect to the line θ =. The Pole: Since the graph is symmetric with respect to both the polar axis and the line θ =, the graph is also symmetric with respect to the pole. (It turns out that the test for symmetry with respect to the pole is satisfied. Replace θ with θ + and simplify to verify this.) Since the graph is symmetric with respect to the polar axis, the line θ =, and the pole, we only need to assign values of θ from 0 to. Below is the part of the graph on 0 θ. θ r = cos(θ) 0 (1) = ( 1 6 ) = 1 (0) = 0 4 ( 1) = 1 3 ( 1) = Since the complete graph is symmetric with respect to the line θ =, let s reflect the graph shown above with respect to this line. The graph shown below is the part of the complete graph on 0 θ. 9

10 Finally, to obtain the complete graph, we ll use the fact that it is symmetric with respect to the polar axis. Reflecting the graph shown right above with respect to the polar axis yields the complete graph (on 0 θ ). This curve is called a rose (with 4 petals). 10

11 Definition: Rose curves are characterized by the equations of the form r = a cos(nθ) r = a sin(nθ) where a 0. If n 0 is even, the rose has n petals. If n ±1 is odd, the rose has n petals. Example 6: Graph the equation r = 4 sin(θ) on the xy-plane. Solution: Note that the tests for symmetry with respect to the polar axis and the line θ = both fail (Verify this). Since replacing r with r results in the same equation, the graph is symmetric with respect to the pole. With symmetry with respect to the pole, we would ordinarily consider values of θ on [0, ]. However, we must be careful here since the equation includes r (instead of just r) which is never negative. Thus, we must only consider values of θ for which 4 sin(θ) 0. Since the period of sin(θ) is, sin(θ) < 0 when θ lies on (, ) (Quadrant II). Therefore, we ll consider values of θ that lie on [0, ] and then use the symmetry with respect to the origin to obtain the complete graph. θ r = 4 sin(θ) r 0 4(0) = 0 0 ( 3 ) 3.5 ± (1) = 4 ± 4 4( 3 3 ) 3.5 ± 1.9 4(0) = 0 0 This curve is called a lemniscate (from the Greek word ribbon ). It is also referred to as a figure eight. 11

12 Definition: Lemniscates are characterized by equations of the form r = a sin(θ) r = a cos(θ) where a 0 and have graphs that are propeller shaped. Example 7: Graph the equation r = e θ/15 on the xy-plane. Solution: The tests for symmetry discussed in this section all fail. Moreover, since the exponential function is never zero, there is no value of θ such that r = 0, which means that the graph neither touches nor crosses the pole. Observe that r increases exponentially as θ increases and r approaches 0 as θ approaches. The following figure illustrates the part of the graph on 4 θ 4. This curve is called a logarithmic spiral since its equation may be written as θ = 15 ln r and it spirals infinitely both toward the pole and away from it. 1