618 CHAPTER 9 Polar Coordinates; Vectors Graphing a Polar Equation (Limaçon without an Inner Loop) Graph the equation: r 3 2cosθ = + Solution EXAMPLE 9 Figure 32 θ = + r 3 2cos x O y 2 4 u 5 0 u 5 p u 5 1 3 p– 2 u 5 3p –– 2 u 5 7p –– 4 u 5 p– 4 u 5 3p –– 4 u 5 5p –– 4 5 (5, 0) (2, ) (4.73, ) p– 6 (1.27, ) (1, p) 5p –– 6 2p –– 3 (3, )p– 2 (4, )p– 3 1 2 3 4 5 (a) (b) Check for symmetry first. Polar Axis: Replace θ by .θ− The result is θ θ ( ) = + − = + r 3 2 cos 3 2 cos cos cos θ θ ( ) − = The test is satisfied, so the graph is symmetric with respect to the polar axis. The Line θ π = 2 : Replace θ by . π θ − The result is π θ π θ π θ θ ( ) ( ) = + − = + + = − r 3 2 cos 3 2 cos cos sin sin 3 2 cos The test fails. Replace r by r− and θ by .θ− The result is θ θ θ ( ) − = + − − = + = − − r r r 3 2 cos 3 2 cos 3 2 cos cos cos θ θ ( ) − = This test also fails, so the graph may or may not be symmetric with respect to the line 2 . θ π = The Pole: Replace r by r. − The test fails. Replace θ by . θ π + This test also fails, so the graph may or may not be symmetric with respect to the pole. Next, identify points on the graph by assigning values to the angle θ and calculating the corresponding values of r. Due to the periodicity of the cosine function and the symmetry with respect to the polar axis, just assign values to θ from 0 to ,π as given in Table 2. Now plot the points r, θ ( ) from Table 2 and trace out the graph, beginning at the point 5, 0 ( ) and ending at the point 1, . π ( ) Then reflect this portion of the graph about the polar axis (the x -axis) to obtain the complete graph. Figure 32(a) shows the graph drawn by hand. Figure 32(b) shows the graph using GeoGebra. DEFINITION Limaçons without an Inner Loop Limaçons without an inner loop are characterized by equations of the form • θ = + r a bcos • θ = − r a bcos • θ = + r a bsin • θ = − r a bsin where a b 0. > > The graph of a limaçon without an inner loop does not pass through the pole. The curve in Figure 32 is an example of a limaçon (a French word for snail ) without an inner loop . Now Work PROBLEM 45 θ θ = + r 3 2cos 0 + ⋅ = 3 2 1 5 π 6 + ⋅ ≈ 3 2 3 2 4.73 π 3 + ⋅ = 3 2 1 2 4 π 2 + ⋅ = 3 2 0 3 π2 3 ( ) + − = 3 2 1 2 2 π5 6 + − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ ≈ 3 2 3 2 1.27 π ( ) + − = 3 2 1 1 Table 2 Exploration Graph θ = − r 3 2cos . 1 Clear the screen and graph θ = + r 3 2sin . 1 Clear the screen and graph θ = − r 3 2sin . 1 Do you see a pattern?
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