620 CHAPTER 9 Polar Coordinates; Vectors The curve in Figure 34(b) or (c) is called a rose with four petals. Graphing a Polar Equation (Rose) Graph the equation: θ ( ) = r 2 cos 2 EXAMPLE 11 Figure 34 (a) x u 5 0 u 5 p u 5 p– 2 u 5 3p –– 2 u 5 7p –– 4 u 5 p– 4 u 5 3p –– 4 u 5 5p –– 4 1 2 4 5 y 3 21, p– 3 1, p– 6 0, p– 4 (2, 0) ( ) ( ) ( ) (b) r 5 2 cos (2u) x u 5 0 u 5 p u 5 p– 2 u 5 3p –– 2 u 5 7p –– 4 u 5 p– 4 u 5 3p –– 4 u 5 5p –– 4 4 5 y 3 (2, 0) 2 1, p– 6 ( ) 21, p– 3 ( ) 22, p– 2 ( ) 22, p– 2 ( ) 3 4 5 3 4 5 22.5 (c) 24 4 2.5 θ θ ( ) = r 2cos 2 0 ⋅ = 2 1 2 π 6 ⋅ = 2 1 2 1 π 4 ⋅ = 2 0 0 π 3 ( ) − = − 2 1 2 1 π 2 ( ) − = − 2 1 2 Table 4 Exploration Graph θ ( ) = r 2cos 4 ; 1 clear the screen and graph θ ( ) = r 2cos 6 . 1 How many petals did each of these graphs have? Clear the screen and graph, in order, each on a clear screen, θ ( ) = r 2cos 3 , 1 θ ( ) = r 2cos 5 , 1 and θ ( ) = r 2cos 7 . 1 What do you notice about the number of petals? Solution Check for symmetry. Polar Axis : Replace θ by .θ− The result is θ θ [ ( )] ( ) = − = r 2 cos 2 2 cos 2 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 θ ( ) [ ] ( ) ( ) = − = − = r 2cos 2 2 cos 2 2 2 cos 2 The test is satisfied, so the graph is symmetric with respect to the line 2 . θ π = The Pole : Since the graph is symmetric with respect to both the polar axis and the line 2 , θ π = it must be symmetric with respect to the pole. Next, construct Table 4. Because of the periodicity of the cosine function and the symmetry with respect to the polar axis, the line 2 , θ π = and the pole, only list values of θ from 0 to 2 . π Plot and connect these points as shown in Figure 34(a). Finally, because of symmetry, reflect this portion of the graph first about the polar axis (the x -axis) and then about the line 2 θ π = (the y -axis) to obtain the complete graph. See Figure 34(b). Figure 34(c) shows the graph using a TI-84 Plus CE with min 0, max 2 , θ θ π = = and step 24 . θ π = Now Work PROBLEM 51 DEFINITION Rose Rose curves are characterized by equations of the form • r a n cos θ ( ) = • r a n a sin 0 θ ( ) = ≠ and have graphs that are rose shaped. If n 0 ≠ is even, the rose has n2 petals; if n 1 ≠ ± is odd, the rose has n petals.
RkJQdWJsaXNoZXIy NjM5ODQ=