SECTION 9.2 Polar Equations and Graphs 621 The curve in Figure 35(b) or (c) is an example of a lemniscate (from the Greek word for ribbon ). Graphing a Polar Equation (Lemniscate) Graph the equation: θ ( ) = r 4 sin 2 2 EXAMPLE 12 Solution We leave it to you to verify that the graph is symmetric with respect to the pole. Because of the symmetry with respect to the pole, only list those values of between 0 and . θ θ θ π = = Note that for 2 π θ π < < (quadrant II) there are no points on the graph since r 0 2 < for such values. Table 5 lists points on the graph for values of 0 θ = through 2 . θ π = The points from Table 5 where r 0 ≥ are plotted in Figure 35(a). The remaining points on the graph may be obtained by using symmetry. Figure 35(b) shows the final graph drawn. Figure 35(c) shows the graph using Desmos. θ θ ( ) = r 4sin 2 2 r 0 ⋅ = 4 0 0 0 π 6 ⋅ = 4 3 2 2 3 ±1.9 π 4 ⋅ = 4 1 4 ±2 π 3 ⋅ = 4 3 2 2 3 ±1.9 π 2 ⋅ = 4 0 0 0 Table 5 Figure 35 (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 y 1.9, p– 6 1.9, p– 3 2, p– 4 ( ( ) ) ( ) 1.9, ( ) p– 6 (b) r 2 = 4 sin (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 1 2 y ( ) 1.9, (0, 0) (0, 0) p– 3 ( ) 2, p– 4 (c) DEFINITION Lemniscates Lemniscates are characterized by equations of the form • r a sin 2 2 2 θ ( ) = • r a cos 2 2 2 θ ( ) = where a 0, ≠ and have graphs that are propeller shaped. Now Work PROBLEM 55 Graphing a Polar Equation (Spiral) Graph the equation: r e 5 = θ EXAMPLE 13 Solution The tests for symmetry with respect to the pole, the polar axis, and the line 2 θ π = fail. Furthermore, there is no number θ for which r 0, = so the graph does not pass through the pole. From the equation, note that r is positive for all ,θ r increases as θ increases, r 0 → as , θ →−∞ and r →∞ as θ →∞. (continued)
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