{"id":2257,"date":"2025-08-12T16:34:53","date_gmt":"2025-08-12T16:34:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2257"},"modified":"2025-08-13T17:03:34","modified_gmt":"2025-08-13T17:03:34","slug":"graphing-in-polar-coordinates-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphing-in-polar-coordinates-learn-it-5\/","title":{"raw":"Graphing in Polar Coordinates: Learn It 5","rendered":"Graphing in Polar Coordinates: Learn It 5"},"content":{"raw":"

Investigating Lemniscates<\/h2>\r\nThe lemniscate is a polar curve resembling the infinity symbol [latex]\\infty [\/latex] or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition.\r\n\r\n
\r\n

lemniscates<\/h3>\r\nThe formulas that generate the graph of a lemniscate<\/strong> are given by [latex]{r}^{2}={a}^{2}\\cos 2\\theta [\/latex] and [latex]{r}^{2}={a}^{2}\\sin 2\\theta [\/latex] where [latex]a\\ne 0[\/latex]. The formula [latex]{r}^{2}={a}^{2}\\sin 2\\theta [\/latex] is symmetric with respect to the pole. The formula [latex]{r}^{2}={a}^{2}\\cos 2\\theta [\/latex] is symmetric with respect to the pole, the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], and the polar axis. See Figure 13\u00a0for the graphs.\r\n\r\n\"Four\r\n\r\n<\/section>
Sketch the graph of [latex]{r}^{2}=4\\cos 2\\theta [\/latex].[reveal-answer q=\"721985\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"721985\"]The equation exhibits symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, and the pole.\r\n\r\nLet\u2019s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution [latex]u=2\\theta [\/latex].\r\n

[latex]\\begin{align}&0=4\\cos 2\\theta \\\\ &0=4\\cos u \\\\ &0=\\cos u \\\\ &{\\cos }^{-1}0=\\frac{\\pi }{2} \\\\ &u=\\frac{\\pi }{2}&& \\text{Substitute }2\\theta \\text{ back in for }u. \\\\ &2\\theta =\\frac{\\pi }{2} \\\\ &\\theta =\\frac{\\pi }{4} \\end{align}[\/latex]<\/p>\r\nSo, the point [latex]\\left(0,\\frac{\\pi }{4}\\right)[\/latex] is a zero of the equation.\r\n\r\nNow let\u2019s find the maximum value. Since the maximum of [latex]\\cos u=1[\/latex] when [latex]u=0[\/latex], the maximum [latex]\\cos 2\\theta =1[\/latex] when [latex]2\\theta =0[\/latex]. Thus,\r\n

[latex]\\begin{gathered}{r}^{2}=4\\cos \\left(0\\right)\\\\ {r}^{2}=4\\left(1\\right)=4\\\\ r=\\pm \\sqrt{4}=2\\end{gathered}[\/latex]<\/p>\r\nWe have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], and the polar axis, we only need to plot points in the first quadrant.\r\n\r\nMake a table similar to the table below.\r\n\r\n\r\n\r\n\r\n
[latex]\\theta [\/latex] <\/strong><\/td>\r\n0<\/td>\r\n[latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n[latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]r[\/latex] <\/strong><\/td>\r\n2<\/td>\r\n[latex]\\sqrt{2}[\/latex]<\/td>\r\n0<\/td>\r\n[latex]\\sqrt{2}[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot the points on the graph.\r\n\r\n\"Graph\r\n

Analysis of the Solution<\/h4>\r\nMaking a substitution such as [latex]u=2\\theta [\/latex] is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown.\r\n\r\nSome of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of [latex]r[\/latex]. This is because there are no real square roots for these values of [latex]\\theta [\/latex]. In other words, the corresponding r<\/em>-values of [latex]\\sqrt{4\\cos \\left(2\\theta \\right)}[\/latex]\r\nare complex numbers because there is a negative number under the radical.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
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Investigating Rose Curves<\/h2>\r\n<\/div>\r\nThe next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.\r\n\r\n
\r\n

rose curves<\/h3>\r\nThe formulas that generate the graph of a rose curve<\/strong> are given by [latex]r=a\\cos n\\theta [\/latex] and [latex]r=a\\sin n\\theta [\/latex] where [latex]a\\ne 0[\/latex]. If [latex]n[\/latex] is even, the curve has [latex]2n[\/latex] petals. If [latex]n[\/latex] is odd, the curve has [latex]n[\/latex] petals.\r\n\r\n\"Graph\r\n\r\n<\/section>
Sketch the graph of [latex]r=2\\cos 4\\theta [\/latex].[reveal-answer q=\"667305\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"667305\"]Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] and the pole.\r\n\r\nNow we will find the zeros. First make the substitution [latex]u=4\\theta [\/latex].\r\n

[latex]\\begin{gathered}0=2\\cos 4\\theta \\\\ 0=\\cos 4\\theta \\\\ 0=\\cos u\\\\ {\\cos }^{-1}0=u\\\\ u=\\frac{\\pi }{2}\\\\ 4\\theta =\\frac{\\pi }{2}\\\\ \\theta =\\frac{\\pi }{8}\\end{gathered}[\/latex]<\/p>\r\nThe zero is [latex]\\theta =\\frac{\\pi }{8}[\/latex]. The point [latex]\\left(0,\\frac{\\pi }{8}\\right)[\/latex] is on the curve.\r\n\r\nNext, we find the maximum [latex]|r|[\/latex]. We know that the maximum value of [latex]\\cos u=1[\/latex] when [latex]\\theta =0[\/latex]. Thus,\r\n

[latex]\\begin{gathered}r=2\\cos \\left(4\\cdot 0\\right) \\\\ r=2\\cos \\left(0\\right) \\\\ r=2\\left(1\\right)=2 \\end{gathered}[\/latex]<\/p>\r\nThe point [latex]\\left(2,0\\right)[\/latex] is on the curve.\r\n\r\nThe graph of the rose curve has unique properties, which are revealed in the table below.\r\n\r\n\r\n\r\n\r\n
[latex]\\theta [\/latex] <\/strong><\/td>\r\n0<\/td>\r\n[latex]\\frac{\\pi }{8}[\/latex]<\/td>\r\n[latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n[latex]\\frac{3\\pi }{8}[\/latex]<\/td>\r\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n[latex]\\frac{5\\pi }{8}[\/latex]<\/td>\r\n[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]r[\/latex] <\/strong><\/td>\r\n2<\/td>\r\n0<\/td>\r\n\u22122<\/td>\r\n0<\/td>\r\n2<\/td>\r\n0<\/td>\r\n\u22122<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAs [latex]r=0[\/latex] when [latex]\\theta =\\frac{\\pi }{8}[\/latex], it makes sense to divide values in the table by [latex]\\frac{\\pi }{8}[\/latex] units. A definite pattern emerges. Look at the range of r<\/em>-values: 2, 0, \u22122, 0, 2, 0, \u22122, and so on. This represents the development of the curve one petal at a time. Starting at [latex]r=0[\/latex], each petal extends out a distance of [latex]r=2[\/latex], and then turns back to zero [latex]2n[\/latex] times for a total of eight petals.\r\n\r\n\"Sketch\r\n

Analysis of the Solution<\/h4>\r\nWhen these curves are drawn, it is best to plot the points in order, as in the table of Example 8's solution. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
\r\n
\r\n\r\nSketch the graph of [latex]r=4\\sin \\left(2\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"661678\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"661678\"]\r\n\r\nThe graph is a rose curve, [latex]n[\/latex] even\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
Sketch the graph of [latex]r=2\\sin \\left(5\\theta \\right)[\/latex].[reveal-answer q=\"482872\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"482872\"]The graph of the equation shows symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex]. Next, find the zeros and maximum. We will want to make the substitution [latex]u=5\\theta [\/latex].\r\n

[latex]\\begin{gathered}0=2\\sin \\left(5\\theta \\right)\\\\ 0=\\sin u\\\\ {\\sin }^{-1}0=0\\\\ u=0\\\\ 5\\theta =0\\\\ \\theta =0\\end{gathered}[\/latex]<\/p>\r\nThe maximum value is calculated at the angle where [latex]\\sin \\theta [\/latex] is a maximum. Therefore,\r\n

[latex]\\begin{gathered} r=2\\sin \\left(5\\cdot \\frac{\\pi }{2}\\right) \\\\ r=2\\left(1\\right)=2 \\end{gathered}[\/latex]<\/p>\r\nThus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for [latex]n[\/latex] odd yields the same number of petals as [latex]n[\/latex], there will be five petals on the graph.\r\n\r\nCreate a table of values similar to the table below.\r\n\r\n\r\n\r\n\r\n
[latex]\\theta [\/latex] <\/strong><\/td>\r\n0<\/td>\r\n[latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\r\n[latex]\\pi [\/latex]<\/td>\r\n<\/tr>\r\n
[latex]r[\/latex] <\/strong><\/td>\r\n0<\/td>\r\n1<\/td>\r\n\u22121.73<\/td>\r\n2<\/td>\r\n\u22121.73<\/td>\r\n1<\/td>\r\n0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
\r\n
\r\n\r\nSketch the graph of [latex]r=3\\cos \\left(3\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"214154\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"214154\"]\r\n\r\nRose curve, [latex]n[\/latex] odd\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
[ohm_question hide_question_numbers=1]149355[\/ohm_question]<\/section>","rendered":"

Investigating Lemniscates<\/h2>\n

The lemniscate is a polar curve resembling the infinity symbol [latex]\\infty[\/latex] or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition.<\/p>\n

\n

lemniscates<\/h3>\n

The formulas that generate the graph of a lemniscate<\/strong> are given by [latex]{r}^{2}={a}^{2}\\cos 2\\theta[\/latex] and [latex]{r}^{2}={a}^{2}\\sin 2\\theta[\/latex] where [latex]a\\ne 0[\/latex]. The formula [latex]{r}^{2}={a}^{2}\\sin 2\\theta[\/latex] is symmetric with respect to the pole. The formula [latex]{r}^{2}={a}^{2}\\cos 2\\theta[\/latex] is symmetric with respect to the pole, the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], and the polar axis. See Figure 13\u00a0for the graphs.<\/p>\n

\"Four<\/p>\n<\/section>\n

Sketch the graph of [latex]{r}^{2}=4\\cos 2\\theta[\/latex].<\/p>\n