{"id":1018,"date":"2025-06-20T17:29:15","date_gmt":"2025-06-20T17:29:15","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1018"},"modified":"2025-09-10T18:15:18","modified_gmt":"2025-09-10T18:15:18","slug":"understanding-polar-coordinates-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/understanding-polar-coordinates-learn-it-3\/","title":{"raw":"Understanding Polar Coordinates: Learn It 3","rendered":"Understanding Polar Coordinates: Learn It 3"},"content":{"raw":"

Polar Curves<\/h2>\r\n

Just as we graph functions [latex]y=f(x)[\/latex] in the rectangular coordinate system, we can graph functions [latex]r=f(\\theta)[\/latex] in the polar coordinate system to create curves.<\/p>\r\n

The process for graphing polar curves follows the same fundamental approach as rectangular graphing. You start with values for the independent variable [latex]\\theta[\/latex], calculate corresponding values of the dependent variable [latex]r[\/latex], then plot and connect the resulting points.<\/p>\r\n

Since many polar functions are periodic, you often need to evaluate only a limited range of [latex]\\theta[\/latex] values to capture the complete curve pattern.<\/p>\r\n\r\n

\r\n

Problem-Solving Strategy: Plotting a Curve in Polar Coordinates<\/strong><\/p>\r\n\r\n

    \r\n \t
  1. Create a table with columns for [latex]\\theta[\/latex] and [latex]r[\/latex]<\/li>\r\n \t
  2. Choose appropriate [latex]\\theta[\/latex] values (consider the function's period)<\/li>\r\n \t
  3. Calculate the corresponding [latex]r[\/latex] values for each [latex]\\theta [\/latex].<\/li>\r\n \t
  4. Plot each ordered pair [latex]\\left(r,\\theta \\right)[\/latex] on the coordinate axes.<\/li>\r\n \t
  5. Connect the points and identify any patterns in the curve<\/li>\r\n<\/ol>\r\n<\/section>
    \r\n
    \r\n

    Graph the curve defined by the function [latex]r=4\\sin\\theta [\/latex]. Identify the curve and rewrite the equation in rectangular coordinates.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n

    \r\n

    Because the function is a multiple of a sine function, it is periodic with period [latex]2\\pi [\/latex], so use values for [latex]\\theta [\/latex] between 0 and [latex]2\\pi [\/latex]. The result of steps 1\u20133 appear in the following table. Figure 5 shows the graph based on this table.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]\\theta [\/latex]<\/th>\r\n[latex]r=4\\sin\\theta [\/latex]<\/th>\r\n<\/th>\r\n[latex]\\theta [\/latex]<\/th>\r\n[latex]r=4\\sin\\theta [\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    [latex]0[\/latex]<\/td>\r\n[latex]0[\/latex]<\/td>\r\n<\/td>\r\n[latex]\\pi [\/latex]<\/td>\r\n[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n[latex]\\frac{7\\pi }{6}[\/latex]<\/td>\r\n[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n[latex]2\\sqrt{2}\\approx 2.8[\/latex]<\/td>\r\n[latex]\\frac{5\\pi }{4}[\/latex]<\/td>\r\n[latex]-2\\sqrt{2}\\approx -2.8[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n[latex]2\\sqrt{3}\\approx 3.4[\/latex]<\/td>\r\n[latex]\\frac{4\\pi }{3}[\/latex]<\/td>\r\n[latex]-2\\sqrt{3}\\approx -3.4[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n[latex]4[\/latex]<\/td>\r\n[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n[latex]2\\sqrt{3}\\approx 3.4[\/latex]<\/td>\r\n[latex]\\frac{5\\pi }{3}[\/latex]<\/td>\r\n[latex]-2\\sqrt{3}\\approx -3.4[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\frac{3\\pi }{4}[\/latex]<\/td>\r\n[latex]2\\sqrt{2}\\approx 2.8[\/latex]<\/td>\r\n[latex]\\frac{7\\pi }{4}[\/latex]<\/td>\r\n[latex]-2\\sqrt{2}\\approx -2.8[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\frac{5\\pi }{6}[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n[latex]\\frac{11\\pi }{6}[\/latex]<\/td>\r\n[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n
    <\/td>\r\n<\/td>\r\n[latex]2\\pi [\/latex]<\/td>\r\n[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
    [caption id=\"\" align=\"aligncenter\" width=\"417\"]\"On Figure 5. The graph of the function [latex]r=4\\sin\\theta [\/latex] is a circle.[\/caption]<\/figure>\r\n

    This is the graph of a circle. The equation [latex]r=4\\sin\\theta [\/latex] can be converted into rectangular coordinates by first multiplying both sides by [latex]r[\/latex]. This gives the equation [latex]{r}^{2}=4r\\sin\\theta [\/latex]. Next use the facts that [latex]{r}^{2}={x}^{2}+{y}^{2}[\/latex] and [latex]y=r\\sin\\theta [\/latex]. This gives [latex]{x}^{2}+{y}^{2}=4y[\/latex]. To put this equation into standard form, subtract [latex]4y[\/latex] from both sides of the equation and complete the square:<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill {x}^{2}+{y}^{2}-4y& =\\hfill & 0\\hfill \\\\ \\hfill {x}^{2}+\\left({y}^{2}-4y\\right)& =\\hfill & 0\\hfill \\\\ \\hfill {x}^{2}+\\left({y}^{2}-4y+4\\right)& =\\hfill & 0+4\\hfill \\\\ \\hfill {x}^{2}+{\\left(y - 2\\right)}^{2}& =\\hfill & 4.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

    This is the equation of a circle with radius 2 and center [latex]\\left(0,2\\right)[\/latex] in the rectangular coordinate system.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    Watch the following video to see the worked solution to the example above.