{"id":1958,"date":"2023-10-12T00:36:13","date_gmt":"2023-10-12T00:36:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/polar-coordinates\/"},"modified":"2024-04-29T18:47:41","modified_gmt":"2024-04-29T18:47:41","slug":"polar-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/polar-coordinates\/","title":{"raw":"Polar Coordinates","rendered":"Polar Coordinates"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Plot points using polar coordinates.<\/li>\r\n \t<li>Convert from polar coordinates to rectangular coordinates.<\/li>\r\n \t<li>Convert from rectangular coordinates to polar coordinates.<\/li>\r\n \t<li>Transform equations between polar and rectangular forms.<\/li>\r\n \t<li>Identify and graph polar equations by converting to rectangular equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\nOver 12 kilometers from port, a sailboat encounters rough weather and is blown off course by a 16-knot wind. How can the sailor indicate his location to the Coast Guard? In this section, we will investigate a method of representing location that is different from a standard coordinate grid.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165344\/CNX_Precalc_Figure_08_03_0012.jpg\" alt=\"An illustration of a boat on the polar grid.\" width=\"487\" height=\"402\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\n&nbsp;\r\n<h2>Plotting Points Using Polar Coordinates<\/h2>\r\nWhen we think about plotting points in the plane, we usually think of <strong>rectangular coordinates<\/strong> [latex]\\left(x,y\\right)[\/latex] in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to <strong>polar coordinates<\/strong>, which are points labeled [latex]\\left(r,\\theta \\right)[\/latex] and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the <strong>pole<\/strong>, or the origin of the coordinate plane.\r\n\r\nThe <strong>polar grid<\/strong> is scaled as the unit circle with the positive <em>x-<\/em>axis now viewed as the <strong>polar axis<\/strong> and the origin as the pole. The first coordinate [latex]r[\/latex] is the radius or length of the directed line segment from the pole. The angle [latex]\\theta [\/latex], measured in radians, indicates the direction of [latex]r[\/latex]. We move counterclockwise from the polar axis by an angle of [latex]\\theta [\/latex], and measure a directed line segment the length of [latex]r[\/latex] in the direction of [latex]\\theta [\/latex]. Even though we measure [latex]\\theta [\/latex] first and then [latex]r[\/latex], the polar point is written with the <em>r<\/em>-coordinate first. For example, to plot the point [latex]\\left(2,\\frac{\\pi }{4}\\right)[\/latex], we would move [latex]\\frac{\\pi }{4}[\/latex] units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in\u00a0Figure 2.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165346\/CNX_Precalc_Figure_08_03_0022.jpg\" alt=\"Polar grid with point (2, pi\/4) plotted.\" width=\"487\" height=\"398\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Plotting a Point on the Polar Grid<\/h3>\r\nPlot the point [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] on the polar grid.\r\n\r\n[reveal-answer q=\"457738\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457738\"]\r\n\r\nThe angle [latex]\\frac{\\pi }{2}[\/latex] is found by sweeping in a counterclockwise direction 90\u00b0 from the polar axis. The point is located at a length of 3 units from the pole in the [latex]\\frac{\\pi }{2}[\/latex] direction, as shown in Figure 3.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165348\/CNX_Precalc_Figure_08_03_0032.jpg\" alt=\"Polar grid with point (3, pi\/2) plotted.\" width=\"487\" height=\"369\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nPlot the point [latex]\\left(2,\\frac{\\pi }{3}\\right)[\/latex] in the <strong>polar grid<\/strong>.\r\n\r\n[reveal-answer q=\"632864\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"632864\"]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165426\/CNX_Precalc_Figure_08_03_0042.jpg\" alt=\"Polar grid with point (2, pi\/3) plotted.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Plotting a Point in the Polar Coordinate System with a Negative Component<\/h3>\r\nPlot the point [latex]\\left(-2,\\frac{\\pi }{6}\\right)[\/latex] on the polar grid.\r\n\r\n[reveal-answer q=\"158861\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"158861\"]\r\n\r\nWe know that [latex]\\frac{\\pi }{6}[\/latex] is located in the first quadrant. However, [latex]r=-2[\/latex]. We can approach plotting a point with a negative [latex]r[\/latex] in two ways:\r\n<ol>\r\n \t<li>Plot the point [latex]\\left(2,\\frac{\\pi }{6}\\right)[\/latex] by moving [latex]\\frac{\\pi }{6}[\/latex] in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant;<\/li>\r\n \t<li>Move [latex]\\frac{\\pi }{6}[\/latex] in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant.<\/li>\r\n<\/ol>\r\nSee\u00a0Figure 4(a). Compare this to the graph of the polar coordinate [latex]\\left(2,\\frac{\\pi }{6}\\right)[\/latex] shown in Figure 4(b).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165351\/CNX_Precalc_Figure_08_03_0052.jpg\" alt=\"Two polar grids. Points (2, pi\/6) and (-2, pi\/6) are plotted. They are reflections across the origin in Q1 and Q3. \" width=\"731\" height=\"403\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nPlot the points [latex]\\left(3,-\\frac{\\pi }{6}\\right)[\/latex] and [latex]\\left(2,\\frac{9\\pi }{4}\\right)[\/latex] on the same polar grid.\r\n\r\n[reveal-answer q=\"625444\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"625444\"]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165428\/CNX_Precalc_Figure_08_03_0062.jpg\" alt=\"Points (2, 9pi\/4) and (3, -pi\/6) are plotted in the polar grid.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174888[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Equivalent Representations of Points in Polar Coordinates<\/h2>\r\nUnlike Cartesian coordinates where (a, b) and (c, d) represent the same point if and only if a = c and b = d, a point can be represented by infinitely many polar coordinate pairs. We explore this notion more in the following example.\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/h3>\r\nFor each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has r &gt; 0 and the other with r &lt; 0.\r\n\r\n<strong>1. P (2, 240\u00b0)<\/strong>\r\n\r\n<strong>Solution:<\/strong>\r\n\r\nWhether we move 2 units along the polar axis and then rotate 240\u00b0 or rotate 240\u00b0 then move out 2 units from the pole, we plot P (2, 240\u00b0) below.\r\n\r\n<img class=\"alignnone size-full wp-image-1663\" src=\"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-content\/uploads\/sites\/5870\/2023\/10\/polar-1-1.png\" alt=\"\" width=\"1908\" height=\"420\" \/>\r\n\r\nWe now set about finding alternate descriptions (r, \u03b8) for the point P. Since P is 2 units from the pole, r = \u00b12. Next, we choose angles \u03b8 for each of the r values. The given representation for P is (2, 240\u00b0) so the angle \u03b8 we choose for the r= 2 case must be coterminal with 240\u00b0. (Can you see why?) One such angle is \u03b8 = \u2212120\u00b0 so one answer for this case is (2, \u2212120\u00b0). For the case r = \u22122, we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate \u03b8 = 60\u00b0 to arrive at location coterminal with 240\u00b0. Hence, our answer here is (\u22122, 60\u00b0). We check our answers by plotting them.\r\n\r\n<img class=\"alignnone size-full wp-image-1664\" src=\"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-content\/uploads\/sites\/5870\/2023\/10\/polar-2.png\" alt=\"\" width=\"2210\" height=\"438\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question sameseed=1 hide_question_numbers=1]149415[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Converting Between Polar Coordinates to Rectangular Coordinates<\/h2>\r\nWhen given a set of <strong>polar coordinates<\/strong>, we may need to convert them to <strong>rectangular coordinates<\/strong>. To do so, we can recall the relationships that exist among the variables [latex]x,y,r[\/latex], and [latex]\\theta [\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered} \\cos \\theta =\\frac{x}{r}\\to x=r\\cos \\theta \\\\ \\sin \\theta =\\frac{y}{r}\\to y=r\\sin \\theta \\end{gathered}[\/latex]<\/div>\r\nDropping a perpendicular from the point in the plane to the <em>x-<\/em>axis forms a right triangle, as illustrated in Figure 5. An easy way to remember the equations above is to think of [latex]\\cos \\theta [\/latex] as the adjacent side over the hypotenuse and [latex]\\sin \\theta [\/latex] as the opposite side over the hypotenuse.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165353\/CNX_Precalc_Figure_08_03_0072.jpg\" alt=\"Comparison between polar coordinates and rectangular coordinates. There is a right triangle plotted on the x,y axis. The sides are a horizontal line on the x-axis of length x, a vertical line extending from thex-axis to some point in quadrant 1, and a hypotenuse r extending from the origin to that same point in quadrant 1. The vertices are at the origin (0,0), some point along the x-axis at (x,0), and that point in quadrant 1. This last point is (x,y) or (r, theta), depending which system of coordinates you use.\" width=\"487\" height=\"290\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Converting from Polar Coordinates to Rectangular Coordinates<\/h3>\r\nTo convert polar coordinates [latex]\\left(r,\\theta \\right)[\/latex] to rectangular coordinates [latex]\\left(x,y\\right)[\/latex], let\r\n<p style=\"text-align: center;\">[latex]\\cos \\theta =\\frac{x}{r}\\to x=r\\cos \\theta [\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\sin \\theta =\\frac{y}{r}\\to y=r\\sin \\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given polar coordinates, convert to rectangular coordinates.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Given the polar coordinate [latex]\\left(r,\\theta \\right)[\/latex], write [latex]x=r\\cos \\theta [\/latex] and [latex]y=r\\sin \\theta [\/latex].<\/li>\r\n \t<li>Evaluate [latex]\\cos \\theta [\/latex] and [latex]\\sin \\theta [\/latex].<\/li>\r\n \t<li>Multiply [latex]\\cos \\theta [\/latex] by [latex]r[\/latex] to find the <em>x-<\/em>coordinate of the rectangular form.<\/li>\r\n \t<li>Multiply [latex]\\sin \\theta [\/latex] by [latex]r[\/latex] to find the <em>y-<\/em>coordinate of the rectangular form.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Writing Polar Coordinates as Rectangular Coordinates<\/h3>\r\nWrite the polar coordinates [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] as rectangular coordinates.\r\n\r\n[reveal-answer q=\"788608\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788608\"]\r\n\r\nUse the equivalent relationships.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;x=r\\cos \\theta \\\\ &amp;x=3\\cos \\frac{\\pi }{2}=0 \\\\ &amp;y=r\\sin \\theta \\\\ &amp;y=3\\sin \\frac{\\pi }{2}=3 \\end{align}[\/latex]<\/p>\r\nThe rectangular coordinates are [latex]\\left(0,3\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165355\/CNX_Precalc_Figure_08_03_0082.jpg\" alt=\"Illustration of (3, pi\/2) in polar coordinates and (0,3) in rectangular coordinates - they are the same point!\" width=\"975\" height=\"404\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Writing Polar Coordinates as Rectangular Coordinates<\/h3>\r\nWrite the polar coordinates [latex]\\left(-2,0\\right)[\/latex] as rectangular coordinates.\r\n\r\n[reveal-answer q=\"635694\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"635694\"]\r\n\r\nSee Figure 7. Writing the polar coordinates as rectangular, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;x=r\\cos \\theta \\\\ &amp;x=-2\\cos \\left(0\\right)=-2 \\\\ \\text{ } \\\\ &amp;y=r\\sin \\theta \\\\ &amp;y=-2\\sin \\left(0\\right)=0 \\end{align}[\/latex]<\/p>\r\nThe rectangular coordinates are also [latex]\\left(-2,0\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165358\/CNX_Precalc_Figure_08_03_0092.jpg\" alt=\"Illustration of (-2, 0) in polar coordinates and (-2,0) in rectangular coordinates - they are the same point!\" width=\"731\" height=\"375\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the polar coordinates [latex]\\left(-1,\\frac{2\\pi }{3}\\right)[\/latex] as rectangular coordinates.\r\n\r\n[reveal-answer q=\"68654\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"68654\"]\r\n\r\n[latex]\\left(x,y\\right)=\\left(\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]133878[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Converting from Rectangular Coordinates to Polar Coordinates<\/h2>\r\nTo convert <strong>rectangular coordinates<\/strong> to<strong> polar coordinates<\/strong>, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Converting from Rectangular Coordinates to Polar Coordinates<\/h3>\r\nConverting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in Figure 8.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\theta =\\frac{x}{r}\\text{ or }x=r\\cos \\theta \\\\ \\sin \\theta =\\frac{y}{r}\\text{ or }y=r\\sin \\theta \\\\ {r}^{2}={x}^{2}+{y}^{2} \\\\ \\tan \\theta =\\frac{y}{x} \\end{gathered}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165400\/CNX_Precalc_Figure_08_03_010new2.jpg\" alt=\"A right triangle with sides x, y, and r on a graph. The side x runs along the x-axis, and the point of the triangle opposite side x is the point (x, y), (r, theta). The side opposite the right angle of the triangle is labeled r. The side y is opposite the angle theta, and the vertex of angle theta is the point (0,0).\" width=\"487\" height=\"298\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Writing Rectangular Coordinates as Polar Coordinates<\/h3>\r\nConvert the rectangular coordinates [latex]\\left(3,3\\right)[\/latex] to polar coordinates.\r\n\r\n[reveal-answer q=\"532928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"532928\"]\r\n\r\nWe see that the original point [latex]\\left(3,3\\right)[\/latex] is in the first quadrant. To find [latex]\\theta [\/latex], use the formula [latex]\\tan \\theta =\\frac{y}{x}[\/latex]. This gives\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\tan \\theta =\\frac{3}{3} \\\\ &amp;\\tan \\theta =1 \\\\ &amp;{\\tan }^{-1}\\left(1\\right)=\\frac{\\pi }{4} \\end{align}[\/latex]<\/p>\r\nTo find [latex]r[\/latex], we substitute the values for [latex]x[\/latex] and [latex]y[\/latex] into the formula [latex]r=\\sqrt{{x}^{2}+{y}^{2}}[\/latex]. We know that [latex]r[\/latex] must be positive, as [latex]\\frac{\\pi }{4}[\/latex] is in the first quadrant. Thus\r\n<p style=\"text-align: center;\">[latex]\\begin{align} r&amp;=\\sqrt{{3}^{2}+{3}^{2}} \\\\ r&amp;=\\sqrt{9+9} \\\\ r&amp;=\\sqrt{18}=3\\sqrt{2} \\end{align}[\/latex]<\/p>\r\nSo, [latex]r=3\\sqrt{2}[\/latex] and [latex]\\theta \\text{=}\\frac{\\pi }{4}[\/latex], giving us the polar point [latex]\\left(3\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165403\/CNX_Precalc_Figure_08_03_0112.jpg\" alt=\"Illustration of (3rad2, pi\/4) in polar coordinates and (3,3) in rectangular coordinates - they are the same point!\" width=\"975\" height=\"375\" \/> <b>Figure 9<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nThere are other sets of polar coordinates that will be the same as our first solution. For example, the points [latex]\\left(-3\\sqrt{2},\\frac{5\\pi }{4}\\right)[\/latex] and [latex]\\left(3\\sqrt{2},-\\frac{7\\pi }{4}\\right)[\/latex] will coincide with the original solution of [latex]\\left(3\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex]. The point [latex]\\left(-3\\sqrt{2},\\frac{5\\pi }{4}\\right)[\/latex] indicates a move further counterclockwise by [latex]\\pi [\/latex], which is directly opposite [latex]\\frac{\\pi }{4}[\/latex]. The radius is expressed as [latex]-3\\sqrt{2}[\/latex]. However, the angle [latex]\\frac{5\\pi }{4}[\/latex] is located in the third quadrant and, as [latex]r[\/latex] is negative, we extend the directed line segment in the opposite direction, into the first quadrant. This is the same point as [latex]\\left(3\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex]. The point [latex]\\left(3\\sqrt{2},-\\frac{7\\pi }{4}\\right)[\/latex] is a move further clockwise by [latex]-\\frac{7\\pi }{4}[\/latex], from [latex]\\frac{\\pi }{4}[\/latex]. The radius, [latex]3\\sqrt{2}[\/latex], is the same.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<h2>Transforming Equations between Polar and Rectangular Forms<\/h2>\r\nWe can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation in polar form, graph it using a graphing calculator.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Change the <strong>MODE<\/strong> to <strong>POL<\/strong>, representing polar form.<\/li>\r\n \t<li>Press the <strong>Y= <\/strong>button to bring up a screen allowing the input of six equations: [latex]{r}_{1},{r}_{2},...,{r}_{6}[\/latex].<\/li>\r\n \t<li>Enter the polar equation, set equal to [latex]r[\/latex].<\/li>\r\n \t<li>Press <strong>GRAPH<\/strong>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Writing a Cartesian Equation in Polar Form<\/h3>\r\nWrite the Cartesian equation [latex]{x}^{2}+{y}^{2}=9[\/latex] in polar form.\r\n\r\n[reveal-answer q=\"229215\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"229215\"]\r\n\r\nThe goal is to eliminate [latex]x[\/latex] and [latex]y[\/latex] from the equation and introduce [latex]r[\/latex] and [latex]\\theta [\/latex]. Ideally, we would write the equation [latex]r[\/latex] as a function of [latex]\\theta [\/latex]. To obtain the polar form, we will use the relationships between [latex]\\left(x,y\\right)[\/latex] and [latex]\\left(r,\\theta \\right)[\/latex]. Since [latex]x=r\\cos \\theta [\/latex] and [latex]y=r\\sin \\theta [\/latex], we can substitute and solve for [latex]r[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{\\left(r\\cos \\theta \\right)}^{2}+{\\left(r\\sin \\theta \\right)}^{2}=9 \\\\ &amp;{r}^{2}{\\cos }^{2}\\theta +{r}^{2}{\\sin }^{2}\\theta =9 \\\\ &amp;{r}^{2}\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)=9 \\\\ &amp;{r}^{2}\\left(1\\right)=9&amp;&amp; {\\text{Substitute cos}}^{2}\\theta +{\\sin }^{2}\\theta =1. \\\\ &amp;r=\\pm 3&amp;&amp; \\text{Use the square root property}. \\end{align}[\/latex]<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165405\/CNX_Precalc_Figure_08_03_0162.jpg\" alt=\"Plotting a circle of radius 3 with center at the origin in polar and rectangular coordinates. It is the same in both systems.\" width=\"731\" height=\"360\" \/> <b>Figure 10.<\/b> (a) Cartesian form [latex]{x}^{2}+{y}^{2}=9[\/latex] (b) Polar form [latex]r=3[\/latex][\/caption]Thus, [latex]{x}^{2}+{y}^{2}=9,r=3[\/latex], and [latex]r=-3[\/latex] should generate the same graph.To graph a circle in rectangular form, we must first solve for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} {x}^{2}+{y}^{2}=9 \\\\ {y}^{2}=9-{x}^{2} \\\\ y=\\pm \\sqrt{9-{x}^{2}}\\end{gathered}[\/latex]<\/p>\r\nNote that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form [latex]{Y}_{1}=\\sqrt{9-{x}^{2}}[\/latex] and [latex]{Y}_{2}=-\\sqrt{9-{x}^{2}}[\/latex]. Press <strong>GRAPH.<\/strong>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Rewriting a Cartesian Equation as a Polar Equation<\/h3>\r\nRewrite the <strong>Cartesian equation<\/strong> [latex]{x}^{2}+{y}^{2}=6y[\/latex] as a polar equation.\r\n\r\n[reveal-answer q=\"483629\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"483629\"]\r\n\r\nThis equation appears similar to the previous example, but it requires different steps to convert the equation.\r\n\r\nWe can still follow the same procedures we have already learned and make the following substitutions:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{r}^{2}=6y&amp;&amp; \\text{Use }{x}^{2}+{y}^{2}={r}^{2}. \\\\ &amp;{r}^{2}=6r\\sin \\theta&amp;&amp; \\text{Substitute}y=r\\sin \\theta . \\\\ &amp;{r}^{2}-6r\\sin \\theta =0&amp;&amp; \\text{Set equal to 0}. \\\\ &amp;r\\left(r - 6\\sin \\theta \\right)=0&amp;&amp; \\text{Factor and solve}. \\\\ &amp;r=0&amp;&amp; \\text{We reject }r=0,\\text{as it only represents one point, }\\left(0,0\\right). \\\\ &amp;\\text{or }r=6\\sin \\theta \\end{align}[\/latex]<\/p>\r\nTherefore, the equations [latex]{x}^{2}+{y}^{2}=6y[\/latex] and [latex]r=6\\sin \\theta [\/latex] should give us the same graph.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165407\/CNX_Precalc_Figure_08_03_0122.jpg\" alt=\"Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are circles.\" width=\"975\" height=\"328\" \/> <b>Figure 11.<\/b> (a) Cartesian form [latex]{x}^{2}+{y}^{2}=6y[\/latex] (b) polar form [latex]r=6\\sin \\theta [\/latex][\/caption]The Cartesian or <strong>rectangular equation<\/strong> is plotted on the rectangular grid, and the <strong>polar equation<\/strong> is plotted on the polar grid. Clearly, the graphs are identical.[\/hidden-answer]<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Rewriting a Cartesian Equation in Polar Form<\/h3>\r\nRewrite the Cartesian equation [latex]y=3x+2[\/latex] as a polar equation.\r\n\r\n[reveal-answer q=\"530698\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530698\"]\r\n\r\nWe will use the relationships [latex]x=r\\cos \\theta [\/latex] and [latex]y=r\\sin \\theta [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=3x+2 \\\\ &amp;r\\sin \\theta =3r\\cos \\theta +2 \\\\ &amp;r\\sin \\theta -3r\\cos \\theta =2 \\\\ &amp;r\\left(\\sin \\theta -3\\cos \\theta \\right)=2&amp;&amp; \\text{Isolate }r. \\\\ &amp;r=\\frac{2}{\\sin \\theta -3\\cos \\theta }&amp;&amp; \\text{Solve for }r. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]<span style=\"text-align: center; font-size: 0.9em;\">\u00a0<\/span>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nRewrite the Cartesian equation [latex]{y}^{2}=3-{x}^{2}[\/latex] in polar form.\r\n\r\n[reveal-answer q=\"187427\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"187427\"]\r\n\r\n[latex]r=\\sqrt{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149349[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>\u00a0Identify and Graph Polar Equations by Converting to Rectangular Equations<\/h2>\r\nWe have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Graphing a Polar Equation by Converting to a Rectangular Equation<\/h3>\r\nCovert the polar equation [latex]r=2\\sec \\theta [\/latex] to a rectangular equation, and draw its corresponding graph.\r\n\r\n[reveal-answer q=\"584712\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"584712\"]\r\n\r\nThe conversion is\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}r=2\\sec \\theta \\\\ r=\\frac{2}{\\cos \\theta } \\\\ r\\cos \\theta =2 \\\\ x=2 \\end{gathered}[\/latex]<\/p>\r\nNotice that the equation [latex]r=2\\sec \\theta [\/latex] drawn on the polar grid is clearly the same as the vertical line [latex]x=2[\/latex] drawn on the rectangular grid. Just as [latex]x=c[\/latex] is the standard form for a vertical line in rectangular form, [latex]r=c\\sec \\theta [\/latex] is the standard form for a vertical line in polar form.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165410\/CNX_Precalc_Figure_08_03_0132.jpg\" alt=\"Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are lines.\" width=\"731\" height=\"408\" \/> <b>Figure 12.<\/b> (a) Polar grid (b) Rectangular coordinate system[\/caption]\r\n\r\nA similar discussion would demonstrate that the graph of the function [latex]r=2\\csc \\theta [\/latex] will be the horizontal line [latex]y=2[\/latex]. In fact, [latex]r=c\\csc \\theta [\/latex] is the standard form for a horizontal line in polar form, corresponding to the rectangular form [latex]y=c[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Rewriting a Polar Equation in Cartesian Form<\/h3>\r\nRewrite the polar equation [latex]r=\\frac{3}{1 - 2\\cos \\theta }[\/latex] as a Cartesian equation.\r\n\r\n[reveal-answer q=\"165840\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"165840\"]\r\n\r\nThe goal is to eliminate [latex]\\theta [\/latex] and [latex]r[\/latex], and introduce [latex]x[\/latex] and [latex]y[\/latex]. We clear the fraction, and then use substitution. In order to replace [latex]r[\/latex] with [latex]x[\/latex] and [latex]y[\/latex], we must use the expression [latex]{x}^{2}+{y}^{2}={r}^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;r=\\frac{3}{1 - 2\\cos \\theta } \\\\ &amp;r\\left(1 - 2\\cos \\theta \\right)=3 \\\\ &amp;r\\left(1 - 2\\left(\\frac{x}{r}\\right)\\right)=3&amp;&amp; \\text{Use }\\cos \\theta =\\frac{x}{r}\\text{ to eliminate }\\theta . \\\\ &amp;r - 2x=3 \\\\ &amp;r=3+2x&amp;&amp; \\text{Isolate }r. \\\\ &amp;{r}^{2}={\\left(3+2x\\right)}^{2}&amp;&amp; \\text{Square both sides}. \\\\ &amp;{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2}&amp;&amp; \\text{Use }{x}^{2}+{y}^{2}={r}^{2}. \\end{align}[\/latex]<\/p>\r\nThe Cartesian equation is [latex]{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2}[\/latex]. However, to graph it, especially using a graphing calculator or computer program, we want to isolate [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2} \\\\ {y}^{2}={\\left(3+2x\\right)}^{2}-{x}^{2} \\\\ y=\\pm \\sqrt{{\\left(3+2x\\right)}^{2}-{x}^{2}} \\end{gathered}[\/latex]<\/p>\r\nWhen our entire equation has been changed from [latex]r[\/latex] and [latex]\\theta [\/latex] to [latex]x[\/latex] and [latex]y[\/latex], we can stop, unless asked to solve for [latex]y[\/latex] or simplify.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165413\/CNX_Precalc_Figure_08_03_0152.jpg\" alt=\"Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are hyperbolas.\" width=\"975\" height=\"481\" \/> <b>Figure 13<\/b>[\/caption]\r\n\r\nThe \"hour-glass\" shape of the graph is called a <em>hyperbola<\/em>. Hyperbolas have many interesting geometric features and applications, which we will investigate further in <a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/introduction-to-analytic-geometry\/\" target=\"_blank\" rel=\"noopener\">Analytic Geometry<\/a>.\r\n<h4>Analysis of the Solution<\/h4>\r\nIn this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola\u2019s standard form. To do this, we can start with the initial equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2} \\\\ &amp;{x}^{2}+{y}^{2}-{\\left(3+2x\\right)}^{2}=0 \\\\ &amp;{x}^{2}+{y}^{2}-\\left(9+12x+4{x}^{2}\\right)=0 \\\\ &amp;{x}^{2}+{y}^{2}-9 - 12x - 4{x}^{2}=0 \\\\ &amp;-3{x}^{2}-12x+{y}^{2}=9&amp;&amp; \\text{Multiply through by }-1. \\\\ &amp;3{x}^{2}+12x-{y}^{2}=-9 \\\\ &amp;3\\left({x}^{2}+4x+\\right)-{y}^{2}=-9&amp;&amp; \\text{Organize terms to complete the square for}x. \\\\ &amp;3\\left({x}^{2}+4x+4\\right)-{y}^{2}=-9+12 \\\\ &amp;3{\\left(x+2\\right)}^{2}-{y}^{2}=3 \\\\ &amp;{\\left(x+2\\right)}^{2}-\\frac{{y}^{2}}{3}=1 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nRewrite the polar equation [latex]r=2\\sin \\theta [\/latex] in Cartesian form.\r\n\r\n[reveal-answer q=\"407542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"407542\"]\r\n\r\n[latex]{x}^{2}+{y}^{2}=2y[\/latex] or, in the standard form for a circle, [latex]{x}^{2}+{\\left(y - 1\\right)}^{2}=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173797[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Rewriting a Polar Equation in Cartesian Form<\/h3>\r\nRewrite the polar equation [latex]r=\\sin \\left(2\\theta \\right)[\/latex] in Cartesian form.\r\n\r\n[reveal-answer q=\"810563\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"810563\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;r=\\sin \\left(2\\theta \\right)&amp;&amp; \\text{Use the double angle identity for sine}. \\\\ &amp;r=2\\sin \\theta \\cos \\theta&amp;&amp; \\text{Use }\\cos \\theta =\\frac{x}{r}\\text{ and }\\sin \\theta =\\frac{y}{r}. \\\\ &amp;r=2\\left(\\frac{x}{r}\\right)\\left(\\frac{y}{r}\\right)&amp;&amp; \\text{Simplify}. \\\\ &amp;r=\\frac{2xy}{{r}^{2}}&amp;&amp; \\text{ Multiply both sides by }{r}^{2}. \\\\ &amp;{r}^{3}=2xy \\\\ &amp;{\\left(\\sqrt{{x}^{2}+{y}^{2}}\\right)}^{3}=2xy&amp;&amp; \\text{As }{x}^{2}+{y}^{2}={r}^{2},r=\\sqrt{{x}^{2}+{y}^{2}}. \\end{align}[\/latex]<\/p>\r\nThis equation can also be written as\r\n<p style=\"text-align: center;\">[latex]{\\left({x}^{2}+{y}^{2}\\right)}^{\\frac{3}{2}}=2xy\\text{ or }{x}^{2}+{y}^{2}={\\left(2xy\\right)}^{\\frac{2}{3}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table id=\"eip-id3984733\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Conversion formulas<\/td>\r\n<td>[latex]\\begin{align}&amp; \\cos \\theta =\\frac{x}{r}\\to x=r\\cos \\theta \\\\ &amp; \\sin \\theta =\\frac{y}{r}\\to y=r\\sin \\theta \\\\ &amp; {r}^{2}={x}^{2}+{y}^{2} \\\\ &amp; \\tan \\theta =\\frac{y}{x} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>The polar grid is represented as a series of concentric circles radiating out from the pole, or origin.<\/li>\r\n \t<li>To plot a point in the form [latex]\\left(r,\\theta \\right),\\theta &gt;0[\/latex], move in a counterclockwise direction from the polar axis by an angle of [latex]\\theta [\/latex], and then extend a directed line segment from the pole the length of [latex]r[\/latex] in the direction of [latex]\\theta [\/latex]. If [latex]\\theta [\/latex] is negative, move in a clockwise direction, and extend a directed line segment the length of [latex]r[\/latex] in the direction of [latex]\\theta [\/latex].<\/li>\r\n \t<li>If [latex]r[\/latex] is negative, extend the directed line segment in the opposite direction of [latex]\\theta [\/latex].<\/li>\r\n \t<li>To convert from polar coordinates to rectangular coordinates, use the formulas [latex]x=r\\cos \\theta [\/latex] and [latex]y=r\\sin \\theta [\/latex].<\/li>\r\n \t<li>To convert from rectangular coordinates to polar coordinates, use one or more of the formulas: [latex]\\cos \\theta =\\frac{x}{r},\\sin \\theta =\\frac{y}{r},\\tan \\theta =\\frac{y}{x}[\/latex], and [latex]r=\\sqrt{{x}^{2}+{y}^{2}}[\/latex].<\/li>\r\n \t<li>Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations.<\/li>\r\n \t<li>Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135439812\" class=\"definition\">\r\n \t<dt>polar axis<\/dt>\r\n \t<dd id=\"fs-id1165134391590\">on the polar grid, the equivalent of the positive <em>x-<\/em>axis on the rectangular grid<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134391599\" class=\"definition\">\r\n \t<dt>polar coordinates<\/dt>\r\n \t<dd id=\"fs-id1165134041274\">on the polar grid, the coordinates of a point labeled [latex]\\left(r,\\theta \\right)[\/latex], where [latex]\\theta [\/latex] indicates the angle of rotation from the polar axis and [latex]r[\/latex] represents the radius, or the distance of the point from the pole in the direction of [latex]\\theta [\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134149787\" class=\"definition\">\r\n \t<dt>pole<\/dt>\r\n \t<dd id=\"fs-id1165133307606\">the origin of the polar grid<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Plot points using polar coordinates.<\/li>\n<li>Convert from polar coordinates to rectangular coordinates.<\/li>\n<li>Convert from rectangular coordinates to polar coordinates.<\/li>\n<li>Transform equations between polar and rectangular forms.<\/li>\n<li>Identify and graph polar equations by converting to rectangular equations.<\/li>\n<\/ul>\n<\/div>\n<p>Over 12 kilometers from port, a sailboat encounters rough weather and is blown off course by a 16-knot wind. How can the sailor indicate his location to the Coast Guard? In this section, we will investigate a method of representing location that is different from a standard coordinate grid.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165344\/CNX_Precalc_Figure_08_03_0012.jpg\" alt=\"An illustration of a boat on the polar grid.\" width=\"487\" height=\"402\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Plotting Points Using Polar Coordinates<\/h2>\n<p>When we think about plotting points in the plane, we usually think of <strong>rectangular coordinates<\/strong> [latex]\\left(x,y\\right)[\/latex] in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to <strong>polar coordinates<\/strong>, which are points labeled [latex]\\left(r,\\theta \\right)[\/latex] and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the <strong>pole<\/strong>, or the origin of the coordinate plane.<\/p>\n<p>The <strong>polar grid<\/strong> is scaled as the unit circle with the positive <em>x-<\/em>axis now viewed as the <strong>polar axis<\/strong> and the origin as the pole. The first coordinate [latex]r[\/latex] is the radius or length of the directed line segment from the pole. The angle [latex]\\theta[\/latex], measured in radians, indicates the direction of [latex]r[\/latex]. We move counterclockwise from the polar axis by an angle of [latex]\\theta[\/latex], and measure a directed line segment the length of [latex]r[\/latex] in the direction of [latex]\\theta[\/latex]. Even though we measure [latex]\\theta[\/latex] first and then [latex]r[\/latex], the polar point is written with the <em>r<\/em>-coordinate first. For example, to plot the point [latex]\\left(2,\\frac{\\pi }{4}\\right)[\/latex], we would move [latex]\\frac{\\pi }{4}[\/latex] units in the counterclockwise direction and then a length of 2 from the pole. This point is plotted on the grid in\u00a0Figure 2.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165346\/CNX_Precalc_Figure_08_03_0022.jpg\" alt=\"Polar grid with point (2, pi\/4) plotted.\" width=\"487\" height=\"398\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Plotting a Point on the Polar Grid<\/h3>\n<p>Plot the point [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] on the polar grid.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q457738\">Show Solution<\/span><\/p>\n<div id=\"q457738\" class=\"hidden-answer\" style=\"display: none\">\n<p>The angle [latex]\\frac{\\pi }{2}[\/latex] is found by sweeping in a counterclockwise direction 90\u00b0 from the polar axis. The point is located at a length of 3 units from the pole in the [latex]\\frac{\\pi }{2}[\/latex] direction, as shown in Figure 3.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165348\/CNX_Precalc_Figure_08_03_0032.jpg\" alt=\"Polar grid with point (3, pi\/2) plotted.\" width=\"487\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Plot the point [latex]\\left(2,\\frac{\\pi }{3}\\right)[\/latex] in the <strong>polar grid<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q632864\">Show Solution<\/span><\/p>\n<div id=\"q632864\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165426\/CNX_Precalc_Figure_08_03_0042.jpg\" alt=\"Polar grid with point (2, pi\/3) plotted.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Plotting a Point in the Polar Coordinate System with a Negative Component<\/h3>\n<p>Plot the point [latex]\\left(-2,\\frac{\\pi }{6}\\right)[\/latex] on the polar grid.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q158861\">Show Solution<\/span><\/p>\n<div id=\"q158861\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know that [latex]\\frac{\\pi }{6}[\/latex] is located in the first quadrant. However, [latex]r=-2[\/latex]. We can approach plotting a point with a negative [latex]r[\/latex] in two ways:<\/p>\n<ol>\n<li>Plot the point [latex]\\left(2,\\frac{\\pi }{6}\\right)[\/latex] by moving [latex]\\frac{\\pi }{6}[\/latex] in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant;<\/li>\n<li>Move [latex]\\frac{\\pi }{6}[\/latex] in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant.<\/li>\n<\/ol>\n<p>See\u00a0Figure 4(a). Compare this to the graph of the polar coordinate [latex]\\left(2,\\frac{\\pi }{6}\\right)[\/latex] shown in Figure 4(b).<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165351\/CNX_Precalc_Figure_08_03_0052.jpg\" alt=\"Two polar grids. Points (2, pi\/6) and (-2, pi\/6) are plotted. They are reflections across the origin in Q1 and Q3.\" width=\"731\" height=\"403\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Plot the points [latex]\\left(3,-\\frac{\\pi }{6}\\right)[\/latex] and [latex]\\left(2,\\frac{9\\pi }{4}\\right)[\/latex] on the same polar grid.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q625444\">Show Solution<\/span><\/p>\n<div id=\"q625444\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165428\/CNX_Precalc_Figure_08_03_0062.jpg\" alt=\"Points (2, 9pi\/4) and (3, -pi\/6) are plotted in the polar grid.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174888\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174888&theme=oea&iframe_resize_id=ohm174888\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Equivalent Representations of Points in Polar Coordinates<\/h2>\n<p>Unlike Cartesian coordinates where (a, b) and (c, d) represent the same point if and only if a = c and b = d, a point can be represented by infinitely many polar coordinate pairs. We explore this notion more in the following example.<\/p>\n<div class=\"textbox examples\">\n<h3>Examples<\/h3>\n<p>For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has r &gt; 0 and the other with r &lt; 0.<\/p>\n<p><strong>1. P (2, 240\u00b0)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Whether we move 2 units along the polar axis and then rotate 240\u00b0 or rotate 240\u00b0 then move out 2 units from the pole, we plot P (2, 240\u00b0) below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1663\" src=\"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-content\/uploads\/sites\/5870\/2023\/10\/polar-1-1.png\" alt=\"\" width=\"1908\" height=\"420\" \/><\/p>\n<p>We now set about finding alternate descriptions (r, \u03b8) for the point P. Since P is 2 units from the pole, r = \u00b12. Next, we choose angles \u03b8 for each of the r values. The given representation for P is (2, 240\u00b0) so the angle \u03b8 we choose for the r= 2 case must be coterminal with 240\u00b0. (Can you see why?) One such angle is \u03b8 = \u2212120\u00b0 so one answer for this case is (2, \u2212120\u00b0). For the case r = \u22122, we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate \u03b8 = 60\u00b0 to arrive at location coterminal with 240\u00b0. Hence, our answer here is (\u22122, 60\u00b0). We check our answers by plotting them.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1664\" src=\"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-content\/uploads\/sites\/5870\/2023\/10\/polar-2.png\" alt=\"\" width=\"2210\" height=\"438\" \/><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149415\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149415&theme=oea&iframe_resize_id=ohm149415&sameseed=1\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Converting Between Polar Coordinates to Rectangular Coordinates<\/h2>\n<p>When given a set of <strong>polar coordinates<\/strong>, we may need to convert them to <strong>rectangular coordinates<\/strong>. To do so, we can recall the relationships that exist among the variables [latex]x,y,r[\/latex], and [latex]\\theta[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered} \\cos \\theta =\\frac{x}{r}\\to x=r\\cos \\theta \\\\ \\sin \\theta =\\frac{y}{r}\\to y=r\\sin \\theta \\end{gathered}[\/latex]<\/div>\n<p>Dropping a perpendicular from the point in the plane to the <em>x-<\/em>axis forms a right triangle, as illustrated in Figure 5. An easy way to remember the equations above is to think of [latex]\\cos \\theta[\/latex] as the adjacent side over the hypotenuse and [latex]\\sin \\theta[\/latex] as the opposite side over the hypotenuse.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165353\/CNX_Precalc_Figure_08_03_0072.jpg\" alt=\"Comparison between polar coordinates and rectangular coordinates. There is a right triangle plotted on the x,y axis. The sides are a horizontal line on the x-axis of length x, a vertical line extending from thex-axis to some point in quadrant 1, and a hypotenuse r extending from the origin to that same point in quadrant 1. The vertices are at the origin (0,0), some point along the x-axis at (x,0), and that point in quadrant 1. This last point is (x,y) or (r, theta), depending which system of coordinates you use.\" width=\"487\" height=\"290\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Converting from Polar Coordinates to Rectangular Coordinates<\/h3>\n<p>To convert polar coordinates [latex]\\left(r,\\theta \\right)[\/latex] to rectangular coordinates [latex]\\left(x,y\\right)[\/latex], let<\/p>\n<p style=\"text-align: center;\">[latex]\\cos \\theta =\\frac{x}{r}\\to x=r\\cos \\theta[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\sin \\theta =\\frac{y}{r}\\to y=r\\sin \\theta[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given polar coordinates, convert to rectangular coordinates.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Given the polar coordinate [latex]\\left(r,\\theta \\right)[\/latex], write [latex]x=r\\cos \\theta[\/latex] and [latex]y=r\\sin \\theta[\/latex].<\/li>\n<li>Evaluate [latex]\\cos \\theta[\/latex] and [latex]\\sin \\theta[\/latex].<\/li>\n<li>Multiply [latex]\\cos \\theta[\/latex] by [latex]r[\/latex] to find the <em>x-<\/em>coordinate of the rectangular form.<\/li>\n<li>Multiply [latex]\\sin \\theta[\/latex] by [latex]r[\/latex] to find the <em>y-<\/em>coordinate of the rectangular form.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Writing Polar Coordinates as Rectangular Coordinates<\/h3>\n<p>Write the polar coordinates [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] as rectangular coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788608\">Show Solution<\/span><\/p>\n<div id=\"q788608\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the equivalent relationships.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &x=r\\cos \\theta \\\\ &x=3\\cos \\frac{\\pi }{2}=0 \\\\ &y=r\\sin \\theta \\\\ &y=3\\sin \\frac{\\pi }{2}=3 \\end{align}[\/latex]<\/p>\n<p>The rectangular coordinates are [latex]\\left(0,3\\right)[\/latex].<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165355\/CNX_Precalc_Figure_08_03_0082.jpg\" alt=\"Illustration of (3, pi\/2) in polar coordinates and (0,3) in rectangular coordinates - they are the same point!\" width=\"975\" height=\"404\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Writing Polar Coordinates as Rectangular Coordinates<\/h3>\n<p>Write the polar coordinates [latex]\\left(-2,0\\right)[\/latex] as rectangular coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q635694\">Show Solution<\/span><\/p>\n<div id=\"q635694\" class=\"hidden-answer\" style=\"display: none\">\n<p>See Figure 7. Writing the polar coordinates as rectangular, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&x=r\\cos \\theta \\\\ &x=-2\\cos \\left(0\\right)=-2 \\\\ \\text{ } \\\\ &y=r\\sin \\theta \\\\ &y=-2\\sin \\left(0\\right)=0 \\end{align}[\/latex]<\/p>\n<p>The rectangular coordinates are also [latex]\\left(-2,0\\right)[\/latex].<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165358\/CNX_Precalc_Figure_08_03_0092.jpg\" alt=\"Illustration of (-2, 0) in polar coordinates and (-2,0) in rectangular coordinates - they are the same point!\" width=\"731\" height=\"375\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write the polar coordinates [latex]\\left(-1,\\frac{2\\pi }{3}\\right)[\/latex] as rectangular coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q68654\">Show Solution<\/span><\/p>\n<div id=\"q68654\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(x,y\\right)=\\left(\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm133878\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=133878&theme=oea&iframe_resize_id=ohm133878\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Converting from Rectangular Coordinates to Polar Coordinates<\/h2>\n<p>To convert <strong>rectangular coordinates<\/strong> to<strong> polar coordinates<\/strong>, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Converting from Rectangular Coordinates to Polar Coordinates<\/h3>\n<p>Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in Figure 8.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\theta =\\frac{x}{r}\\text{ or }x=r\\cos \\theta \\\\ \\sin \\theta =\\frac{y}{r}\\text{ or }y=r\\sin \\theta \\\\ {r}^{2}={x}^{2}+{y}^{2} \\\\ \\tan \\theta =\\frac{y}{x} \\end{gathered}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165400\/CNX_Precalc_Figure_08_03_010new2.jpg\" alt=\"A right triangle with sides x, y, and r on a graph. The side x runs along the x-axis, and the point of the triangle opposite side x is the point (x, y), (r, theta). The side opposite the right angle of the triangle is labeled r. The side y is opposite the angle theta, and the vertex of angle theta is the point (0,0).\" width=\"487\" height=\"298\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Writing Rectangular Coordinates as Polar Coordinates<\/h3>\n<p>Convert the rectangular coordinates [latex]\\left(3,3\\right)[\/latex] to polar coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q532928\">Show Solution<\/span><\/p>\n<div id=\"q532928\" class=\"hidden-answer\" style=\"display: none\">\n<p>We see that the original point [latex]\\left(3,3\\right)[\/latex] is in the first quadrant. To find [latex]\\theta[\/latex], use the formula [latex]\\tan \\theta =\\frac{y}{x}[\/latex]. This gives<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\tan \\theta =\\frac{3}{3} \\\\ &\\tan \\theta =1 \\\\ &{\\tan }^{-1}\\left(1\\right)=\\frac{\\pi }{4} \\end{align}[\/latex]<\/p>\n<p>To find [latex]r[\/latex], we substitute the values for [latex]x[\/latex] and [latex]y[\/latex] into the formula [latex]r=\\sqrt{{x}^{2}+{y}^{2}}[\/latex]. We know that [latex]r[\/latex] must be positive, as [latex]\\frac{\\pi }{4}[\/latex] is in the first quadrant. Thus<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} r&=\\sqrt{{3}^{2}+{3}^{2}} \\\\ r&=\\sqrt{9+9} \\\\ r&=\\sqrt{18}=3\\sqrt{2} \\end{align}[\/latex]<\/p>\n<p>So, [latex]r=3\\sqrt{2}[\/latex] and [latex]\\theta \\text{=}\\frac{\\pi }{4}[\/latex], giving us the polar point [latex]\\left(3\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex].<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165403\/CNX_Precalc_Figure_08_03_0112.jpg\" alt=\"Illustration of (3rad2, pi\/4) in polar coordinates and (3,3) in rectangular coordinates - they are the same point!\" width=\"975\" height=\"375\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>There are other sets of polar coordinates that will be the same as our first solution. For example, the points [latex]\\left(-3\\sqrt{2},\\frac{5\\pi }{4}\\right)[\/latex] and [latex]\\left(3\\sqrt{2},-\\frac{7\\pi }{4}\\right)[\/latex] will coincide with the original solution of [latex]\\left(3\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex]. The point [latex]\\left(-3\\sqrt{2},\\frac{5\\pi }{4}\\right)[\/latex] indicates a move further counterclockwise by [latex]\\pi[\/latex], which is directly opposite [latex]\\frac{\\pi }{4}[\/latex]. The radius is expressed as [latex]-3\\sqrt{2}[\/latex]. However, the angle [latex]\\frac{5\\pi }{4}[\/latex] is located in the third quadrant and, as [latex]r[\/latex] is negative, we extend the directed line segment in the opposite direction, into the first quadrant. This is the same point as [latex]\\left(3\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex]. The point [latex]\\left(3\\sqrt{2},-\\frac{7\\pi }{4}\\right)[\/latex] is a move further clockwise by [latex]-\\frac{7\\pi }{4}[\/latex], from [latex]\\frac{\\pi }{4}[\/latex]. The radius, [latex]3\\sqrt{2}[\/latex], is the same.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<h2>Transforming Equations between Polar and Rectangular Forms<\/h2>\n<p>We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an equation in polar form, graph it using a graphing calculator.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Change the <strong>MODE<\/strong> to <strong>POL<\/strong>, representing polar form.<\/li>\n<li>Press the <strong>Y= <\/strong>button to bring up a screen allowing the input of six equations: [latex]{r}_{1},{r}_{2},...,{r}_{6}[\/latex].<\/li>\n<li>Enter the polar equation, set equal to [latex]r[\/latex].<\/li>\n<li>Press <strong>GRAPH<\/strong>.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Writing a Cartesian Equation in Polar Form<\/h3>\n<p>Write the Cartesian equation [latex]{x}^{2}+{y}^{2}=9[\/latex] in polar form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q229215\">Show Solution<\/span><\/p>\n<div id=\"q229215\" class=\"hidden-answer\" style=\"display: none\">\n<p>The goal is to eliminate [latex]x[\/latex] and [latex]y[\/latex] from the equation and introduce [latex]r[\/latex] and [latex]\\theta[\/latex]. Ideally, we would write the equation [latex]r[\/latex] as a function of [latex]\\theta[\/latex]. To obtain the polar form, we will use the relationships between [latex]\\left(x,y\\right)[\/latex] and [latex]\\left(r,\\theta \\right)[\/latex]. Since [latex]x=r\\cos \\theta[\/latex] and [latex]y=r\\sin \\theta[\/latex], we can substitute and solve for [latex]r[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{\\left(r\\cos \\theta \\right)}^{2}+{\\left(r\\sin \\theta \\right)}^{2}=9 \\\\ &{r}^{2}{\\cos }^{2}\\theta +{r}^{2}{\\sin }^{2}\\theta =9 \\\\ &{r}^{2}\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)=9 \\\\ &{r}^{2}\\left(1\\right)=9&& {\\text{Substitute cos}}^{2}\\theta +{\\sin }^{2}\\theta =1. \\\\ &r=\\pm 3&& \\text{Use the square root property}. \\end{align}[\/latex]<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165405\/CNX_Precalc_Figure_08_03_0162.jpg\" alt=\"Plotting a circle of radius 3 with center at the origin in polar and rectangular coordinates. It is the same in both systems.\" width=\"731\" height=\"360\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10.<\/b> (a) Cartesian form [latex]{x}^{2}+{y}^{2}=9[\/latex] (b) Polar form [latex]r=3[\/latex]<\/p>\n<\/div>\n<p>Thus, [latex]{x}^{2}+{y}^{2}=9,r=3[\/latex], and [latex]r=-3[\/latex] should generate the same graph.To graph a circle in rectangular form, we must first solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} {x}^{2}+{y}^{2}=9 \\\\ {y}^{2}=9-{x}^{2} \\\\ y=\\pm \\sqrt{9-{x}^{2}}\\end{gathered}[\/latex]<\/p>\n<p>Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form [latex]{Y}_{1}=\\sqrt{9-{x}^{2}}[\/latex] and [latex]{Y}_{2}=-\\sqrt{9-{x}^{2}}[\/latex]. Press <strong>GRAPH.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Rewriting a Cartesian Equation as a Polar Equation<\/h3>\n<p>Rewrite the <strong>Cartesian equation<\/strong> [latex]{x}^{2}+{y}^{2}=6y[\/latex] as a polar equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q483629\">Show Solution<\/span><\/p>\n<div id=\"q483629\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation appears similar to the previous example, but it requires different steps to convert the equation.<\/p>\n<p>We can still follow the same procedures we have already learned and make the following substitutions:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{r}^{2}=6y&& \\text{Use }{x}^{2}+{y}^{2}={r}^{2}. \\\\ &{r}^{2}=6r\\sin \\theta&& \\text{Substitute}y=r\\sin \\theta . \\\\ &{r}^{2}-6r\\sin \\theta =0&& \\text{Set equal to 0}. \\\\ &r\\left(r - 6\\sin \\theta \\right)=0&& \\text{Factor and solve}. \\\\ &r=0&& \\text{We reject }r=0,\\text{as it only represents one point, }\\left(0,0\\right). \\\\ &\\text{or }r=6\\sin \\theta \\end{align}[\/latex]<\/p>\n<p>Therefore, the equations [latex]{x}^{2}+{y}^{2}=6y[\/latex] and [latex]r=6\\sin \\theta[\/latex] should give us the same graph.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165407\/CNX_Precalc_Figure_08_03_0122.jpg\" alt=\"Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are circles.\" width=\"975\" height=\"328\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11.<\/b> (a) Cartesian form [latex]{x}^{2}+{y}^{2}=6y[\/latex] (b) polar form [latex]r=6\\sin \\theta [\/latex]<\/p>\n<\/div>\n<p>The Cartesian or <strong>rectangular equation<\/strong> is plotted on the rectangular grid, and the <strong>polar equation<\/strong> is plotted on the polar grid. Clearly, the graphs are identical.<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Rewriting a Cartesian Equation in Polar Form<\/h3>\n<p>Rewrite the Cartesian equation [latex]y=3x+2[\/latex] as a polar equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530698\">Show Solution<\/span><\/p>\n<div id=\"q530698\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will use the relationships [latex]x=r\\cos \\theta[\/latex] and [latex]y=r\\sin \\theta[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=3x+2 \\\\ &r\\sin \\theta =3r\\cos \\theta +2 \\\\ &r\\sin \\theta -3r\\cos \\theta =2 \\\\ &r\\left(\\sin \\theta -3\\cos \\theta \\right)=2&& \\text{Isolate }r. \\\\ &r=\\frac{2}{\\sin \\theta -3\\cos \\theta }&& \\text{Solve for }r. \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><span style=\"text-align: center; font-size: 0.9em;\">\u00a0<\/span><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Rewrite the Cartesian equation [latex]{y}^{2}=3-{x}^{2}[\/latex] in polar form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q187427\">Show Solution<\/span><\/p>\n<div id=\"q187427\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]r=\\sqrt{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149349\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149349&theme=oea&iframe_resize_id=ohm149349\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<h2>\u00a0Identify and Graph Polar Equations by Converting to Rectangular Equations<\/h2>\n<p>We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 9: Graphing a Polar Equation by Converting to a Rectangular Equation<\/h3>\n<p>Covert the polar equation [latex]r=2\\sec \\theta[\/latex] to a rectangular equation, and draw its corresponding graph.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q584712\">Show Solution<\/span><\/p>\n<div id=\"q584712\" class=\"hidden-answer\" style=\"display: none\">\n<p>The conversion is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}r=2\\sec \\theta \\\\ r=\\frac{2}{\\cos \\theta } \\\\ r\\cos \\theta =2 \\\\ x=2 \\end{gathered}[\/latex]<\/p>\n<p>Notice that the equation [latex]r=2\\sec \\theta[\/latex] drawn on the polar grid is clearly the same as the vertical line [latex]x=2[\/latex] drawn on the rectangular grid. Just as [latex]x=c[\/latex] is the standard form for a vertical line in rectangular form, [latex]r=c\\sec \\theta[\/latex] is the standard form for a vertical line in polar form.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165410\/CNX_Precalc_Figure_08_03_0132.jpg\" alt=\"Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are lines.\" width=\"731\" height=\"408\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12.<\/b> (a) Polar grid (b) Rectangular coordinate system<\/p>\n<\/div>\n<p>A similar discussion would demonstrate that the graph of the function [latex]r=2\\csc \\theta[\/latex] will be the horizontal line [latex]y=2[\/latex]. In fact, [latex]r=c\\csc \\theta[\/latex] is the standard form for a horizontal line in polar form, corresponding to the rectangular form [latex]y=c[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 10: Rewriting a Polar Equation in Cartesian Form<\/h3>\n<p>Rewrite the polar equation [latex]r=\\frac{3}{1 - 2\\cos \\theta }[\/latex] as a Cartesian equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q165840\">Show Solution<\/span><\/p>\n<div id=\"q165840\" class=\"hidden-answer\" style=\"display: none\">\n<p>The goal is to eliminate [latex]\\theta[\/latex] and [latex]r[\/latex], and introduce [latex]x[\/latex] and [latex]y[\/latex]. We clear the fraction, and then use substitution. In order to replace [latex]r[\/latex] with [latex]x[\/latex] and [latex]y[\/latex], we must use the expression [latex]{x}^{2}+{y}^{2}={r}^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&r=\\frac{3}{1 - 2\\cos \\theta } \\\\ &r\\left(1 - 2\\cos \\theta \\right)=3 \\\\ &r\\left(1 - 2\\left(\\frac{x}{r}\\right)\\right)=3&& \\text{Use }\\cos \\theta =\\frac{x}{r}\\text{ to eliminate }\\theta . \\\\ &r - 2x=3 \\\\ &r=3+2x&& \\text{Isolate }r. \\\\ &{r}^{2}={\\left(3+2x\\right)}^{2}&& \\text{Square both sides}. \\\\ &{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2}&& \\text{Use }{x}^{2}+{y}^{2}={r}^{2}. \\end{align}[\/latex]<\/p>\n<p>The Cartesian equation is [latex]{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2}[\/latex]. However, to graph it, especially using a graphing calculator or computer program, we want to isolate [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2} \\\\ {y}^{2}={\\left(3+2x\\right)}^{2}-{x}^{2} \\\\ y=\\pm \\sqrt{{\\left(3+2x\\right)}^{2}-{x}^{2}} \\end{gathered}[\/latex]<\/p>\n<p>When our entire equation has been changed from [latex]r[\/latex] and [latex]\\theta[\/latex] to [latex]x[\/latex] and [latex]y[\/latex], we can stop, unless asked to solve for [latex]y[\/latex] or simplify.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165413\/CNX_Precalc_Figure_08_03_0152.jpg\" alt=\"Plots of the equations stated above - the plots are the same in both rectangular and polar coordinates. They are hyperbolas.\" width=\"975\" height=\"481\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 13<\/b><\/p>\n<\/div>\n<p>The &#8220;hour-glass&#8221; shape of the graph is called a <em>hyperbola<\/em>. Hyperbolas have many interesting geometric features and applications, which we will investigate further in <a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/introduction-to-analytic-geometry\/\" target=\"_blank\" rel=\"noopener\">Analytic Geometry<\/a>.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>In this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola\u2019s standard form. To do this, we can start with the initial equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2} \\\\ &{x}^{2}+{y}^{2}-{\\left(3+2x\\right)}^{2}=0 \\\\ &{x}^{2}+{y}^{2}-\\left(9+12x+4{x}^{2}\\right)=0 \\\\ &{x}^{2}+{y}^{2}-9 - 12x - 4{x}^{2}=0 \\\\ &-3{x}^{2}-12x+{y}^{2}=9&& \\text{Multiply through by }-1. \\\\ &3{x}^{2}+12x-{y}^{2}=-9 \\\\ &3\\left({x}^{2}+4x+\\right)-{y}^{2}=-9&& \\text{Organize terms to complete the square for}x. \\\\ &3\\left({x}^{2}+4x+4\\right)-{y}^{2}=-9+12 \\\\ &3{\\left(x+2\\right)}^{2}-{y}^{2}=3 \\\\ &{\\left(x+2\\right)}^{2}-\\frac{{y}^{2}}{3}=1 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Rewrite the polar equation [latex]r=2\\sin \\theta[\/latex] in Cartesian form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q407542\">Show Solution<\/span><\/p>\n<div id=\"q407542\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{x}^{2}+{y}^{2}=2y[\/latex] or, in the standard form for a circle, [latex]{x}^{2}+{\\left(y - 1\\right)}^{2}=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173797\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173797&theme=oea&iframe_resize_id=ohm173797\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 11: Rewriting a Polar Equation in Cartesian Form<\/h3>\n<p>Rewrite the polar equation [latex]r=\\sin \\left(2\\theta \\right)[\/latex] in Cartesian form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q810563\">Show Solution<\/span><\/p>\n<div id=\"q810563\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&r=\\sin \\left(2\\theta \\right)&& \\text{Use the double angle identity for sine}. \\\\ &r=2\\sin \\theta \\cos \\theta&& \\text{Use }\\cos \\theta =\\frac{x}{r}\\text{ and }\\sin \\theta =\\frac{y}{r}. \\\\ &r=2\\left(\\frac{x}{r}\\right)\\left(\\frac{y}{r}\\right)&& \\text{Simplify}. \\\\ &r=\\frac{2xy}{{r}^{2}}&& \\text{ Multiply both sides by }{r}^{2}. \\\\ &{r}^{3}=2xy \\\\ &{\\left(\\sqrt{{x}^{2}+{y}^{2}}\\right)}^{3}=2xy&& \\text{As }{x}^{2}+{y}^{2}={r}^{2},r=\\sqrt{{x}^{2}+{y}^{2}}. \\end{align}[\/latex]<\/p>\n<p>This equation can also be written as<\/p>\n<p style=\"text-align: center;\">[latex]{\\left({x}^{2}+{y}^{2}\\right)}^{\\frac{3}{2}}=2xy\\text{ or }{x}^{2}+{y}^{2}={\\left(2xy\\right)}^{\\frac{2}{3}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Equations<\/h2>\n<table id=\"eip-id3984733\" summary=\"..\">\n<tbody>\n<tr>\n<td>Conversion formulas<\/td>\n<td>[latex]\\begin{align}& \\cos \\theta =\\frac{x}{r}\\to x=r\\cos \\theta \\\\ & \\sin \\theta =\\frac{y}{r}\\to y=r\\sin \\theta \\\\ & {r}^{2}={x}^{2}+{y}^{2} \\\\ & \\tan \\theta =\\frac{y}{x} \\end{align}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>The polar grid is represented as a series of concentric circles radiating out from the pole, or origin.<\/li>\n<li>To plot a point in the form [latex]\\left(r,\\theta \\right),\\theta >0[\/latex], move in a counterclockwise direction from the polar axis by an angle of [latex]\\theta[\/latex], and then extend a directed line segment from the pole the length of [latex]r[\/latex] in the direction of [latex]\\theta[\/latex]. If [latex]\\theta[\/latex] is negative, move in a clockwise direction, and extend a directed line segment the length of [latex]r[\/latex] in the direction of [latex]\\theta[\/latex].<\/li>\n<li>If [latex]r[\/latex] is negative, extend the directed line segment in the opposite direction of [latex]\\theta[\/latex].<\/li>\n<li>To convert from polar coordinates to rectangular coordinates, use the formulas [latex]x=r\\cos \\theta[\/latex] and [latex]y=r\\sin \\theta[\/latex].<\/li>\n<li>To convert from rectangular coordinates to polar coordinates, use one or more of the formulas: [latex]\\cos \\theta =\\frac{x}{r},\\sin \\theta =\\frac{y}{r},\\tan \\theta =\\frac{y}{x}[\/latex], and [latex]r=\\sqrt{{x}^{2}+{y}^{2}}[\/latex].<\/li>\n<li>Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations.<\/li>\n<li>Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135439812\" class=\"definition\">\n<dt>polar axis<\/dt>\n<dd id=\"fs-id1165134391590\">on the polar grid, the equivalent of the positive <em>x-<\/em>axis on the rectangular grid<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134391599\" class=\"definition\">\n<dt>polar coordinates<\/dt>\n<dd id=\"fs-id1165134041274\">on the polar grid, the coordinates of a point labeled [latex]\\left(r,\\theta \\right)[\/latex], where [latex]\\theta[\/latex] indicates the angle of rotation from the polar axis and [latex]r[\/latex] represents the radius, or the distance of the point from the pole in the direction of [latex]\\theta[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134149787\" class=\"definition\">\n<dt>pole<\/dt>\n<dd id=\"fs-id1165133307606\">the origin of the polar grid<\/dd>\n<\/dl>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1958\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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