{"id":2230,"date":"2025-08-11T17:24:57","date_gmt":"2025-08-11T17:24:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2230"},"modified":"2025-08-13T16:58:56","modified_gmt":"2025-08-13T16:58:56","slug":"polar-coordinates-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polar-coordinates-learn-it-3\/","title":{"raw":"Polar Coordinates: Learn It 3","rendered":"Polar Coordinates: Learn It 3"},"content":{"raw":"

Converting Between Polar Coordinates to Rectangular Coordinates<\/h2>\r\nWhen given a set of polar coordinates<\/strong>, we may need to convert them to 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
[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 x-<\/em>axis forms a right triangle. 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\"Comparison\r\n\r\n
\r\n

converting polar 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

[latex]\\cos \\theta =\\frac{x}{r}\\to x=r\\cos \\theta [\/latex]<\/p>\r\n

[latex]\\sin \\theta =\\frac{y}{r}\\to y=r\\sin \\theta [\/latex]<\/p>\r\n\r\n<\/section>

How To: Given polar coordinates, convert to rectangular coordinates.\r\n<\/strong>\r\n
    \r\n \t
  1. 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
  2. Evaluate [latex]\\cos \\theta [\/latex] and [latex]\\sin \\theta [\/latex].<\/li>\r\n \t
  3. Multiply [latex]\\cos \\theta [\/latex] by [latex]r[\/latex] to find the x-<\/em>coordinate of the rectangular form.<\/li>\r\n \t
  4. Multiply [latex]\\sin \\theta [\/latex] by [latex]r[\/latex] to find the y-<\/em>coordinate of the rectangular form.<\/li>\r\n<\/ol>\r\n<\/section> \r\n\r\n
    Write the polar coordinates [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] as rectangular coordinates.[reveal-answer q=\"788608\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788608\"]\r\n\r\nUse the equivalent relationships.\r\n

    [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>\r\nThe rectangular coordinates are [latex]\\left(0,3\\right)[\/latex].\r\n\r\n\"Illustration\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

    Write the polar coordinates [latex]\\left(-2,0\\right)[\/latex] as rectangular coordinates.[reveal-answer q=\"635694\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"635694\"]\r\n\r\nWriting the polar coordinates as rectangular, we have\r\n

    [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>\r\nThe rectangular coordinates are also [latex]\\left(-2,0\\right)[\/latex].\r\n\r\n\"Illustration\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

    Write the polar coordinates [latex]\\left(-1,\\frac{2\\pi }{3}\\right)[\/latex] as rectangular coordinates.[reveal-answer q=\"68654\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"68654\"][latex]\\left(x,y\\right)=\\left(\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/section>
    [ohm_question hide_question_numbers=1]133878[\/ohm_question]<\/section>
    \r\n

    Converting from Rectangular Coordinates to Polar Coordinates<\/h2>\r\nTo convert rectangular coordinates<\/strong> to 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\r\n
    \r\n

    converting rectangular to polar coordinates<\/h3>\r\nConverting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships:\r\n

    [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<\/section>

    Convert the rectangular coordinates [latex]\\left(3,3\\right)[\/latex] to polar coordinates.[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

    [latex]\\begin{align}&\\tan \\theta =\\frac{3}{3} \\\\ &\\tan \\theta =1 \\\\ &{\\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

    [latex]\\begin{align} r&=\\sqrt{{3}^{2}+{3}^{2}} \\\\ r&=\\sqrt{9+9} \\\\ r&=\\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\"Illustration\r\n

    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<\/section><\/section>","rendered":"

    Converting Between Polar Coordinates to Rectangular Coordinates<\/h2>\n

    When given a set of polar coordinates<\/strong>, we may need to convert them to 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

    [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

    Dropping a perpendicular from the point in the plane to the x-<\/em>axis forms a right triangle. 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

    \"Comparison<\/p>\n

    \n

    converting polar to rectangular coordinates<\/h3>\n

    To convert polar coordinates [latex]\\left(r,\\theta \\right)[\/latex] to rectangular coordinates [latex]\\left(x,y\\right)[\/latex], let<\/p>\n

    [latex]\\cos \\theta =\\frac{x}{r}\\to x=r\\cos \\theta[\/latex]<\/p>\n

    [latex]\\sin \\theta =\\frac{y}{r}\\to y=r\\sin \\theta[\/latex]<\/p>\n<\/section>\n

    How To: Given polar coordinates, convert to rectangular coordinates.
    \n<\/strong><\/p>\n
      \n
    1. 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
    2. Evaluate [latex]\\cos \\theta[\/latex] and [latex]\\sin \\theta[\/latex].<\/li>\n
    3. Multiply [latex]\\cos \\theta[\/latex] by [latex]r[\/latex] to find the x-<\/em>coordinate of the rectangular form.<\/li>\n
    4. Multiply [latex]\\sin \\theta[\/latex] by [latex]r[\/latex] to find the y-<\/em>coordinate of the rectangular form.<\/li>\n<\/ol>\n<\/section>\n

       <\/p>\n

      Write the polar coordinates [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] as rectangular coordinates.<\/p>\n