{"id":823,"date":"2021-08-27T20:07:06","date_gmt":"2021-08-27T20:07:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=823"},"modified":"2022-10-29T00:53:01","modified_gmt":"2022-10-29T00:53:01","slug":"the-normal-and-binormal-vectors","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/the-normal-and-binormal-vectors\/","title":{"raw":"The Normal and Binormal Vectors","rendered":"The Normal and Binormal Vectors"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the meaning of the normal and binormal vectors of a curve in space.<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe have seen that the derivative [latex]{\\bf{r}}'\\,(t)[\/latex] of a vector-valued function is a tangent vector to the curve defined by [latex]{\\bf{r}}\\,(t)[\/latex], and the unit tangent vector [latex]{\\bf{T}}\\,(t)[\/latex] can be calculated by dividing [latex]{\\bf{r}}'\\,(t)[\/latex] by its magnitude. When studying motion in three dimensions, two other vectors are useful in describing the motion of a particle along a path in space: the\u00a0<strong>principal unit vector<\/strong> and the\u00a0<strong>binomial vector<\/strong>.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]C[\/latex] be a three-dimensional\u00a0<strong>smooth<\/strong> curve represented [latex]{\\bf{r}}[\/latex] over an open interval [latex]I[\/latex]. If [latex]{\\bf{T}}'\\,(t)\\neq{0}[\/latex], then the principal unit normal vector at [latex]t[\/latex] is defined to be\r\n<div style=\"text-align: center;\">[latex]{\\bf{N}}\\,(t)=\\frac{{\\bf{T}}'\\,(t)}{\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert}[\/latex].<\/div>\r\nThe binomial vector at [latex]t[\/latex] is defined as\r\n<div style=\"text-align: center;\">[latex]{\\bf{B}}\\,(t)={\\bf{T}}\\,(t)\\,\\times\\,{\\bf{N}}\\,(t)[\/latex],<\/div>\r\nwhere [latex]{\\bf{T}}(t)[\/latex] is the unit tangent vector.\r\n\r\n<\/div>\r\nNote that, by definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector. Furthermore, [latex]{\\bf{B}}\\,(t)[\/latex] is always a unit vector. This cam be shown using the formula for the magnitude of a cross product\r\n<p style=\"text-align: center;\">[latex]\\left\\Vert{\\bf{B}}\\,(t)\\right\\Vert=\\left\\Vert{\\bf{T}}\\,(t)\\,\\times\\,{\\bf{N}}\\,(t)\\right\\Vert=\\left\\Vert{\\bf{T}}\\,(t)\\right\\Vert\\,\\left\\Vert{\\bf{N}}\\,(t)\\right\\Vert\\sin{\\theta},[\/latex]<\/p>\r\nwhere [latex]\\theta[\/latex] is the angle between [latex]{\\bf{T}}\\,(t)[\/latex] and [latex]{\\bf{N}}\\,(t)[\/latex]. Since [latex]{\\bf{N}}\\,(t)[\/latex] is the derivative of a unit vector, property 7 of the derivative of a vector-valued function tells us that [latex]{\\bf{T}}(t)[\/latex] and [latex]{\\bf{N}}(t)[\/latex] are orthogonal to each other, so [latex]\\theta=\\pi\/2[\/latex]. Furthermore, they are both unit vectors, so their magnitude is [latex]1[\/latex]. Therefore, [latex]\\left\\Vert{\\bf{T}}\\,(t)\\right\\Vert\\,\\left\\Vert{\\bf{N}}\\,(t)\\right\\Vert\\sin{\\theta}=(1)(1)\\sin{(\\pi\/2)}=1[\/latex] and [latex]{\\bf{B}}\\,(t)[\/latex] is a unit vector.\r\n\r\nThe principal unit normal vector can be challenging to calculate because the unit tangent vector involves a quotient, and this quotient often has a square root in the denominator. In the three-dimensional case, finding the cross product of the unit tangent vector and the unit normal vector can be even more cumbersome. Fortunately, we have alternative formulas for finding these two vectors, and they are presented in Motion in Space.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the principal unit normal vector and binormal vector<\/h3>\r\nFor each of the following vector-valued functions, find the principal unit normal vector. Then, if possible, find the binormal vector.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. [latex]{\\bf{r}}\\,(t)=4\\cos{t{\\bf{i}}}-4\\sin{t{\\bf{j}}}[\/latex]<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. [latex]{\\bf{r}}\\,(t)=(6t+2)\\,{\\bf{i}}+5t^{2}\\,{\\bf{j}}-8t\\,{\\bf{k}}[\/latex]<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793461764\" class=\"exercise\">[reveal-answer q=\"fs-id1167795056166\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167795056166\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>This function describes a circle.\r\n<div>\r\n\r\n[caption id=\"attachment_913\" align=\"aligncenter\" width=\"448\"]<img class=\"size-full wp-image-913\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160405\/3-3-4.jpeg\" alt=\"This figure is the graph of a circle centered at the origin with radius of 2. The orientation of the circle is clockwise. It represents the vector-valued function r(t) = 4costi \u2013 4 sintj.\" width=\"448\" height=\"422\" \/> Figure 1.[\/caption]\r\n\r\n<\/div>\r\nTo find the principal unit normal vector, we first must find the unit tangent vector [latex]{\\bf{T}}\\,(t)[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)} &amp;=\\hfill&amp;{\\frac{{\\bf{r}}'\\,(t)}{\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-4\\sin{t{\\bf{i}}}-4\\cos{t{\\bf{j}}}}{\\sqrt{(-4\\sin{t})^{2}+(-4\\cos{t})^{2}}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-4\\sin{t{\\bf{i}}}-4\\cos{t{\\bf{j}}}}{\\sqrt{16\\sin{^{2}t}+16\\cos{^{2}t}}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-4\\sin{t{\\bf{i}}}-4\\cos{t{\\bf{j}}}}{\\sqrt{16(\\sin{^{2}t}+\\cos{^{2}t})}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-4\\sin{t{\\bf{i}}}-4\\cos{t{\\bf{j}}}}{4}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\sin{t{\\bf{i}}}-\\cos{t{\\bf{j}}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext, we use the first equation from the definition\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{N}}\\,(t)} &amp;=\\hfill&amp;{\\frac{{\\bf{T}}'\\,(t)}{\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-\\cos{t{\\bf{i}}}+\\sin{t{\\bf{j}}}}{\\sqrt{(-\\cos{t})^{2}+(\\sin{t})^{2}}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-\\cos{t{\\bf{i}}}+\\sin{t{\\bf{j}}}}{\\sqrt{\\cos{^{2}t}+\\sin{^{2}t}}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\cos{t{\\bf{i}}}+\\sin{t{\\bf{j}}}}\\hfill \\\\ \\hfill \\end{array}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nNotice that the unit tangent vector and the principal unit normal vector are orthogonal to each other for all rules of [latex]t[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)\\cdot{\\bf{N}}\\,(t)} &amp;=\\hfill&amp;{\\langle{-}\\sin{t},-\\cos{t}\\rangle\\cdot\\langle{-}\\cos{t},\\sin{t}\\rangle}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sin{t}\\cos{t}-\\cos{t}\\sin{t}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {0.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nFurthermore, the principal unit normal vector points toward the center of the circle from every point on the circle. Since [latex]{\\bf{r}}\\,(t)[\/latex] defines a curve in tow dimensions, we cannot calculate the binormal vector\r\n\r\n[caption id=\"attachment_914\" align=\"aligncenter\" width=\"438\"]<img class=\"size-full wp-image-914\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160457\/3-3-5.jpeg\" alt=\"This figure is the graph of a circle centered at the origin with radius of 2. The orientation of the circle is clockwise. It represents the vector-valued function r(t) = 4costi \u2013 4 sintj. On the circle in the first quadrant is a vector pointing inward. It is labeled \u201cprincipal unit normal vector\u201d.\" width=\"438\" height=\"272\" \/> Figure 2.[\/caption]<\/li>\r\n \t<li>This function looks like this:\r\n<div>\r\n\r\n[caption id=\"attachment_915\" align=\"aligncenter\" width=\"420\"]<img class=\"size-full wp-image-915\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160538\/3-3-6.jpeg\" alt=\"This figure is a curve in 3 dimensions. It is inside of a box. The box represents the first octant. The curve starts at the bottom right of the box and curves through the box in a parabolic curve to the top.\" width=\"420\" height=\"504\" \/> Figure 3.[\/caption]\r\n\r\n<\/div>\r\nTo find the principal unit normal vector, we first find the unit tangent vector [latex]{\\bf{T}}\\,(t)[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)} &amp;=\\hfill &amp; {\\frac{{\\bf{r}}'\\,(t)}{\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{6\\,{\\bf{i}}+10t\\,{\\bf{j}}-8\\,{\\bf{k}}}{\\sqrt{6^{2}+(10t)^{2}+(-8)^{2}}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{6\\,{\\bf{i}}+10t\\,{\\bf{j}}-8\\,{\\bf{k}}}{\\sqrt{36+100t^{2}+64}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{6\\,{\\bf{i}}+10t\\,{\\bf{j}}-8\\,{\\bf{k}}}{\\sqrt{100(t^{2}+1)}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{3\\,{\\bf{i}}+5t\\,{\\bf{j}}-4\\,{\\bf{k}}}{5\\,\\sqrt{t^{2}+1}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{3}{5}\\,(t^{2}+1)^{-1\/2}{\\bf{i}}+t\\,(t^{2}+1)^{-1\/2}{\\bf{j}}-\\frac{4}{5}\\,(t^{2}+1)^{-1\/2}{\\bf{k}}.}\\hfill \\\\ \\hfill\\end{array}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nNext, we calculate [latex]{\\bf{T}}'\\,(t)[\/latex] and [latex]\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)} &amp;=\\hfill &amp; {\\frac{3}{5}\\,(-\\frac{1}{2})\\,(t^{2}+1)^{-3\/2}(2t)\\,{\\bf{i}}+\\big((t^{2}+1)^{-1\/2}-t\\,(\\frac{1}{2})\\,(t^{2}+1)^{-3\/2}(2t)\\big)\\,{\\bf{j}}-\\frac{4}{5}\\,(-\\frac{1}{2})\\,(t^{2}+1)^{-3\/2}(2t)\\,{\\bf{k}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\frac{3t}{5(t^{2}+1)^{3\/2}}\\,{\\bf{i}}+\\frac{1}{(t^{2}+1)^{3\/2}}\\,{\\bf{j}}+\\frac{4}{5(t^{2}+1)^{3\/2}}\\,{\\bf{k}}} \\hfill \\\\ \\hfill {\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert}&amp; =\\hfill &amp; {\\sqrt{\\bigg(-\\frac{3t}{5(t^{2}+1)^{3\/2}}\\bigg)^{2}+\\bigg(-\\frac{1}{(t^{2}+1)^{3\/2}}\\bigg)^{2}+\\bigg(\\frac{4t}{5(t^{2}+1)^{3\/2}}\\bigg)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{\\frac{9t^{2}}{25(t^{2}+1)^{3}}+\\frac{1}{(t^{2}+1)^{3}}+\\frac{16t^{2}}{25(t^{2}+1)^{3}}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{\\frac{25t^{2}+25}{25(t^{2}+1)^{3}}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{\\frac{1}{(t^{2}+1)^{2}}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{1}{t^{2}+1}.}\\hfill \\\\ \\hfill\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore, according to the second equation from the above definition:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{N}}\\,(t)} &amp;=\\hfill &amp; {\\frac{{\\bf{T}}'\\,(t)}{\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\bigg(-\\frac{3t}{5(t^{2}+1)^{3\/2}}{\\bf{i}}+\\frac{1}{(t^{2}+1)^{3\/2}}{\\bf{j}}+\\frac{4t}{5(t^{2}+1)^{3\/2}}{\\bf{k}}\\bigg)(t^{2}+1)} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\frac{3t}{5(t^{2}+1)^{1\/2}}{\\bf{i}}+\\frac{1}{(t^{2}+1)^{1\/2}}{\\bf{j}}+\\frac{4t}{5(t^{2}+1)^{1\/2}}{\\bf{k}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\frac{3t\\,{\\bf{i}}-5\\,{\\bf{j}}-4t\\,{\\bf{k}}}{5\\sqrt{t^{2}+1}}}\\hfill \\\\ \\hfill\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nOnce again, the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of [latex]t[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)\\cdot{\\bf{N}}\\,(t)} &amp;=\\hfill &amp; {\\bigg(\\frac{3\\,{\\bf{i}}+5t\\,{\\bf{j}}-4\\,{\\bf{k}}}{5\\sqrt{t^{2}+1}}\\bigg)\\cdot\\bigg(-\\frac{3t\\,{\\bf{i}}-5\\,{\\bf{j}}-4t\\,{\\bf{k}}}{5\\sqrt{t^{2}+1}}\\bigg)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{3(-3)-5t(-5)-4(4t)}{5\\sqrt{t^{2}+1}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-9t+25t-16t}{5\\sqrt{t^{2}+1}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {0.}\\hfill \\\\ \\hfill\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nLast, since [latex]{\\bf{r}}\\,(t)[\/latex] represents a three-dimensional curve, we can calculate the binormal vector using the first equation from the above definition:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{B}}\\,(t)} &amp;=\\hfill &amp; {{\\bf{T}}\\,(t)\\,\\times\\,{\\bf{N}}\\,(t)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}}\\\\\\frac{3}{5\\sqrt{t^{2}+1}}&amp;+\\frac{5t}{5\\sqrt{t^{2}+1}}&amp;-\\frac{4}{5\\sqrt{t^{2}+1}}\\\\ -\\frac{3t}{5\\sqrt{t^{2}+1}}&amp;+\\frac{5}{5\\sqrt{t^{2}+1}}&amp;\\frac{4t}{5\\sqrt{t^{2}+1}}\\end{vmatrix}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\bigg(\\Big(+\\frac{5t}{5\\sqrt{t^{2}+1}}\\Big)\\Big(\\frac{4t}{5\\sqrt{t^{2}+1}}\\Big)-\\Big(-\\frac{4}{5\\sqrt{t^{2}+1}}\\Big)\\Big(-\\frac{5}{5\\sqrt{t^{2}+1}}\\Big)\\bigg)\\,{\\bf{i}} \\\\ +\\bigg(\\Big(\\frac{3}{5\\sqrt{t^{2}+1}}\\Big)\\Big(\\frac{4t}{5\\sqrt{t^{2}+1}}\\Big)-\\Big(-\\frac{4}{5\\sqrt{t^{2}+1}}\\Big)\\Big(-\\frac{3t}{5\\sqrt{t^{2}+1}}\\Big)\\bigg)\\,{\\bf{j}} \\\\ +\\bigg(\\Big(\\frac{3}{5\\sqrt{t^{2}+1}}\\Big)\\Big(+\\frac{5}{5\\sqrt{t^{2}+1}}\\Big)-\\Big(+\\frac{5t}{5\\sqrt{t^{2}+1}}\\Big)\\Big(-\\frac{3t}{5\\sqrt{t^{2}+1}}\\Big)\\bigg){\\bf{k}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\Big(\\frac{20t^{2}+20}{25(t^{2}+1)}\\Big)\\,{\\bf{i}}+\\Big(\\frac{-15-15t^{2}}{25(t^{2}+1)}\\Big)\\,{\\bf{k}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {20\\,\\Big(\\frac{t^{2}+1}{25(t^{2}+1)}\\Big)\\,{\\bf{i}}-15\\,\\Big(\\frac{t^{2}+1}{25(t^{2}+1)}\\Big)\\,{\\bf{k}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{4}{5}\\,{\\bf{i}}-\\frac{3}{5}\\,{\\bf{k}}.}\\hfill \\\\ \\hfill\\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the unit normal vector for the vector-valued function [latex]{\\bf{r}}\\,(t)=(t^{2}-3)\\,{\\bf{I}}+(4t+1)\\,{\\bf{j}}[\/latex] and evaluate it at [latex]t=2[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1167794933264\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933264\"]\r\n<div style=\"text-align: center;\">[latex]\\kappa=\\frac{6}{101^{3\/2}}\\approx{0.0059}[\/latex]<\/div>\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]244857[\/ohm_question]\r\n\r\n<\/div>\r\nFor any smooth curve in three dimensions that is defined by a vector-valued function, we now have formulas for the unit tangent vector\u00a0[latex]{\\bf{T}}[\/latex], the unit normal vector\u00a0[latex]{\\bf{N}}[\/latex], and the binormal vector\u00a0[latex]\\bf{B}[\/latex]. The unit normal vector and the binormal vector form a plane that is perpendicular to the curve at any point on the curve, called the <strong><span id=\"term136\" data-type=\"term\">normal plane<\/span><\/strong>. In addition, these three vectors form a frame of reference in three-dimensional space called the <strong><span id=\"term137\" data-type=\"term\">Frenet frame of reference<\/span><\/strong> (also called the <strong data-effect=\"bold\">TNB<\/strong> frame) (Figure 7). Lat, the plane determined by the vectors\u00a0[latex]\\bf{T}[\/latex] and\u00a0[latex]\\bf{N}[\/latex]\u00a0forms the osculating plane of\u00a0[latex]C[\/latex] at any point\u00a0[latex]P[\/latex] on the curve.\r\n\r\n[caption id=\"attachment_916\" align=\"aligncenter\" width=\"677\"]<img class=\"size-full wp-image-916\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160628\/3-3-7.jpeg\" alt=\"This figure is the graph of a curve increasing and decreasing. Along the curve at 4 different points are 3 vectors at each point. The first vector is labeled \u201cT\u201d and is tangent to the curve at the point. The second vector is labeled \u201cN\u201d and is normal to the curve at the point. The third vector is labeled \u201cB\u201d and is orthogonal to T and N.\" width=\"677\" height=\"321\" \/> Figure 4.\u00a0<span class=\"os-caption\">This figure depicts a Frenet frame of reference. At every point\u00a0[latex]P[\/latex]\u00a0on a three-dimensional curve, the unit tangent, unit normal, and binormal vectors form a three-dimensional frame of reference.<\/span>[\/caption]Suppose we form a circle in the osculating plane of\u00a0[latex]C[\/latex] at point\u00a0[latex]P[\/latex]\u00a0on the curve. Assume that the circle has the same curvature as the curve does at point\u00a0[latex]P[\/latex] and let the circle have radius [latex]r[\/latex]<em data-effect=\"italics\">.<\/em> Then, the curvature of the circle is given by [latex]1\/r[\/latex].\u00a0We call [latex]r[\/latex] the <span id=\"term138\" data-type=\"term\">radius of curvature<\/span> of the curve, and it is equal to the reciprocal of the curvature. If this circle lies on the concave side of the curve and is tangent to the curve at point [latex]P[\/latex]<em data-effect=\"italics\">,<\/em> then this circle is called the <span id=\"term139\" data-type=\"term\">osculating circle<\/span> of [latex]C[\/latex] at [latex]P[\/latex], as shown in the following figure.\r\n\r\n[caption id=\"attachment_918\" align=\"aligncenter\" width=\"721\"]<img class=\"size-full wp-image-918\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160750\/3-3-8.jpeg\" alt=\"This figure is the graph of a curve with a circle in the middle. The bottom of the circle is the same as part of the curve. Inside of the circle is a vector labeled \u201cr\u201d. It starts at point \u201cP\u201d on the circle and points towards the radius. There is also a line segment perpendicular to the radius and tangent to point P.\" width=\"721\" height=\"305\" \/> Figure 5.\u00a0In this osculating circle, the circle is tangent to curve [latex]C[\/latex]\u00a0at point\u00a0[latex]P[\/latex]\u00a0and shares the same curvature.[\/caption]&nbsp;\r\n<div class=\"textbox tryit\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Interactive<\/h3>\r\nFor more information on osculating circles, see this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle1\" target=\"_blank\" rel=\"noopener nofollow\">demonstration<\/a> on curvature and torsion, this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle3\" target=\"_blank\" rel=\"noopener nofollow\">article<\/a> on osculating circles, and this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle2\" target=\"_blank\" rel=\"noopener nofollow\">discussion<\/a> of Serret formulas.\r\n\r\n<\/div>\r\nTo find the equation of an osculating circle in two dimensions, we need find only the center and radius of the circle.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the Equation of an Osculating Circle<\/h3>\r\nFind the equation of the osculating circle of the helix defined by the function [latex]y=x^{3}-3x+1[\/latex] at [latex]x=1[\/latex].\r\n<div id=\"fs-id1167793461764\" class=\"exercise\">[reveal-answer q=\"fs-id1167705065166\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167705065166\"]\r\n<p style=\"text-align: left;\">Figure 9 shows the graph of [latex]y=x^{3}-3x+1[\/latex].<\/p>\r\n\r\n<div>[caption id=\"attachment_919\" align=\"aligncenter\" width=\"342\"]<img class=\"size-full wp-image-919\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160915\/3-3-9.jpeg\" alt=\"This figure is the graph of a cubic function y = x^3-3x+1. The curve increases, reaches a maximum at x=-1, decreases passing through the y-axis at 1, then reaching a minimum at x =1 before increasing again.\" width=\"342\" height=\"347\" \/> Figure 6.\u00a0We want to find the osculating circle of this graph at the point where\u00a0[latex]t=1[\/latex].[\/caption]<\/div>\r\nFirst, let's calculate the curvature at [latex]x=1[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\LARGE{\\kappa=\\frac{|f''\\,(x)|}{(1+[f'\\,(x)]^{2})^{3\/2}}=\\frac{|6x|}{(1+[3x^{2}-3]^{2})^{3\/2}}}.[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThis gives [latex]\\kappa=6[\/latex].\u00a0Therefore, the radius of the osculating circle is given by [latex]R=\\frac{1}{\\kappa}=\\frac{1}{6}[\/latex].\u00a0Next, we then calculate the coordinates of the center of the circle. When [latex]x=1[\/latex],\u00a0the slope of the tangent line is zero. Therefore, the center of the osculating circle is directly above the point on the graph with coordinates [latex](1,\\ -1)[\/latex]. The center is located at [latex](1,\\ -\\frac{5}{6})[\/latex]. The formula for a circle with radius [latex]r[\/latex] and center [latex](h,\\ k)[\/latex] is given by [latex](x-h)^{2}+(y-k)^{2}=r^{2}[\/latex]. Therefore, the equation of the osculating circle is [latex](x-1)^{2}+(y+\\frac{5}{6})^{2}=\\frac{1}{36}[\/latex]. The graph and its osculating circle appears in the following graph.\r\n\r\n[caption id=\"attachment_920\" align=\"aligncenter\" width=\"342\"]<img class=\"size-full wp-image-920\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21161011\/3-3-10.jpeg\" alt=\"This figure is the graph of a cubic function y = x^3-3x+1. The curve increases, reaches a maximum at x=-1, decreases passing through the y-axis at 1, then reaching a minimum at x =1 before increasing again. There is a small circle inside of the bend of the cure at x = 1.\" width=\"342\" height=\"347\" \/> Figure 8.\u00a0The osculating circle has radius\u00a0[latex]R=\\frac{1}{6}[\/latex][\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the equation of the osculating circle of the curve defined by the vector-valued function [latex]y=2x^{2}-4x+5[\/latex] at [latex]x=1[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1167794934264\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794934264\"]\r\n<p style=\"text-align: left;\">[latex]\\kappa=\\frac{4}{[1+(4x-4)^{2}]^{3\/2}}[\/latex] At the point [latex]x=1[\/latex], the curvature is equal to [latex]4[\/latex]. Therefore, the radius of the osculating circle is [latex]\\frac{1}{4}[\/latex]. A graph of this functions appears next:<\/p>\r\n\r\n\r\n[caption id=\"attachment_921\" align=\"aligncenter\" width=\"362\"]<img class=\"size-full wp-image-921\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21161115\/3-3-tryitans.jpeg\" alt=\"This figure is the graph of the function y = 2x^2-4x+5. The curve is a parabola opening up with vertex at (1, 3).\" width=\"362\" height=\"347\" \/> Figure 9.[\/caption]\r\n\r\nThe vertex of this parabola is located at the point [latex](1,\\ 3)[\/latex]. Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are [latex](1,\\ \\frac{13}{4})[\/latex]. The equation of the osculating circle is [latex](x-1)^{2}+(y-\\frac{13}{4})^{2}=\\frac{1}{16}[\/latex].\r\n<div style=\"text-align: center;\">[latex]{\\bf{v}}\\,(t)={\\bf{r}}'\\,(t)=(2t-3)\\,{\\bf{i}}+2\\,{\\bf{j}}+{\\bf{k}}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949625&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=11CyZk8-Fa0&amp;video_target=tpm-plugin-h4q06w32-11CyZk8-Fa0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.11_transcript.html\">transcript for \u201cCP 3.13\u201d here (opens in new window).<\/a><\/center>&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the meaning of the normal and binormal vectors of a curve in space.<\/li>\n<\/ul>\n<\/div>\n<p>We have seen that the derivative [latex]{\\bf{r}}'\\,(t)[\/latex] of a vector-valued function is a tangent vector to the curve defined by [latex]{\\bf{r}}\\,(t)[\/latex], and the unit tangent vector [latex]{\\bf{T}}\\,(t)[\/latex] can be calculated by dividing [latex]{\\bf{r}}'\\,(t)[\/latex] by its magnitude. When studying motion in three dimensions, two other vectors are useful in describing the motion of a particle along a path in space: the\u00a0<strong>principal unit vector<\/strong> and the\u00a0<strong>binomial vector<\/strong>.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]C[\/latex] be a three-dimensional\u00a0<strong>smooth<\/strong> curve represented [latex]{\\bf{r}}[\/latex] over an open interval [latex]I[\/latex]. If [latex]{\\bf{T}}'\\,(t)\\neq{0}[\/latex], then the principal unit normal vector at [latex]t[\/latex] is defined to be<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{N}}\\,(t)=\\frac{{\\bf{T}}'\\,(t)}{\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert}[\/latex].<\/div>\n<p>The binomial vector at [latex]t[\/latex] is defined as<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{B}}\\,(t)={\\bf{T}}\\,(t)\\,\\times\\,{\\bf{N}}\\,(t)[\/latex],<\/div>\n<p>where [latex]{\\bf{T}}(t)[\/latex] is the unit tangent vector.<\/p>\n<\/div>\n<p>Note that, by definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector. Furthermore, [latex]{\\bf{B}}\\,(t)[\/latex] is always a unit vector. This cam be shown using the formula for the magnitude of a cross product<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\Vert{\\bf{B}}\\,(t)\\right\\Vert=\\left\\Vert{\\bf{T}}\\,(t)\\,\\times\\,{\\bf{N}}\\,(t)\\right\\Vert=\\left\\Vert{\\bf{T}}\\,(t)\\right\\Vert\\,\\left\\Vert{\\bf{N}}\\,(t)\\right\\Vert\\sin{\\theta},[\/latex]<\/p>\n<p>where [latex]\\theta[\/latex] is the angle between [latex]{\\bf{T}}\\,(t)[\/latex] and [latex]{\\bf{N}}\\,(t)[\/latex]. Since [latex]{\\bf{N}}\\,(t)[\/latex] is the derivative of a unit vector, property 7 of the derivative of a vector-valued function tells us that [latex]{\\bf{T}}(t)[\/latex] and [latex]{\\bf{N}}(t)[\/latex] are orthogonal to each other, so [latex]\\theta=\\pi\/2[\/latex]. Furthermore, they are both unit vectors, so their magnitude is [latex]1[\/latex]. Therefore, [latex]\\left\\Vert{\\bf{T}}\\,(t)\\right\\Vert\\,\\left\\Vert{\\bf{N}}\\,(t)\\right\\Vert\\sin{\\theta}=(1)(1)\\sin{(\\pi\/2)}=1[\/latex] and [latex]{\\bf{B}}\\,(t)[\/latex] is a unit vector.<\/p>\n<p>The principal unit normal vector can be challenging to calculate because the unit tangent vector involves a quotient, and this quotient often has a square root in the denominator. In the three-dimensional case, finding the cross product of the unit tangent vector and the unit normal vector can be even more cumbersome. Fortunately, we have alternative formulas for finding these two vectors, and they are presented in Motion in Space.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding the principal unit normal vector and binormal vector<\/h3>\n<p>For each of the following vector-valued functions, find the principal unit normal vector. Then, if possible, find the binormal vector.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\"> <\/ol>\n<ol style=\"list-style-type: lower-alpha;\"> <\/ol>\n<\/li>\n<\/ol>\n<div id=\"fs-id1167793461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167795056166\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167795056166\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>This function describes a circle.\n<div>\n<div id=\"attachment_913\" style=\"width: 458px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-913\" class=\"size-full wp-image-913\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160405\/3-3-4.jpeg\" alt=\"This figure is the graph of a circle centered at the origin with radius of 2. The orientation of the circle is clockwise. It represents the vector-valued function r(t) = 4costi \u2013 4 sintj.\" width=\"448\" height=\"422\" \/><\/p>\n<p id=\"caption-attachment-913\" class=\"wp-caption-text\">Figure 1.<\/p>\n<\/div>\n<\/div>\n<p>To find the principal unit normal vector, we first must find the unit tangent vector [latex]{\\bf{T}}\\,(t)[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)} &=\\hfill&{\\frac{{\\bf{r}}'\\,(t)}{\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{-4\\sin{t{\\bf{i}}}-4\\cos{t{\\bf{j}}}}{\\sqrt{(-4\\sin{t})^{2}+(-4\\cos{t})^{2}}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{-4\\sin{t{\\bf{i}}}-4\\cos{t{\\bf{j}}}}{\\sqrt{16\\sin{^{2}t}+16\\cos{^{2}t}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{-4\\sin{t{\\bf{i}}}-4\\cos{t{\\bf{j}}}}{\\sqrt{16(\\sin{^{2}t}+\\cos{^{2}t})}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{-4\\sin{t{\\bf{i}}}-4\\cos{t{\\bf{j}}}}{4}}\\hfill \\\\ \\hfill & =\\hfill & {-\\sin{t{\\bf{i}}}-\\cos{t{\\bf{j}}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next, we use the first equation from the definition<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{N}}\\,(t)} &=\\hfill&{\\frac{{\\bf{T}}'\\,(t)}{\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{-\\cos{t{\\bf{i}}}+\\sin{t{\\bf{j}}}}{\\sqrt{(-\\cos{t})^{2}+(\\sin{t})^{2}}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{-\\cos{t{\\bf{i}}}+\\sin{t{\\bf{j}}}}{\\sqrt{\\cos{^{2}t}+\\sin{^{2}t}}}}\\hfill \\\\ \\hfill & =\\hfill & {-\\cos{t{\\bf{i}}}+\\sin{t{\\bf{j}}}}\\hfill \\\\ \\hfill \\end{array}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Notice that the unit tangent vector and the principal unit normal vector are orthogonal to each other for all rules of [latex]t[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)\\cdot{\\bf{N}}\\,(t)} &=\\hfill&{\\langle{-}\\sin{t},-\\cos{t}\\rangle\\cdot\\langle{-}\\cos{t},\\sin{t}\\rangle}\\hfill \\\\ \\hfill & =\\hfill & {\\sin{t}\\cos{t}-\\cos{t}\\sin{t}} \\hfill \\\\ \\hfill & =\\hfill & {0.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Furthermore, the principal unit normal vector points toward the center of the circle from every point on the circle. Since [latex]{\\bf{r}}\\,(t)[\/latex] defines a curve in tow dimensions, we cannot calculate the binormal vector<\/p>\n<div id=\"attachment_914\" style=\"width: 448px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-914\" class=\"size-full wp-image-914\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160457\/3-3-5.jpeg\" alt=\"This figure is the graph of a circle centered at the origin with radius of 2. The orientation of the circle is clockwise. It represents the vector-valued function r(t) = 4costi \u2013 4 sintj. On the circle in the first quadrant is a vector pointing inward. It is labeled \u201cprincipal unit normal vector\u201d.\" width=\"438\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-914\" class=\"wp-caption-text\">Figure 2.<\/p>\n<\/div>\n<\/li>\n<li>This function looks like this:\n<div>\n<div id=\"attachment_915\" style=\"width: 430px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-915\" class=\"size-full wp-image-915\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160538\/3-3-6.jpeg\" alt=\"This figure is a curve in 3 dimensions. It is inside of a box. The box represents the first octant. The curve starts at the bottom right of the box and curves through the box in a parabolic curve to the top.\" width=\"420\" height=\"504\" \/><\/p>\n<p id=\"caption-attachment-915\" class=\"wp-caption-text\">Figure 3.<\/p>\n<\/div>\n<\/div>\n<p>To find the principal unit normal vector, we first find the unit tangent vector [latex]{\\bf{T}}\\,(t)[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)} &=\\hfill & {\\frac{{\\bf{r}}'\\,(t)}{\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{6\\,{\\bf{i}}+10t\\,{\\bf{j}}-8\\,{\\bf{k}}}{\\sqrt{6^{2}+(10t)^{2}+(-8)^{2}}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{6\\,{\\bf{i}}+10t\\,{\\bf{j}}-8\\,{\\bf{k}}}{\\sqrt{36+100t^{2}+64}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{6\\,{\\bf{i}}+10t\\,{\\bf{j}}-8\\,{\\bf{k}}}{\\sqrt{100(t^{2}+1)}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{3\\,{\\bf{i}}+5t\\,{\\bf{j}}-4\\,{\\bf{k}}}{5\\,\\sqrt{t^{2}+1}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{3}{5}\\,(t^{2}+1)^{-1\/2}{\\bf{i}}+t\\,(t^{2}+1)^{-1\/2}{\\bf{j}}-\\frac{4}{5}\\,(t^{2}+1)^{-1\/2}{\\bf{k}}.}\\hfill \\\\ \\hfill\\end{array}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Next, we calculate [latex]{\\bf{T}}'\\,(t)[\/latex] and [latex]\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)} &=\\hfill & {\\frac{3}{5}\\,(-\\frac{1}{2})\\,(t^{2}+1)^{-3\/2}(2t)\\,{\\bf{i}}+\\big((t^{2}+1)^{-1\/2}-t\\,(\\frac{1}{2})\\,(t^{2}+1)^{-3\/2}(2t)\\big)\\,{\\bf{j}}-\\frac{4}{5}\\,(-\\frac{1}{2})\\,(t^{2}+1)^{-3\/2}(2t)\\,{\\bf{k}}}\\hfill \\\\ \\hfill & =\\hfill & {-\\frac{3t}{5(t^{2}+1)^{3\/2}}\\,{\\bf{i}}+\\frac{1}{(t^{2}+1)^{3\/2}}\\,{\\bf{j}}+\\frac{4}{5(t^{2}+1)^{3\/2}}\\,{\\bf{k}}} \\hfill \\\\ \\hfill {\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert}& =\\hfill & {\\sqrt{\\bigg(-\\frac{3t}{5(t^{2}+1)^{3\/2}}\\bigg)^{2}+\\bigg(-\\frac{1}{(t^{2}+1)^{3\/2}}\\bigg)^{2}+\\bigg(\\frac{4t}{5(t^{2}+1)^{3\/2}}\\bigg)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{\\frac{9t^{2}}{25(t^{2}+1)^{3}}+\\frac{1}{(t^{2}+1)^{3}}+\\frac{16t^{2}}{25(t^{2}+1)^{3}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{\\frac{25t^{2}+25}{25(t^{2}+1)^{3}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{\\frac{1}{(t^{2}+1)^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{1}{t^{2}+1}.}\\hfill \\\\ \\hfill\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore, according to the second equation from the above definition:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{N}}\\,(t)} &=\\hfill & {\\frac{{\\bf{T}}'\\,(t)}{\\left\\Vert{\\bf{T}}'\\,(t)\\right\\Vert}}\\hfill \\\\ \\hfill & =\\hfill & {\\bigg(-\\frac{3t}{5(t^{2}+1)^{3\/2}}{\\bf{i}}+\\frac{1}{(t^{2}+1)^{3\/2}}{\\bf{j}}+\\frac{4t}{5(t^{2}+1)^{3\/2}}{\\bf{k}}\\bigg)(t^{2}+1)} \\hfill \\\\ \\hfill & =\\hfill & {-\\frac{3t}{5(t^{2}+1)^{1\/2}}{\\bf{i}}+\\frac{1}{(t^{2}+1)^{1\/2}}{\\bf{j}}+\\frac{4t}{5(t^{2}+1)^{1\/2}}{\\bf{k}}}\\hfill \\\\ \\hfill & =\\hfill & {-\\frac{3t\\,{\\bf{i}}-5\\,{\\bf{j}}-4t\\,{\\bf{k}}}{5\\sqrt{t^{2}+1}}}\\hfill \\\\ \\hfill\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Once again, the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of [latex]t[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{T}}\\,(t)\\cdot{\\bf{N}}\\,(t)} &=\\hfill & {\\bigg(\\frac{3\\,{\\bf{i}}+5t\\,{\\bf{j}}-4\\,{\\bf{k}}}{5\\sqrt{t^{2}+1}}\\bigg)\\cdot\\bigg(-\\frac{3t\\,{\\bf{i}}-5\\,{\\bf{j}}-4t\\,{\\bf{k}}}{5\\sqrt{t^{2}+1}}\\bigg)}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{3(-3)-5t(-5)-4(4t)}{5\\sqrt{t^{2}+1}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{-9t+25t-16t}{5\\sqrt{t^{2}+1}}}\\hfill \\\\ \\hfill & =\\hfill & {0.}\\hfill \\\\ \\hfill\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Last, since [latex]{\\bf{r}}\\,(t)[\/latex] represents a three-dimensional curve, we can calculate the binormal vector using the first equation from the above definition:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{B}}\\,(t)} &=\\hfill & {{\\bf{T}}\\,(t)\\,\\times\\,{\\bf{N}}\\,(t)}\\hfill \\\\ \\hfill & =\\hfill & {\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}}\\\\\\frac{3}{5\\sqrt{t^{2}+1}}&+\\frac{5t}{5\\sqrt{t^{2}+1}}&-\\frac{4}{5\\sqrt{t^{2}+1}}\\\\ -\\frac{3t}{5\\sqrt{t^{2}+1}}&+\\frac{5}{5\\sqrt{t^{2}+1}}&\\frac{4t}{5\\sqrt{t^{2}+1}}\\end{vmatrix}} \\hfill \\\\ \\hfill & =\\hfill & {\\bigg(\\Big(+\\frac{5t}{5\\sqrt{t^{2}+1}}\\Big)\\Big(\\frac{4t}{5\\sqrt{t^{2}+1}}\\Big)-\\Big(-\\frac{4}{5\\sqrt{t^{2}+1}}\\Big)\\Big(-\\frac{5}{5\\sqrt{t^{2}+1}}\\Big)\\bigg)\\,{\\bf{i}} \\\\ +\\bigg(\\Big(\\frac{3}{5\\sqrt{t^{2}+1}}\\Big)\\Big(\\frac{4t}{5\\sqrt{t^{2}+1}}\\Big)-\\Big(-\\frac{4}{5\\sqrt{t^{2}+1}}\\Big)\\Big(-\\frac{3t}{5\\sqrt{t^{2}+1}}\\Big)\\bigg)\\,{\\bf{j}} \\\\ +\\bigg(\\Big(\\frac{3}{5\\sqrt{t^{2}+1}}\\Big)\\Big(+\\frac{5}{5\\sqrt{t^{2}+1}}\\Big)-\\Big(+\\frac{5t}{5\\sqrt{t^{2}+1}}\\Big)\\Big(-\\frac{3t}{5\\sqrt{t^{2}+1}}\\Big)\\bigg){\\bf{k}}}\\hfill \\\\ \\hfill & =\\hfill & {\\Big(\\frac{20t^{2}+20}{25(t^{2}+1)}\\Big)\\,{\\bf{i}}+\\Big(\\frac{-15-15t^{2}}{25(t^{2}+1)}\\Big)\\,{\\bf{k}}}\\hfill \\\\ \\hfill & =\\hfill & {20\\,\\Big(\\frac{t^{2}+1}{25(t^{2}+1)}\\Big)\\,{\\bf{i}}-15\\,\\Big(\\frac{t^{2}+1}{25(t^{2}+1)}\\Big)\\,{\\bf{k}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{4}{5}\\,{\\bf{i}}-\\frac{3}{5}\\,{\\bf{k}}.}\\hfill \\\\ \\hfill\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the unit normal vector for the vector-valued function [latex]{\\bf{r}}\\,(t)=(t^{2}-3)\\,{\\bf{I}}+(4t+1)\\,{\\bf{j}}[\/latex] and evaluate it at [latex]t=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794933264\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794933264\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\kappa=\\frac{6}{101^{3\/2}}\\approx{0.0059}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm244857\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=244857&theme=oea&iframe_resize_id=ohm244857&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>For any smooth curve in three dimensions that is defined by a vector-valued function, we now have formulas for the unit tangent vector\u00a0[latex]{\\bf{T}}[\/latex], the unit normal vector\u00a0[latex]{\\bf{N}}[\/latex], and the binormal vector\u00a0[latex]\\bf{B}[\/latex]. The unit normal vector and the binormal vector form a plane that is perpendicular to the curve at any point on the curve, called the <strong><span id=\"term136\" data-type=\"term\">normal plane<\/span><\/strong>. In addition, these three vectors form a frame of reference in three-dimensional space called the <strong><span id=\"term137\" data-type=\"term\">Frenet frame of reference<\/span><\/strong> (also called the <strong data-effect=\"bold\">TNB<\/strong> frame) (Figure 7). Lat, the plane determined by the vectors\u00a0[latex]\\bf{T}[\/latex] and\u00a0[latex]\\bf{N}[\/latex]\u00a0forms the osculating plane of\u00a0[latex]C[\/latex] at any point\u00a0[latex]P[\/latex] on the curve.<\/p>\n<div id=\"attachment_916\" style=\"width: 687px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-916\" class=\"size-full wp-image-916\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160628\/3-3-7.jpeg\" alt=\"This figure is the graph of a curve increasing and decreasing. Along the curve at 4 different points are 3 vectors at each point. The first vector is labeled \u201cT\u201d and is tangent to the curve at the point. The second vector is labeled \u201cN\u201d and is normal to the curve at the point. The third vector is labeled \u201cB\u201d and is orthogonal to T and N.\" width=\"677\" height=\"321\" \/><\/p>\n<p id=\"caption-attachment-916\" class=\"wp-caption-text\">Figure 4.\u00a0<span class=\"os-caption\">This figure depicts a Frenet frame of reference. At every point\u00a0[latex]P[\/latex]\u00a0on a three-dimensional curve, the unit tangent, unit normal, and binormal vectors form a three-dimensional frame of reference.<\/span><\/p>\n<\/div>\n<p>Suppose we form a circle in the osculating plane of\u00a0[latex]C[\/latex] at point\u00a0[latex]P[\/latex]\u00a0on the curve. Assume that the circle has the same curvature as the curve does at point\u00a0[latex]P[\/latex] and let the circle have radius [latex]r[\/latex]<em data-effect=\"italics\">.<\/em> Then, the curvature of the circle is given by [latex]1\/r[\/latex].\u00a0We call [latex]r[\/latex] the <span id=\"term138\" data-type=\"term\">radius of curvature<\/span> of the curve, and it is equal to the reciprocal of the curvature. If this circle lies on the concave side of the curve and is tangent to the curve at point [latex]P[\/latex]<em data-effect=\"italics\">,<\/em> then this circle is called the <span id=\"term139\" data-type=\"term\">osculating circle<\/span> of [latex]C[\/latex] at [latex]P[\/latex], as shown in the following figure.<\/p>\n<div id=\"attachment_918\" style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-918\" class=\"size-full wp-image-918\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160750\/3-3-8.jpeg\" alt=\"This figure is the graph of a curve with a circle in the middle. The bottom of the circle is the same as part of the curve. Inside of the circle is a vector labeled \u201cr\u201d. It starts at point \u201cP\u201d on the circle and points towards the radius. There is also a line segment perpendicular to the radius and tangent to point P.\" width=\"721\" height=\"305\" \/><\/p>\n<p id=\"caption-attachment-918\" class=\"wp-caption-text\">Figure 5.\u00a0In this osculating circle, the circle is tangent to curve [latex]C[\/latex]\u00a0at point\u00a0[latex]P[\/latex]\u00a0and shares the same curvature.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox tryit\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Interactive<\/h3>\n<p>For more information on osculating circles, see this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle1\" target=\"_blank\" rel=\"noopener nofollow\">demonstration<\/a> on curvature and torsion, this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle3\" target=\"_blank\" rel=\"noopener nofollow\">article<\/a> on osculating circles, and this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle2\" target=\"_blank\" rel=\"noopener nofollow\">discussion<\/a> of Serret formulas.<\/p>\n<\/div>\n<p>To find the equation of an osculating circle in two dimensions, we need find only the center and radius of the circle.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding the Equation of an Osculating Circle<\/h3>\n<p>Find the equation of the osculating circle of the helix defined by the function [latex]y=x^{3}-3x+1[\/latex] at [latex]x=1[\/latex].<\/p>\n<div id=\"fs-id1167793461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167705065166\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167705065166\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">Figure 9 shows the graph of [latex]y=x^{3}-3x+1[\/latex].<\/p>\n<div>\n<div id=\"attachment_919\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-919\" class=\"size-full wp-image-919\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21160915\/3-3-9.jpeg\" alt=\"This figure is the graph of a cubic function y = x^3-3x+1. The curve increases, reaches a maximum at x=-1, decreases passing through the y-axis at 1, then reaching a minimum at x =1 before increasing again.\" width=\"342\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-919\" class=\"wp-caption-text\">Figure 6.\u00a0We want to find the osculating circle of this graph at the point where\u00a0[latex]t=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>First, let&#8217;s calculate the curvature at [latex]x=1[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\LARGE{\\kappa=\\frac{|f''\\,(x)|}{(1+[f'\\,(x)]^{2})^{3\/2}}=\\frac{|6x|}{(1+[3x^{2}-3]^{2})^{3\/2}}}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>This gives [latex]\\kappa=6[\/latex].\u00a0Therefore, the radius of the osculating circle is given by [latex]R=\\frac{1}{\\kappa}=\\frac{1}{6}[\/latex].\u00a0Next, we then calculate the coordinates of the center of the circle. When [latex]x=1[\/latex],\u00a0the slope of the tangent line is zero. Therefore, the center of the osculating circle is directly above the point on the graph with coordinates [latex](1,\\ -1)[\/latex]. The center is located at [latex](1,\\ -\\frac{5}{6})[\/latex]. The formula for a circle with radius [latex]r[\/latex] and center [latex](h,\\ k)[\/latex] is given by [latex](x-h)^{2}+(y-k)^{2}=r^{2}[\/latex]. Therefore, the equation of the osculating circle is [latex](x-1)^{2}+(y+\\frac{5}{6})^{2}=\\frac{1}{36}[\/latex]. The graph and its osculating circle appears in the following graph.<\/p>\n<div id=\"attachment_920\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-920\" class=\"size-full wp-image-920\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21161011\/3-3-10.jpeg\" alt=\"This figure is the graph of a cubic function y = x^3-3x+1. The curve increases, reaches a maximum at x=-1, decreases passing through the y-axis at 1, then reaching a minimum at x =1 before increasing again. There is a small circle inside of the bend of the cure at x = 1.\" width=\"342\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-920\" class=\"wp-caption-text\">Figure 8.\u00a0The osculating circle has radius\u00a0[latex]R=\\frac{1}{6}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the equation of the osculating circle of the curve defined by the vector-valued function [latex]y=2x^{2}-4x+5[\/latex] at [latex]x=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794934264\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794934264\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">[latex]\\kappa=\\frac{4}{[1+(4x-4)^{2}]^{3\/2}}[\/latex] At the point [latex]x=1[\/latex], the curvature is equal to [latex]4[\/latex]. Therefore, the radius of the osculating circle is [latex]\\frac{1}{4}[\/latex]. A graph of this functions appears next:<\/p>\n<div id=\"attachment_921\" style=\"width: 372px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-921\" class=\"size-full wp-image-921\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21161115\/3-3-tryitans.jpeg\" alt=\"This figure is the graph of the function y = 2x^2-4x+5. The curve is a parabola opening up with vertex at (1, 3).\" width=\"362\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-921\" class=\"wp-caption-text\">Figure 9.<\/p>\n<\/div>\n<p>The vertex of this parabola is located at the point [latex](1,\\ 3)[\/latex]. Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are [latex](1,\\ \\frac{13}{4})[\/latex]. The equation of the osculating circle is [latex](x-1)^{2}+(y-\\frac{13}{4})^{2}=\\frac{1}{16}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{v}}\\,(t)={\\bf{r}}'\\,(t)=(2t-3)\\,{\\bf{i}}+2\\,{\\bf{j}}+{\\bf{k}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949625&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=11CyZk8-Fa0&amp;video_target=tpm-plugin-h4q06w32-11CyZk8-Fa0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.11_transcript.html\">transcript for \u201cCP 3.13\u201d here (opens in new window).<\/a><\/div>\n<p>&nbsp;<\/p>\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-823\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 3.13. <strong>Authored by<\/strong>: Ryan Melton. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 3.13\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-823","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/823","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":60,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/823\/revisions"}],"predecessor-version":[{"id":5822,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/823\/revisions\/5822"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/823\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=823"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=823"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=823"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=823"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}