{"id":3869,"date":"2022-04-04T16:39:01","date_gmt":"2022-04-04T16:39:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3869"},"modified":"2022-10-29T00:55:02","modified_gmt":"2022-10-29T00:55:02","slug":"motion-vectors-in-the-plane-and-in-space","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/motion-vectors-in-the-plane-and-in-space\/","title":{"raw":"Motion Vectors in the Plane and in Space","rendered":"Motion Vectors in the Plane and in Space"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the velocity and acceleration vectors of a particle moving in space.<\/li>\r\n<\/ul>\r\n<\/div>\r\nOur starting point is using <span id=\"term140\" class=\"no-emphasis\" data-type=\"term\">vector-valued functions<\/span> to represent the position of an object as a function of time. All of the following material can be applied either to curves in the plane or to space curves. For example, when we look at the orbit of the planets, the curves defining these orbits all lie in a plane because they are elliptical. However, a particle traveling along a helix moves on a curve in three dimensions.\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]{\\bf{r}}\\,(t)[\/latex] be a twice-differentiable vector-valued function of the parameter [latex]t[\/latex] that represents\u00a0the position of an object as a function of time. The <strong><span id=\"term141\" data-type=\"term\">velocity vector<\/span><\/strong><span id=\"term141\" data-type=\"term\"> [latex]{\\bf{v}}\\,(t)[\/latex] of the object is given by<\/span>\r\n<p style=\"text-align: center;\">[latex]\\text{Velocity}={\\bf{v}}\\,(t)={\\bf{r}}'\\,(t).[\/latex]<\/p>\r\nThe\u00a0<strong>acceleration vector<\/strong> [latex]{\\bf{a}}\\,(t)[\/latex] is defined to be\r\n<p style=\"text-align: center;\">[latex]\\text{Acceleration}={\\bf{a}}\\,(t)={\\bf{v}}'\\,(t)={\\bf{r}}''\\,(t).[\/latex]<\/p>\r\nThe\u00a0<em>speed<\/em> is defined to be\r\n<p style=\"text-align: center;\">[latex]\\text{Speed}=v\\,(t)=\\left\\Vert{\\bf{v}}\\,(t)\\right\\Vert=\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert=\\frac{ds}{dt}.[\/latex]<\/p>\r\n\r\n<\/div>\r\nSince [latex]{\\bf{r}}\\,(t)[\/latex] can be in either two or three dimensions, these vector-valued functions can have either two or three components. In two dimensions, we define [latex]{\\bf{r}}\\,(t)=x\\,(t)\\,{\\bf{i}}+y\\,(t)\\,{\\bf{j}}[\/latex] and in three dimensions [latex]{\\bf{r}}\\,(t)=x\\,(t)\\,{\\bf{i}}+y\\,(t)\\,{\\bf{j}}+z\\,(t)\\,{\\bf{k}}[\/latex]. Then the velocity, acceleration, and speed can be written as shown in the following table.\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 13.4328%;\"><strong>Quantity<\/strong><\/td>\r\n<td style=\"width: 41.9569%;\"><strong>Two Dimensions<\/strong><\/td>\r\n<td style=\"width: 44.6102%;\"><strong>Three Dimensions<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 13.4328%;\">Position<\/td>\r\n<td style=\"width: 41.9569%;\">[latex]\\large{{\\bf{r}}\\,(t)=x\\,(t)\\,{\\bf{i}}+y\\,(t)\\,{\\bf{j}}}[\/latex]<\/td>\r\n<td style=\"width: 44.6102%;\">[latex]\\large{{\\bf{r}}\\,(t)=x\\,(t)\\,{\\bf{i}}+y\\,(t)\\,{\\bf{j}}+z\\,(t)\\,{\\bf{k}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 13.4328%;\">Velocity<\/td>\r\n<td style=\"width: 41.9569%;\">[latex]\\large{{\\bf{v}}\\,(t)=x'\\,(t)\\,{\\bf{i}}+y'\\,(t)\\,{\\bf{j}}}[\/latex]<\/td>\r\n<td style=\"width: 44.6102%;\">[latex]\\large{{\\bf{v}}\\,(t)=x'\\,(t)\\,{\\bf{i}}+y'\\,(t)\\,{\\bf{j}}+z'\\,(t)\\,{\\bf{k}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 13.4328%;\">Acceleration<\/td>\r\n<td style=\"width: 41.9569%;\">[latex]\\large{{\\bf{a}}\\,(t)=x''\\,(t)\\,{\\bf{i}}+y''\\,(t)\\,{\\bf{j}}}[\/latex]<\/td>\r\n<td style=\"width: 44.6102%;\">[latex]\\large{{\\bf{a}}\\,(t)=x''\\,(t)\\,{\\bf{i}}+y''\\,(t)\\,{\\bf{j}}+z''\\,(t)\\,{\\bf{k}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 13.4328%;\">Speed<\/td>\r\n<td style=\"width: 41.9569%;\">[latex]\\large{v\\,(t)=\\sqrt{(x'\\,(t))^{2}+(y'\\,(t))^{2}}}[\/latex]<\/td>\r\n<td style=\"width: 44.6102%;\">[latex]\\large{v\\,(t)=\\sqrt{(x'\\,(t))^{2}+(y'\\,(t))^{2}+(z'\\,(t))^{2}}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Studying Motion Along a PArabola<\/h3>\r\nA particle moves in a parabolic path defined by the vector-valued function [latex]{\\bf{r}}\\,(t)=t^{2}\\,{\\bf{i}}+\\sqrt{5-t^{2}}\\,{\\bf{j}}[\/latex], where [latex]t[\/latex] measures time in seconds.\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. Find the velocity, acceleration, and speed as functions of time.<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. Sketch the curve along with the velocity vector at time [latex]t=1[\/latex].<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793461764\" class=\"exercise\">[reveal-answer q=\"fs-id1167795055169\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167795055169\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>We use our definition:\r\n<div><\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}\\,(t)} &amp; =\\hfill &amp; {{\\bf{r}}'\\,(t)=2t\\,{\\bf{i}}-\\frac{t}{\\sqrt{5-t^{2}}}\\,{\\bf{j}}} \\hfill \\\\ \\hfill {{\\bf{a}}\\,(t)} &amp; =\\hfill &amp; {{\\bf{v}}'\\,(t)=2\\,{\\bf{i}}-5(5-t^{2})^{-3\/2}\\,{\\bf{j}}} \\hfill \\\\ \\hfill {v\\,(t)} &amp; =\\hfill &amp; {\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{(2t)^{2}+\\Big(-\\frac{t}{\\sqrt{5-t^{2}}}\\Big)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{4t^{2}+\\frac{t^{2}}{5-t^{2}}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{\\frac{21t^{2}-4t^{4}}{5-t^{2}}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>The graph of [latex]{\\bf{r}}\\,(t)=t^{2}\\,{\\bf{I}}+\\sqrt{5-t^{2}}\\,{\\bf{j}}[\/latex] is a portion of a parabola (Figure 1). The velocity vector at [latex]t=1[\/latex] is\r\n<div style=\"text-align: center;\">[latex]{\\bf{v}}\\,(1)={\\bf{r}}'\\,(1)=2\\,(1)\\,{\\bf{i}}-\\frac{1}{\\sqrt{5-(1)^{2}}}\\,{\\bf{j}}=2\\,{\\bf{i}}-\\frac{1}{2}\\,{\\bf{j}}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nand the acceleration vector at [latex]t=1[\/latex] is\r\n<div style=\"text-align: center;\">[latex]{\\bf{a}}\\,(1)={\\bf{v}}'\\,(t)=2\\,{\\bf{i}}-5(5-(1)^{2})^{-3\/2}\\,{\\bf{j}}=2\\,{\\bf{i}}-\\frac{5}{8}\\,{\\bf{j}}.[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNotice that the velocity vector is tangent to the path, as is always the case.\r\n<div>[caption id=\"attachment_923\" align=\"aligncenter\" width=\"342\"]<img class=\"size-full wp-image-923\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21162519\/3-4-1.jpeg\" alt=\"This figure is the graph of a curve in the first quadrant. The curve begins on the y axis slightly above y=2 and decreases to the x-axis at x=5. On the curve is a tangent vector labeled \u201cv(1)\u201d and is pointing towards the x-axis.\" width=\"342\" height=\"346\" \/> Figure 1.\u00a0This graph depicts the velocity vector at time [latex]t=1[\/latex] for a particle moving in a parabolic path.[\/caption]<\/div><\/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\nA particle moves in a path defined by the vector-valued function [latex]{\\bf{r}}\\,(t)=(t^{2}-3t)\\,{\\bf{i}}+(2t-4)\\,{\\bf{j}}+(t+2)\\,{\\bf{k}}[\/latex], where [latex]t[\/latex] measures time in seconds and where distance is measured in feet. Find the velocity, acceleration, and speed as functions of time.\r\n\r\n[reveal-answer q=\"fs-id1167794933164\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933164\"]\r\n\r\n[latex]\\begin{align*}\r\n\r\n\\hspace{5cm} {\\bf{v}}(t)&amp;={\\bf{r}}'(t)=(2t-3)\\,{\\bf{i}}+2\\,{\\bf{j}}+{\\bf{k}} \\\\\r\n\r\n{\\bf{a}}(t)&amp;={\\bf{v}}'(t)=2\\,{\\bf{I}} \\\\\r\n\r\nv(t)&amp;=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=\\sqrt{(2t-3)^{2}+2^{2}+1^{2}}=\\sqrt{4t^{2}-12t+14} \\end{align*} [\/latex]\r\n\r\n&nbsp;\r\n\r\nThe units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.\r\n\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=7949627&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=frfgju2zMDk&amp;video_target=tpm-plugin-oqnq7v8s-frfgju2zMDk\" 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\/CP3.14_transcript.html\">transcript for \u201cCP 3.14\u201d here (opens in new window).<\/a><\/center>To gain a better understanding of the velocity and acceleration vectors, imagine you are driving along a curvy road. If you do not turn the steering wheel, you would continue in a straight line and run off the road. The speed at which you are traveling when you run off the road, coupled with the direction, gives a vector representing your velocity, as illustrated in the following figure.\r\n\r\n[caption id=\"attachment_925\" align=\"aligncenter\" width=\"731\"]<img class=\"size-full wp-image-925\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21162609\/3-4-2.jpeg\" alt=\"This figure represents a curving road. On the road is a car. At the car there are two vectors. The first vector is tangent to the back of the car. The second vector comes out of the front of the car in the direction the car is heading. Both of the vectors are labeled \u201cvelocity vectors\u201d.\" width=\"731\" height=\"306\" \/> Figure 2.\u00a0At each point along a road traveled by a car, the velocity vector of the car is tangent to the path traveled by the car.[\/caption]\r\n\r\nHowever, the fact that you must turn the steering wheel to stay on the road indicates that your velocity is always changing (even if your speed is not) because your <em data-effect=\"italics\">direction<\/em> is constantly changing to keep you on the road. As you turn to the right, your acceleration vector also points to the right. As you turn to the left, your acceleration vector points to the left. This indicates that your velocity and acceleration vectors are constantly changing, regardless of whether your actual speed varies (Figure 3).\r\n\r\n[caption id=\"attachment_926\" align=\"aligncenter\" width=\"561\"]<img class=\"size-full wp-image-926\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21162652\/3-4-3.jpeg\" alt=\"This figure is a car. The path the car is traveling is an increasing curve represented by a dotted line. The center of the car is labeled \u201ctsub0\u201d on the curve. From this point there are two vectors that are orthogonal to each other. The first vector is asubt and the second vector is asubn. In between these two vectors is a vector labeled \u201ca\u201d. It has angle theta between vector a and asubt.\" width=\"561\" height=\"392\" \/> Figure 3.\u00a0The dashed line represents the trajectory of an object (a car, for example). The acceleration vector points toward the inside of the turn at all times.[\/caption]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the velocity and acceleration vectors of a particle moving in space.<\/li>\n<\/ul>\n<\/div>\n<p>Our starting point is using <span id=\"term140\" class=\"no-emphasis\" data-type=\"term\">vector-valued functions<\/span> to represent the position of an object as a function of time. All of the following material can be applied either to curves in the plane or to space curves. For example, when we look at the orbit of the planets, the curves defining these orbits all lie in a plane because they are elliptical. However, a particle traveling along a helix moves on a curve in three dimensions.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]{\\bf{r}}\\,(t)[\/latex] be a twice-differentiable vector-valued function of the parameter [latex]t[\/latex] that represents\u00a0the position of an object as a function of time. The <strong><span id=\"term141\" data-type=\"term\">velocity vector<\/span><\/strong><span id=\"term141\" data-type=\"term\"> [latex]{\\bf{v}}\\,(t)[\/latex] of the object is given by<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\text{Velocity}={\\bf{v}}\\,(t)={\\bf{r}}'\\,(t).[\/latex]<\/p>\n<p>The\u00a0<strong>acceleration vector<\/strong> [latex]{\\bf{a}}\\,(t)[\/latex] is defined to be<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Acceleration}={\\bf{a}}\\,(t)={\\bf{v}}'\\,(t)={\\bf{r}}''\\,(t).[\/latex]<\/p>\n<p>The\u00a0<em>speed<\/em> is defined to be<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Speed}=v\\,(t)=\\left\\Vert{\\bf{v}}\\,(t)\\right\\Vert=\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert=\\frac{ds}{dt}.[\/latex]<\/p>\n<\/div>\n<p>Since [latex]{\\bf{r}}\\,(t)[\/latex] can be in either two or three dimensions, these vector-valued functions can have either two or three components. In two dimensions, we define [latex]{\\bf{r}}\\,(t)=x\\,(t)\\,{\\bf{i}}+y\\,(t)\\,{\\bf{j}}[\/latex] and in three dimensions [latex]{\\bf{r}}\\,(t)=x\\,(t)\\,{\\bf{i}}+y\\,(t)\\,{\\bf{j}}+z\\,(t)\\,{\\bf{k}}[\/latex]. Then the velocity, acceleration, and speed can be written as shown in the following table.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 13.4328%;\"><strong>Quantity<\/strong><\/td>\n<td style=\"width: 41.9569%;\"><strong>Two Dimensions<\/strong><\/td>\n<td style=\"width: 44.6102%;\"><strong>Three Dimensions<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 13.4328%;\">Position<\/td>\n<td style=\"width: 41.9569%;\">[latex]\\large{{\\bf{r}}\\,(t)=x\\,(t)\\,{\\bf{i}}+y\\,(t)\\,{\\bf{j}}}[\/latex]<\/td>\n<td style=\"width: 44.6102%;\">[latex]\\large{{\\bf{r}}\\,(t)=x\\,(t)\\,{\\bf{i}}+y\\,(t)\\,{\\bf{j}}+z\\,(t)\\,{\\bf{k}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 13.4328%;\">Velocity<\/td>\n<td style=\"width: 41.9569%;\">[latex]\\large{{\\bf{v}}\\,(t)=x'\\,(t)\\,{\\bf{i}}+y'\\,(t)\\,{\\bf{j}}}[\/latex]<\/td>\n<td style=\"width: 44.6102%;\">[latex]\\large{{\\bf{v}}\\,(t)=x'\\,(t)\\,{\\bf{i}}+y'\\,(t)\\,{\\bf{j}}+z'\\,(t)\\,{\\bf{k}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 13.4328%;\">Acceleration<\/td>\n<td style=\"width: 41.9569%;\">[latex]\\large{{\\bf{a}}\\,(t)=x''\\,(t)\\,{\\bf{i}}+y''\\,(t)\\,{\\bf{j}}}[\/latex]<\/td>\n<td style=\"width: 44.6102%;\">[latex]\\large{{\\bf{a}}\\,(t)=x''\\,(t)\\,{\\bf{i}}+y''\\,(t)\\,{\\bf{j}}+z''\\,(t)\\,{\\bf{k}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 13.4328%;\">Speed<\/td>\n<td style=\"width: 41.9569%;\">[latex]\\large{v\\,(t)=\\sqrt{(x'\\,(t))^{2}+(y'\\,(t))^{2}}}[\/latex]<\/td>\n<td style=\"width: 44.6102%;\">[latex]\\large{v\\,(t)=\\sqrt{(x'\\,(t))^{2}+(y'\\,(t))^{2}+(z'\\,(t))^{2}}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div>\n<div class=\"textbox exercises\">\n<h3>Example: Studying Motion Along a PArabola<\/h3>\n<p>A particle moves in a parabolic path defined by the vector-valued function [latex]{\\bf{r}}\\,(t)=t^{2}\\,{\\bf{i}}+\\sqrt{5-t^{2}}\\,{\\bf{j}}[\/latex], where [latex]t[\/latex] measures time in seconds.<\/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-id1167795055169\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167795055169\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>We use our definition:\n<div><\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}\\,(t)} & =\\hfill & {{\\bf{r}}'\\,(t)=2t\\,{\\bf{i}}-\\frac{t}{\\sqrt{5-t^{2}}}\\,{\\bf{j}}} \\hfill \\\\ \\hfill {{\\bf{a}}\\,(t)} & =\\hfill & {{\\bf{v}}'\\,(t)=2\\,{\\bf{i}}-5(5-t^{2})^{-3\/2}\\,{\\bf{j}}} \\hfill \\\\ \\hfill {v\\,(t)} & =\\hfill & {\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert} \\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{(2t)^{2}+\\Big(-\\frac{t}{\\sqrt{5-t^{2}}}\\Big)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{4t^{2}+\\frac{t^{2}}{5-t^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{\\frac{21t^{2}-4t^{4}}{5-t^{2}}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>The graph of [latex]{\\bf{r}}\\,(t)=t^{2}\\,{\\bf{I}}+\\sqrt{5-t^{2}}\\,{\\bf{j}}[\/latex] is a portion of a parabola (Figure 1). The velocity vector at [latex]t=1[\/latex] is\n<div style=\"text-align: center;\">[latex]{\\bf{v}}\\,(1)={\\bf{r}}'\\,(1)=2\\,(1)\\,{\\bf{i}}-\\frac{1}{\\sqrt{5-(1)^{2}}}\\,{\\bf{j}}=2\\,{\\bf{i}}-\\frac{1}{2}\\,{\\bf{j}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>and the acceleration vector at [latex]t=1[\/latex] is<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{a}}\\,(1)={\\bf{v}}'\\,(t)=2\\,{\\bf{i}}-5(5-(1)^{2})^{-3\/2}\\,{\\bf{j}}=2\\,{\\bf{i}}-\\frac{5}{8}\\,{\\bf{j}}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Notice that the velocity vector is tangent to the path, as is always the case.<\/p>\n<div>\n<div id=\"attachment_923\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-923\" class=\"size-full wp-image-923\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21162519\/3-4-1.jpeg\" alt=\"This figure is the graph of a curve in the first quadrant. The curve begins on the y axis slightly above y=2 and decreases to the x-axis at x=5. On the curve is a tangent vector labeled \u201cv(1)\u201d and is pointing towards the x-axis.\" width=\"342\" height=\"346\" \/><\/p>\n<p id=\"caption-attachment-923\" class=\"wp-caption-text\">Figure 1.\u00a0This graph depicts the velocity vector at time [latex]t=1[\/latex] for a particle moving in a parabolic path.<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>A particle moves in a path defined by the vector-valued function [latex]{\\bf{r}}\\,(t)=(t^{2}-3t)\\,{\\bf{i}}+(2t-4)\\,{\\bf{j}}+(t+2)\\,{\\bf{k}}[\/latex], where [latex]t[\/latex] measures time in seconds and where distance is measured in feet. Find the velocity, acceleration, and speed as functions of time.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794933164\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794933164\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align*}    \\hspace{5cm} {\\bf{v}}(t)&={\\bf{r}}'(t)=(2t-3)\\,{\\bf{i}}+2\\,{\\bf{j}}+{\\bf{k}} \\\\    {\\bf{a}}(t)&={\\bf{v}}'(t)=2\\,{\\bf{I}} \\\\    v(t)&=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=\\sqrt{(2t-3)^{2}+2^{2}+1^{2}}=\\sqrt{4t^{2}-12t+14} \\end{align*}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.<\/p>\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=7949627&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=frfgju2zMDk&amp;video_target=tpm-plugin-oqnq7v8s-frfgju2zMDk\" 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\/CP3.14_transcript.html\">transcript for \u201cCP 3.14\u201d here (opens in new window).<\/a><\/div>\n<p>To gain a better understanding of the velocity and acceleration vectors, imagine you are driving along a curvy road. If you do not turn the steering wheel, you would continue in a straight line and run off the road. The speed at which you are traveling when you run off the road, coupled with the direction, gives a vector representing your velocity, as illustrated in the following figure.<\/p>\n<div id=\"attachment_925\" style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-925\" class=\"size-full wp-image-925\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21162609\/3-4-2.jpeg\" alt=\"This figure represents a curving road. On the road is a car. At the car there are two vectors. The first vector is tangent to the back of the car. The second vector comes out of the front of the car in the direction the car is heading. Both of the vectors are labeled \u201cvelocity vectors\u201d.\" width=\"731\" height=\"306\" \/><\/p>\n<p id=\"caption-attachment-925\" class=\"wp-caption-text\">Figure 2.\u00a0At each point along a road traveled by a car, the velocity vector of the car is tangent to the path traveled by the car.<\/p>\n<\/div>\n<p>However, the fact that you must turn the steering wheel to stay on the road indicates that your velocity is always changing (even if your speed is not) because your <em data-effect=\"italics\">direction<\/em> is constantly changing to keep you on the road. As you turn to the right, your acceleration vector also points to the right. As you turn to the left, your acceleration vector points to the left. This indicates that your velocity and acceleration vectors are constantly changing, regardless of whether your actual speed varies (Figure 3).<\/p>\n<div id=\"attachment_926\" style=\"width: 571px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-926\" class=\"size-full wp-image-926\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21162652\/3-4-3.jpeg\" alt=\"This figure is a car. The path the car is traveling is an increasing curve represented by a dotted line. The center of the car is labeled \u201ctsub0\u201d on the curve. From this point there are two vectors that are orthogonal to each other. The first vector is asubt and the second vector is asubn. In between these two vectors is a vector labeled \u201ca\u201d. It has angle theta between vector a and asubt.\" width=\"561\" height=\"392\" \/><\/p>\n<p id=\"caption-attachment-926\" class=\"wp-caption-text\">Figure 3.\u00a0The dashed line represents the trajectory of an object (a car, for example). The acceleration vector points toward the inside of the turn at all times.<\/p>\n<\/div>\n<\/div>\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-3869\">\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 3.14. <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":17,"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 3.14\",\"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-3869","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3869","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":7,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3869\/revisions"}],"predecessor-version":[{"id":5894,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3869\/revisions\/5894"}],"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\/3869\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3869"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3869"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3869"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3869"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}