{"id":1000,"date":"2025-10-15T12:57:38","date_gmt":"2025-10-15T12:57:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1000"},"modified":"2025-10-23T12:16:41","modified_gmt":"2025-10-23T12:16:41","slug":"motion-in-space-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/motion-in-space-learn-it-2\/","title":{"raw":"Motion in Space: Learn It 2","rendered":"Motion in Space: Learn It 2"},"content":{"raw":"

Components of the Acceleration Vector<\/h2>\r\nWe can combine concepts from arc length and curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion. Recall that the unit tangent vector [latex]{\\bf{T}}[\/latex] and the unit normal vector [latex]{\\bf{N}}[\/latex] form an osculating plane at any point [latex]P[\/latex] on the curve defined by [latex]{\\bf{r}}(t)[\/latex]. The acceleration vector [latex]{\\bf{a}}(t)[\/latex] always lies in this plane and can be written as a combination of these two unit vectors.\r\n\r\n
\r\n

the plane of the acceleration vector<\/h3>\r\n

The acceleration vector [latex]{\\bf{a}}(t)[\/latex] of an object moving along a curve traced out by a twice-differentiable function [latex]{\\bf{r}}(t)[\/latex] lies in the plane formed by the unit tangent vector [latex]{\\bf{T}}(t)[\/latex] and the principal unit normal vector [latex]{\\bf{N}}(t)[\/latex].\r\n[latex]\\\\[\/latex]<\/p>\r\n

The acceleration vector can be expressed as:<\/p>\r\n

<\/p>\r\n

[latex]{\\bf{a}}(t)=v'(t){\\bf{T}}(t)+[v(t)]^{2}\\kappa{\\bf{N}}(t)[\/latex]<\/p>\r\n

where [latex]v(t)[\/latex] is the speed of the object and [latex]\\kappa[\/latex] is the curvature of the curve traced out by [latex]{\\bf{r}}(t)[\/latex].<\/p>\r\n\r\n<\/section>

\r\n

Proof<\/strong><\/p>\r\n\r\n\r\n

\r\n\r\nBecause [latex]{\\bf{v}}(t)={\\bf{r}}'(t)[\/latex] and [latex]{\\bf{T}}(t)=\\frac{{\\bf{r}}'(t)}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex], we have [latex]{\\bf{v}}(t)=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert{\\bf{T}}(t)=v(t){\\bf{T}}(t)[\/latex]. Now we differentiate this equation:\r\n

[latex]{\\bf{a}}(t)={\\bf{v}}'(t)=\\frac{d}{dt}(v(t){\\bf{T}}(t))=v'(t){\\bf{T}}(t)+v(t){\\bf{T}}'(t).[\/latex]<\/p>\r\nSince [latex]{\\bf{N}}(t)=\\frac{{\\bf{T}}'(t)}{\\left\\Vert{\\bf{T}}'(t)\\right\\Vert}[\/latex], we know [latex]{\\bf{T}}'(t)=\\left\\Vert{\\bf{T}}'(t)\\right\\Vert{\\bf{N}}(t)[\/latex], so\r\n

[latex]{\\bf{a}}(t)=v'(t){\\bf{T}}(t)+v(t)\\left\\Vert{\\bf{T}}'(t)\\right\\Vert{\\bf{N}}(t).[\/latex]<\/p>\r\nA formula for curvature is [latex]\\kappa=\\frac{\\left\\Vert{\\bf{T}}'(t)\\right\\Vert}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex], so [latex]\\left\\Vert{\\bf{T}}'(t)\\right\\Vert=\\kappa\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=\\kappa v(t)[\/latex]. This gives [latex]{\\bf{a}}(t)=v'(t){\\bf{T}}(t)+\\kappa(v(t))^{2}{\\bf{N}}(t)[\/latex].\r\n\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n

The coefficients of [latex]{\\bf{T}}(t)[\/latex] and [latex]{\\bf{N}}(t)[\/latex] in this formula have special names. We call [latex]v'(t)[\/latex] the tangential component of acceleration<\/strong> and write it as [latex]a_{\\bf{T}}[\/latex]. We call [latex][v(t)]^{2}\\kappa[\/latex] the normal component of acceleration<\/strong> and write it as [latex]a_{\\bf{N}}[\/latex].<\/p>\r\n\r\n

\r\n

tangential and normal components of acceleration<\/h3>\r\n

Let [latex]{\\bf{r}}(t)[\/latex] be a vector-valued function that denotes the position of an object as a function of time. Then [latex]{\\bf{a}}(t)={\\bf{r}}''(t)[\/latex] is the acceleration vector. The tangential and normal components of acceleration are:<\/p>\r\n\r\n

    \r\n \t
  • Tangential component:<\/strong> [latex]a_{\\bf{T}}={\\bf{a}}\\cdot{\\bf{T}}=\\frac{{\\bf{v}}\\cdot{\\bf{a}}}{\\left\\Vert{\\bf{v}}\\right\\Vert}[\/latex]<\/li>\r\n \t
  • Normal component:<\/strong> [latex]a_{\\bf{N}}={\\bf{a}}\\cdot{\\bf{N}}=\\frac{\\left\\Vert{\\bf{v}}\\times{\\bf{a}}\\right\\Vert}{\\left\\Vert{\\bf{v}}\\right\\Vert}=\\sqrt{{\\left\\Vert{\\bf{a}}\\right\\Vert}^{2}-a^{2}_{\\bf{T}}}[\/latex]<\/li>\r\n<\/ul>\r\n

    These components are related by:<\/p>\r\n

    [latex]{\\bf{a}}(t)=a_{\\bf{T}}{\\bf{T}}(t)+a_{\\bf{N}}{\\bf{N}}(t)[\/latex]<\/p>\r\n

    where [latex]{\\bf{T}}(t)[\/latex] is the unit tangent vector and [latex]{\\bf{N}}(t)[\/latex] is the unit normal vector to the curve.<\/p>\r\n\r\n<\/section>\r\n

    Understanding Centripetal Acceleration<\/h3>\r\n

    The normal component of acceleration is also called the centripetal component of acceleration<\/strong> or the radial component of acceleration<\/strong>. To understand this concept, imagine traveling in a car on a circular track at constant speed. The acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration.<\/p>\r\n

    The same holds true for noncircular paths. Your body tends to travel in a straight line and resists the force from acceleration that pushes it toward the side. Notice in Figure 1 that at point [latex]B[\/latex], the acceleration vector points backward. This happens because the car is decelerating as it goes into the curve.<\/p>\r\n\r\n\r\n[caption id=\"attachment_928\" align=\"aligncenter\" width=\"640\"]\"This Figure 1. The tangential and normal components of acceleration can be used to describe the acceleration vector.[\/caption]\r\n\r\n

    \r\n

    The tangential and normal unit vectors at any point on the curve provide a frame of reference at that point. Think of the components this way:<\/p>\r\n\r\n

      \r\n \t
    • Tangential component ([latex]a_{\\bf{T}}[\/latex])<\/strong>: Measures how quickly the speed is changing (speeding up or slowing down)<\/li>\r\n \t
    • Normal component ([latex]a_{\\bf{N}}[\/latex])<\/strong>: Measures how quickly the direction is changing (how sharply you're turning)<\/li>\r\n<\/ul>\r\n

      The tangential and normal components of acceleration are simply the projections of the acceleration vector onto [latex]{\\bf{T}}[\/latex] and [latex]{\\bf{N}}[\/latex], respectively.<\/p>\r\n\r\n<\/section>

      A particle moves in a path defined by the vector-valued function [latex]{\\bf{r}}(t)=t^{2}{\\bf{i}}+(2t-3){\\bf{j}}+(3t^{2}-3t){\\bf{k}}[\/latex], where [latex]t[\/latex] measures time in seconds and distance is measured in feet.\r\n
        \r\n \t
      1. \r\n
          a. Find [latex]a_{\\bf{T}}[\/latex] and [latex]a_{\\bf{N}}[\/latex] as functions of [latex]t[\/latex].<\/ol>\r\n
            b. Find [latex]a_{\\bf{T}}[\/latex] and [latex]a_{\\bf{N}}[\/latex] at time [latex]t=2[\/latex].<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
            [reveal-answer q=\"fs-id1167795055160\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167795055160\"]\r\n
              \r\n \t
            1. Let's start with the first equation:\r\n
              <\/div>\r\n
              [latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}(t)} & =\\hfill & {{\\bf{r}}'(t)=2t{\\bf{i}}+2{\\bf{j}}+(6t-3){\\bf{k}}} \\hfill \\\\ \\hfill {{\\bf{a}}(t)} & =\\hfill & {{\\bf{v}}'(t)=2{\\bf{i}}+6{\\bf{k}}} \\hfill \\\\ \\hfill {a_{\\bf{T}}} & =\\hfill & {\\frac{{\\bf{v}}\\cdot{\\bf{a}}}{\\left\\Vert{\\bf{v}}\\right\\Vert}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{(2t{\\bf{i}}+2{\\bf{j}}+(6t-3){\\bf{k}})\\cdot(2{\\bf{i}}+6{\\bf{k}})}{\\left\\Vert2t{\\bf{i}}+2{\\bf{j}}+(6t-3){\\bf{k}}\\right\\Vert}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{4t+6(6t-3)}{\\sqrt{(2t)^{2}+2^{2}+(6t-3)^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{40t-18}{\\sqrt{40t^{2}-36t+13}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nThen we apply the second equation:\r\n
              [latex]\\begin{array}{ccc}\\hfill {a_{\\bf{N}}} & =\\hfill & {\\sqrt{\\left\\Vert{\\bf{a}}\\right\\Vert^{2}-a_{\\bf{T}}}} \\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{\\left\\Vert{2}{\\bf{i}}+6{\\bf{k}}\\right\\Vert^{2}-\\big(\\frac{40t-18}{\\sqrt{40t^{2}-36t+13}}\\big)^{2}}} \\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{4+36-\\frac{(40t-18)^{2}}{40t^{2}-36t+13}}} \\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{\\frac{40(40t^{2}-36t+13)-(1600t^{2}-1440t+324)}{40t^{2}-36t+13}}}\\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{\\frac{196}{40t^{2}-36t+13}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{14}{\\sqrt{40t^{2}-36t+13}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n <\/li>\r\n \t
            2. We must evaluate each of the answers from part a. at [latex]t=2[\/latex]:\r\n
              [latex]\\begin{array}{ccc}\\hfill {a_{\\bf{T}}(2)} & =\\hfill & {\\frac{40(2)-18}{\\sqrt{40(2)^{2}-36(2)+13}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{80-18}{\\sqrt{40(2)^{2}-36(2)+13}}=\\frac{62}{\\sqrt{101}}} \\hfill \\\\ \\hfill {a_{\\bf{N}}(2)}&=\\hfill & {\\frac{14}{\\sqrt{40(2)^{2}-36(2)+13}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{14}{\\sqrt{160-72+13}}=\\frac{14}{\\sqrt{101}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nThe units of acceleration are feet per second squared, as are the units of the normal and tangential components of acceleration.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

              Components of the Acceleration Vector<\/h2>\n

              We can combine concepts from arc length and curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion. Recall that the unit tangent vector [latex]{\\bf{T}}[\/latex] and the unit normal vector [latex]{\\bf{N}}[\/latex] form an osculating plane at any point [latex]P[\/latex] on the curve defined by [latex]{\\bf{r}}(t)[\/latex]. The acceleration vector [latex]{\\bf{a}}(t)[\/latex] always lies in this plane and can be written as a combination of these two unit vectors.<\/p>\n

              \n

              the plane of the acceleration vector<\/h3>\n

              The acceleration vector [latex]{\\bf{a}}(t)[\/latex] of an object moving along a curve traced out by a twice-differentiable function [latex]{\\bf{r}}(t)[\/latex] lies in the plane formed by the unit tangent vector [latex]{\\bf{T}}(t)[\/latex] and the principal unit normal vector [latex]{\\bf{N}}(t)[\/latex].
              \n[latex]\\\\[\/latex]<\/p>\n

              The acceleration vector can be expressed as:<\/p>\n

              \n

              [latex]{\\bf{a}}(t)=v'(t){\\bf{T}}(t)+[v(t)]^{2}\\kappa{\\bf{N}}(t)[\/latex]<\/p>\n

              where [latex]v(t)[\/latex] is the speed of the object and [latex]\\kappa[\/latex] is the curvature of the curve traced out by [latex]{\\bf{r}}(t)[\/latex].<\/p>\n<\/section>\n

              \n

              Proof<\/strong><\/p>\n

              \n

              Because [latex]{\\bf{v}}(t)={\\bf{r}}'(t)[\/latex] and [latex]{\\bf{T}}(t)=\\frac{{\\bf{r}}'(t)}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex], we have [latex]{\\bf{v}}(t)=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert{\\bf{T}}(t)=v(t){\\bf{T}}(t)[\/latex]. Now we differentiate this equation:<\/p>\n

              [latex]{\\bf{a}}(t)={\\bf{v}}'(t)=\\frac{d}{dt}(v(t){\\bf{T}}(t))=v'(t){\\bf{T}}(t)+v(t){\\bf{T}}'(t).[\/latex]<\/p>\n

              Since [latex]{\\bf{N}}(t)=\\frac{{\\bf{T}}'(t)}{\\left\\Vert{\\bf{T}}'(t)\\right\\Vert}[\/latex], we know [latex]{\\bf{T}}'(t)=\\left\\Vert{\\bf{T}}'(t)\\right\\Vert{\\bf{N}}(t)[\/latex], so<\/p>\n

              [latex]{\\bf{a}}(t)=v'(t){\\bf{T}}(t)+v(t)\\left\\Vert{\\bf{T}}'(t)\\right\\Vert{\\bf{N}}(t).[\/latex]<\/p>\n

              A formula for curvature is [latex]\\kappa=\\frac{\\left\\Vert{\\bf{T}}'(t)\\right\\Vert}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex], so [latex]\\left\\Vert{\\bf{T}}'(t)\\right\\Vert=\\kappa\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=\\kappa v(t)[\/latex]. This gives [latex]{\\bf{a}}(t)=v'(t){\\bf{T}}(t)+\\kappa(v(t))^{2}{\\bf{N}}(t)[\/latex].<\/p>\n

              [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

              The coefficients of [latex]{\\bf{T}}(t)[\/latex] and [latex]{\\bf{N}}(t)[\/latex] in this formula have special names. We call [latex]v'(t)[\/latex] the tangential component of acceleration<\/strong> and write it as [latex]a_{\\bf{T}}[\/latex]. We call [latex][v(t)]^{2}\\kappa[\/latex] the normal component of acceleration<\/strong> and write it as [latex]a_{\\bf{N}}[\/latex].<\/p>\n

              \n

              tangential and normal components of acceleration<\/h3>\n

              Let [latex]{\\bf{r}}(t)[\/latex] be a vector-valued function that denotes the position of an object as a function of time. Then [latex]{\\bf{a}}(t)={\\bf{r}}''(t)[\/latex] is the acceleration vector. The tangential and normal components of acceleration are:<\/p>\n

                \n
              • Tangential component:<\/strong> [latex]a_{\\bf{T}}={\\bf{a}}\\cdot{\\bf{T}}=\\frac{{\\bf{v}}\\cdot{\\bf{a}}}{\\left\\Vert{\\bf{v}}\\right\\Vert}[\/latex]<\/li>\n
              • Normal component:<\/strong> [latex]a_{\\bf{N}}={\\bf{a}}\\cdot{\\bf{N}}=\\frac{\\left\\Vert{\\bf{v}}\\times{\\bf{a}}\\right\\Vert}{\\left\\Vert{\\bf{v}}\\right\\Vert}=\\sqrt{{\\left\\Vert{\\bf{a}}\\right\\Vert}^{2}-a^{2}_{\\bf{T}}}[\/latex]<\/li>\n<\/ul>\n

                These components are related by:<\/p>\n

                [latex]{\\bf{a}}(t)=a_{\\bf{T}}{\\bf{T}}(t)+a_{\\bf{N}}{\\bf{N}}(t)[\/latex]<\/p>\n

                where [latex]{\\bf{T}}(t)[\/latex] is the unit tangent vector and [latex]{\\bf{N}}(t)[\/latex] is the unit normal vector to the curve.<\/p>\n<\/section>\n

                Understanding Centripetal Acceleration<\/h3>\n

                The normal component of acceleration is also called the centripetal component of acceleration<\/strong> or the radial component of acceleration<\/strong>. To understand this concept, imagine traveling in a car on a circular track at constant speed. The acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration.<\/p>\n

                The same holds true for noncircular paths. Your body tends to travel in a straight line and resists the force from acceleration that pushes it toward the side. Notice in Figure 1 that at point [latex]B[\/latex], the acceleration vector points backward. This happens because the car is decelerating as it goes into the curve.<\/p>\n

                \"This
                Figure 1. The tangential and normal components of acceleration can be used to describe the acceleration vector.<\/figcaption><\/figure>\n
                \n

                The tangential and normal unit vectors at any point on the curve provide a frame of reference at that point. Think of the components this way:<\/p>\n

                  \n
                • Tangential component ([latex]a_{\\bf{T}}[\/latex])<\/strong>: Measures how quickly the speed is changing (speeding up or slowing down)<\/li>\n
                • Normal component ([latex]a_{\\bf{N}}[\/latex])<\/strong>: Measures how quickly the direction is changing (how sharply you’re turning)<\/li>\n<\/ul>\n

                  The tangential and normal components of acceleration are simply the projections of the acceleration vector onto [latex]{\\bf{T}}[\/latex] and [latex]{\\bf{N}}[\/latex], respectively.<\/p>\n<\/section>\n

                  A particle moves in a path defined by the vector-valued function [latex]{\\bf{r}}(t)=t^{2}{\\bf{i}}+(2t-3){\\bf{j}}+(3t^{2}-3t){\\bf{k}}[\/latex], where [latex]t[\/latex] measures time in seconds and distance is measured in feet.<\/p>\n
                    \n
                  1. \n
                      <\/ol>\n
                        <\/ol>\n<\/li>\n<\/ol>\n
                        \n