Question

Find the tangential and normal components of the acceleration vector.

\({\rm{r(t) = costi + sintj + tk}}\)

Short answer

\(\begin{aligned}{l}{a_T} = 0\;\\{a_N} = 1\end{aligned}\)

Step-by-step solution

Concept Introduction

The insertion of coordinates in the plane or three-dimensional space separates vector spaces from affine geometry. Vector spaces, in other terms, are mathematical objects. They are the central objects of study in linear algebra because they abstractly capture the geometry and algebra of linear equations.

Calculate tangential and normal components 

Using tangential and normal component formula,

\({\bf{r}}(t) = < \cos t,\sin t,t > \)

Find the 1st and 2nd derivatives,

\(\begin{aligned}{l}{{\bf{r}}^\prime }(t) = < - \sin t,\cos t,1 > \;\;\;{\rm{ }}velocity{\rm{ }}\\{{\bf{r}}^{\prime \prime }}(t) = < - \cos t, - \sin t,0 > {\rm{ }}acceleration{\rm{ }}\\\end{aligned}\)

Determine the length of \({{\bf{r}}^\prime }\)

\(\begin{aligned}{l}\left| {{{\bf{r}}^\prime }(t)} \right| = \sqrt {{{( - \sin t)}^2} + {{(\cos t)}^2} + {1^2}} \\ = \sqrt {{{\sin }^2}t + {{\cos }^2}t + 1} \\ = \sqrt {1 + 1} \\ = \sqrt 2 \end{aligned}\)

The tangential acceleration should be 0 because the speed, \(|{r^'}(t)|\), is constant. To be sure, plug into the tangential component formula.

\(\begin{aligned}{l}{a_T} = \frac{{{{\bf{r}}^\prime }(t) \cdot {{\bf{r}}^{\prime \prime }}(t)}}{{\left| {{{\bf{r}}^\prime }(t)} \right|}}\\ = \frac{{( - \sin t)( - \cos t)}}{{\sqrt 2 }}\\ = \frac{{\sin t\cos t - \sin t\cos t}}{{\sqrt 2 }}\\ = 0\end{aligned}\)

Finding the cross product

Find the cross product first for the normal component

\(\begin{aligned}{l}{{\bf{r}}^\prime }(t) \times {{\bf{r}}^{\prime \prime }}(t) = < \cos t(0) - (1)( - \sin t),(1)( - \cos t) - ( - \sin t)(0),( - \sin t)( - \sin t) - (\cos t)( - \cos t) > \\ = < \sin t, - \cos t,{\sin ^2}t + {\cos ^2}t > \\ = < \sin t, - \cos t,1 > \end{aligned}\)

Fill in the component formula as usual:

\(\begin{aligned}{c}{a_N} = \frac{{\left| {{{\bf{r}}^\prime }(t) \times {{\bf{r}}^{\prime \prime }}(t)} \right|}}{{\left| {{{\bf{r}}^\prime }(t)} \right|}}\\ = \frac{{\sqrt {{{(\sin t)}^2} + {{( - \cos t)}^2} + {1^2}} }}{{\sqrt 2 }}\\ = \frac{{\sqrt {{{\sin }^2}t + {{\cos }^2}t + 1} }}{{\sqrt 2 }}\\ = \frac{{\sqrt {1 + 1} }}{{\sqrt 2 }}\\ = 1\end{aligned}\)

Hence, the tangential and normal components of the acceleration vectors are:

\(\begin{aligned}{c}{a_T} = 0\;\\{a_N} = 1\end{aligned}\)