Question

Find the tangential and normal components of the acceleration vector.

\({\rm{r(t) = costi + co}}{{\rm{s}}^{\rm{2}}}{\rm{tj + si}}{{\rm{n}}^{\rm{2}}}{\rm{tk}}\)

Short answer

Tangential and normal components of the acceleration vectors are:

\(\begin{aligned}{c}{a_T} = \frac{{2\sin 4t}}{{\sqrt {1 + 2{{\sin }^2}2t} }}\;\\\;\;{a_N} = \frac{{2\sqrt {2{{\cos }^2}2t} }}{{\sqrt {1 + 2{{\sin }^2}2t} }}\end{aligned}\)

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.

Tangential and normal components of the acceleration vector

Simplify the given values,

\({\bf{r}}(t) = < \left. {t,{{\cos }^2}t,{{\sin }^2}t > } \right\rangle \)

Determine the first and second derivatives,

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

\((2\sin x\cos x = \sin 2x)\)

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

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

Fill in the component formula as usual:

\(\begin{aligned}{c}{a_T} = \frac{{{{\bf{r}}^\prime }(t) \cdot {{\bf{r}}^{\prime \prime }}(t)}}{{\left| {{{\bf{r}}^\prime }(t)} \right|}}\\ = \frac{{1(0) + ( - \sin 2t)( - 2\cos 2t) + \sin 2t(2\cos 2t)}}{{\sqrt {1 + 2{{\sin }^2}2t} }}\end{aligned}\)

\( = \frac{{2\sin 2t\cos 2t + 2\sin 2t\cos 2t}}{{\sqrt {1 + 2{{\sin }^2}2t} }}\)

\(\begin{aligned}{c} = \frac{{2 \cdot 2\sin 2t\cos 2t}}{{\sqrt {1 + 2{{\sin }^2}2t} }}\\ = \frac{{2\sin 4t}}{{\sqrt {1 + 2{{\sin }^2}2t} }}\end{aligned}\)

Finding the cross product

Find the cross product first for the normal component.

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

Now, fill in the component formula as usual.

\(\begin{aligned}{c}{a_N} = \frac{{\left| {{{\bf{r}}^\prime }(t) \times {{\bf{r}}^{\prime \prime }}(t)} \right|}}{{\left| {{{\bf{r}}^\prime }(t)} \right|}}\\ = \frac{{\sqrt {{0^2} + {{( - 2\cos 2t)}^2} + {{( - 2\cos 2t)}^2}} }}{{\sqrt {1 + 2{{\sin }^2}2t} }}\end{aligned}\)

\(\begin{aligned}{c} = \frac{{\sqrt {2 \cdot 4{{\cos }^2}2t} }}{{\sqrt {1 + 2{{\sin }^2}2t} }}\\ = \frac{{2\sqrt {2{{\cos }^2}2t} }}{{\sqrt {1 + 2{{\sin }^2}2t} }}\end{aligned}\)

Rewrite the top using absolute value if you wish to keep the value positive:

\(\frac{{2\sqrt 2 |\cos 2t|}}{{\sqrt {1 + 2{{\sin }^2}2t} }}\)

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

\(\begin{aligned}{c}{a_T} = \frac{{2\sin 4t}}{{\sqrt {1 + 2{{\sin }^2}2t} }}\;\\\;\;{a_N} = \frac{{2\sqrt {2{{\cos }^2}2t} }}{{\sqrt {1 + 2{{\sin }^2}2t} }}\end{aligned}\)