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Section 12.11 : Velocity and Acceleration

2. Determine the tangential and normal components of acceleration for the object whose position is given by \(\vec r\left( t \right) = \left\langle {\cos \left( {2t} \right), - \sin \left( {2t} \right),4t} \right\rangle \).

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Start Solution

First, we need the first and second derivatives of the position function.

\[\vec r'\left( t \right) = \left\langle { - 2sin\left( {2t} \right), - 2\cos \left( {2t} \right),4} \right\rangle \hspace{0.25in}\hspace{0.25in}\vec r''\left( t \right) = \left\langle { - 4\cos \left( {2t} \right),4sin\left( {2t} \right),0} \right\rangle \] Show Step 2

Next, we’ll need the following quantities.

\[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {4{{\sin }^2}\left( {2t} \right) + 4{{\cos }^2}\left( {2t} \right) + 16} = \sqrt {20} = 2\sqrt 5 \] \[\vec r'\left( t \right)\centerdot \vec r''\left( t \right) = 8\sin \left( {2t} \right)\cos \left( {2t} \right) - 8\sin \left( {2t} \right)\cos \left( {2t} \right) + 0 = 0\] \[\begin{align*}\vec r'\left( t \right) \times \vec r''\left( t \right) & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 2sin\left( {2t} \right)}&{ - 2\cos \left( {2t} \right)}&4\\{ - 4\cos \left( {2t} \right)}&{4sin\left( {2t} \right)}&0\end{array}} \right|\\ & = - 16\cos \left( {2t} \right)\vec j - 8{\sin ^2}\left( {2t} \right)\vec k - 8{\cos ^2}\left( {2t} \right)\vec k - 16\sin \left( {2t} \right)\vec i\\ & = - 16\sin \left( {2t} \right)\vec i - 16\cos \left( {2t} \right)\vec j - 8\vec k\end{align*}\] \[\left\| {\vec r'\left( t \right) \times \vec r''\left( t \right)} \right\| = \sqrt {256{{\sin }^2}\left( {2t} \right) + 256{{\cos }^2}\left( {2t} \right) + 64} = \sqrt {320} = 8\sqrt 5 \] Show Step 3

The tangential component of the acceleration is,

\[{a_T} = \frac{{\vec r'\left( t \right)\centerdot \vec r''\left( t \right)}}{{\left\| {\vec r'\left( t \right)} \right\|}} = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]

The normal component of the acceleration is,

\[{a_N} = \frac{{\left\| {\vec r'\left( t \right) \times \vec r''\left( t \right)} \right\|}}{{\left\| {\vec r'\left( t \right)} \right\|}} = \frac{{8\sqrt 5 }}{{2\sqrt 5 }} = \require{bbox} \bbox[2pt,border:1px solid black]{4}\]