Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / Velocity and Acceleration
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 1-11 : Velocity and Acceleration

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

Show All Steps Hide All Steps

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,

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