# find the tangential component of acceleration for the curve

• Jun 16th 2012, 11:48 AM
icelated
find the tangential component of acceleration for the curve
find the tangential component of acceleration for the curve by

$\displaystyle r(t) = ti + \frac{\sqrt6}{2} t^2j + t^3k$

Would i use this formula?

$\displaystyle a_t = \frac {r'(t) \cdot r''(t)}{ ||r'(t)||}$

and

$\displaystyle a_n = \frac {r'(t) \times r''(t)}{ ||r'(t)||}$

so i find the first prime:

$\displaystyle r'(t) = i + \sqrt {6tj} + 3t^2k$

second prime

$\displaystyle r'(t) = \sqrt {6j} + 6tk$

magnitude

$\displaystyle ||r'(t)|| = \sqrt { 1^2 + (\sqrt{6t})^2 +(3t^2)^2 }$

$\displaystyle ||r'(t)|| = \sqrt { 1 + {6t^2} + 9t^4 }$

I am not sure how to further reduce the magnitude..

Am i correct so far? and would i multiple $\displaystyle r' \times r''$ now?

Can you help me find $\displaystyle a_t$ and i will worry about $\displaystyle a_n$
• Jun 16th 2012, 11:56 AM
Reckoner
Re: find the tangential component of acceleration for the curve
Quote:

Originally Posted by icelated
$\displaystyle a_n = \frac {r'(t) \times r''(t)}{ ||r'(t)||}$

No, the numerator should be the magnitude of the cross product.

Your work looks okay so far, except that your sloppy LaTeX makes it a little hard to read (e.g., you've got vectors under radicals). Keep going.
• Jun 16th 2012, 12:31 PM
icelated
Re: find the tangential component of acceleration for the curve
Going further

$\displaystyle r' \cdot r'' = ( ti + \frac{ \sqrt{6}} {2}} t^2j +t^3) \cdot ( \sqrt{6} j + 6tk})$

= $\displaystyle \frac{ \sqrt{36} }{2} t^2 j + 6t^4$

then,

$\displaystyle a_t = \frac {r'(t) \cdot r''(t)}{ ||r'(t)||} = \frac{ \frac{ \sqrt{36} }{2} t^2 j + 6t^4 }{\sqrt{1 + 6t^2 + 9t^4}} }$

Is this correct? and can i reduce this?
• Jun 17th 2012, 09:26 AM
Reckoner
Re: find the tangential component of acceleration for the curve
Quote:

Originally Posted by icelated
$\displaystyle r' \cdot r'' = ( ti + \frac{ \sqrt{6}} {2}} t^2j +t^3) \cdot ( \sqrt{6} j + 6tk})$?

It looks like you did $\displaystyle \mathbf r\cdot\mathbf r'',$ but what we want is $\displaystyle \mathbf r'\cdot\mathbf r''.$