# find the tangential component of acceleration for the curve

• June 16th 2012, 12:48 PM
icelated
find the tangential component of acceleration for the curve
find the tangential component of acceleration for the curve by

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

Would i use this formula?

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

and

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

so i find the first prime:

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

second prime

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

magnitude

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

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

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

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

Can you help me find $a_t$ and i will worry about $a_n$
• June 16th 2012, 12:56 PM
Reckoner
Re: find the tangential component of acceleration for the curve
Quote:

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

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.
• June 16th 2012, 01:31 PM
icelated
Re: find the tangential component of acceleration for the curve
Going further

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

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

then,

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

Is this correct? and can i reduce this?
• June 17th 2012, 10:26 AM
Reckoner
Re: find the tangential component of acceleration for the curve
Quote:

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

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