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 $\displaystyle a_n$

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.

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?

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 $\displaystyle \mathbf r'\cdot\mathbf r''.$