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Math Help - find the tangential component of acceleration for the curve

  1. #1
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    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
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Re: find the tangential component of acceleration for the curve

    Quote Originally Posted by icelated View Post
     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.
    Thanks from icelated
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    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?
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Re: find the tangential component of acceleration for the curve

    Quote Originally Posted by icelated View Post
    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''.
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