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Curvature When t=0

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the curvature of

    ##r(t) = \frac 4 9 (1+t)^ \frac 3 2 i + \frac 4 9 (1-t)^ \frac 3 2 j + \frac 1 3 t \hat k## at t=0

    2. Relevant equations

    K=1/|v| * |dT/dt|

    3. The attempt at a solution

    Found v.

    ##v= \frac 2 3 (1+t)^ \frac 1 2 i - \frac 2 3 (1-t)^ \frac 1 2 j + 1/3 \hat k##
    |v|=1

    v/|v|= <(2/3)(1+t)^(1/2), -(2/3)(1-t)^(3/2), 1/3>

    ##\vec T = \frac v (|v|) ##

    Take derivative of T.

    K(0)=1/1 * |dT/dt|

    ##= \sqrt( \frac 1 3 ^2 + \frac 1 3 ^2 )= \sqrt \frac 2 3 ##

    The answer is ## (\sqrt 2) / 3##. Where did I go wrong?
     
    Last edited: Jun 1, 2014
  2. jcsd
  3. Jun 1, 2014 #2

    Simon Bridge

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    Please show your working for the time-derivative of the j term.
    Did you misplace a minus sign?

    Aside:
    * you used the letter k for two different things: avoid. Curvature is K or \kappa
    The way you wrote it, it looks like r(t) depends on the curvature.
    * it can help to indicate vectors: i,j,k are unit vectors?

    \vec r, \hat\imath, \hat\jmath, \hat k gets you ##\vec r, \hat\imath, \hat\jmath, \hat k##, or you can bold-face them.
     
  4. Jun 1, 2014 #3
    Ok. I tried to use capital K to denote curvature and changed the ##\hat k##

    The problem was copied as is. I fixed the minus signs but since they were squared anyway, it doesn't seem to make a difference.
     
    Last edited: Jun 1, 2014
  5. Jun 1, 2014 #4
    My question is

    1. was my method wrong?

    2. if no to 1 then is there something obviously wrong in any part?
     
    Last edited: Jun 1, 2014
  6. Jun 1, 2014 #5

    ehild

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    You did not square the denominators.

    K(0)=1/1 * |dT/dt|

    ##= \sqrt{ \left(\frac {1} {3}\right) ^2 + \left(\frac {1} {3}\right) ^2 }=\frac{ \sqrt { 2}} {3} ##

    ehild
     
  7. Jun 2, 2014 #6
    Thanks so much!
     
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