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Unit tangent vector (cal 3)

  1. Sep 27, 2005 #1
    positon vectors r(t) find the unit tangent vectors T(t) for the given value of t

    r(t) = (cos5t, sin5t)
    T(pi/4) = ( , )

    r(t) = (t^2, t^3)
    T(1) = ?

    r(t) = e^5t i + e^-1t j + t k
    T(2) = ? i+ ? j+ ? k

    now the to find it i use r'(t)/lr'(t)l
    I did that, but i get wrong answers i dont know what im doing wrong cuz all i c is that is how u find the unit tangent vectors... someone plz help me and explain what u did.. thanks alot!
  2. jcsd
  3. Sep 27, 2005 #2
    the derivative is the tangent vector.
    then divide it by its magnitude to get unit tangent vector.
  4. Sep 27, 2005 #3
    I did that and the damn computer say i got the wrong answers.... like the 1st one i took it derivative then i divid it by its magnitude which is squaring everything and sq rt it right thats what i did..
  5. Sep 27, 2005 #4
    give the answer that you have found for the first one.
  6. Sep 27, 2005 #5
    well i worked out the second one i got (1, 1) i really im doin something stupid
  7. Sep 27, 2005 #6
    i get 2/sq rt of 13, 3/ sq rt of 13
    what is wrong with your steps?
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