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Finding unit normal vectors and normal/tangent components of accelerat

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data
    given r(t) = <t, 1/t,0> find T(t) N(t) aT and aN at t = 1


    2. Relevant equations
    T(t) = r'/||r'||
    N(t) = T'/||T'||
    aT = a . T = (v . a)/||T||
    aN = a . N = ||v x a||/||v|| = sqrt(||a||2 - aT2)


    3. The attempt at a solution
    for my T(t) I get <t2, -1 , 0>/(sqrt(1+t4) (I like keeping things in 3 dimensions even if there is no contribution in the z direction)
    and I am not calculating the normal vector if there isn't some algebra I can use to simplify this greatly , which I am not seeing, t4 + 1 I don't believe I can factor and I can't think of any other way to simplify this one so i'm just moving around through the problem set looking for some r(t) = <cost, sint, t> type of vector that I can simply differentiate and use identities with, desperately trying to avoid those other problems
     
    Last edited: Sep 24, 2013
  2. jcsd
  3. Sep 24, 2013 #2

    jbunniii

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    How did you get that? Your first equation in the "relevant equations" section is ##T(t) = r' / \|r\|##. What is ##r'## in this case?
     
  4. Sep 25, 2013 #3
    r' = <1,-1/t2>
    |r'| = sqrt(1 + 1/t4)
    did I mess something up when differentiating/doing algebra/both?
     
  5. Sep 25, 2013 #4
    |r'| = sqrt((t4+1)/t4) = (1/t2)sqrt(t4+1)
    which should be t2r'/(|r'|)
    which I believe is actually <t2,-1>/(sqrt(t4+1)
    which is the same thing...oops i'm not sure what i'm messing up here.
     
  6. Sep 25, 2013 #5

    jbunniii

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    OK, that looks fine. So what is ##T'##? Just use the quotient rule, it shouldn't be too horrible.
     
  7. Sep 25, 2013 #6
    well that's what i'm saying, are there any ways I can simplify this particular unit tangent vector before differentiating? I know how to differentiate it, but on an exam I will chew up a ton of time making sure I keep everything straight with the differentiation in terms of signs and cancellations.
     
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