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Find T(t) at (1,0,0) when r(t)=<cos(t),sin(t),ln(cos(t))>

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data
    r(t)=<cos(t),sin(t),ln(cos(t))> Find T(t) (and N(t)) at (1,0,0)


    2. Relevant equations



    3. The attempt at a solution
    T(t)=(dr/dt)/ldr/dtl

    dr/dt = <-sin(t),cos(t),-tan(t)>
    ldr/dtl = (sqrt)((-sin(t))^2+(cos(t))^2+(-tan(t))^2) = sec(t)

    therefore T(t)= (<-sin(t),cos(t),-tan(t)>)/sec(t) = <-tan(t), 1, -sin(t)/cos^2(t)>

    The problem I'm having is determining which value for the given point do I use to evaluate T(t) (and thereby N(t) and B(t)). How do I tell which value is best or is there a specific value from the point that is always used? Usually it's easier to tell if one of the values is given in radians, such as pi/2. Thanks for any help you can give me.
     
  2. jcsd
  3. Oct 31, 2011 #2

    LCKurtz

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    You use the t value that gives you r(t) = (1,0,0). What t would that be?
     
  4. Oct 31, 2011 #3
    Ok! That makes a LOT more sense now. So t = 0 so that way the vector shows the point (1,0,0). Thanks a lot. Something small like finding this out makes it make so much more sense now.
     
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