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

  • #1

Homework Statement


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


Homework Equations





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.
 

Answers and Replies

  • #2
LCKurtz
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You use the t value that gives you r(t) = (1,0,0). What t would that be?
 
  • #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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