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Unit tangent vector to a curve at a point

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the unit tangent vector T(t) to the curve r(t) at the point with the given value of the parameter, t.
    r(t)=<e^(2t), t^(-2), 1/(3t)>
    t=1

    2. Relevant equations

    none

    3. The attempt at a solution
    So first I took the derevative to get r'(t) which I got to be <2*e^(2x), -2t^(-3), -3t^(-2)> and plugged in the paramter, 1, so I got <2(e^2), -2,-3> and then I think I should divide that by its own magnitude, which I got to be the square root of (4(e^4) + 13)

    Buttt that's not working and Im not sure which part I went wrong on
    help?

    THANKS
     
  2. jcsd
  3. Sep 23, 2009 #2

    Dick

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    The derivative of 1/(3t) is -(1/3)*t^(-2). Not -3t^(-2).
     
  4. Sep 23, 2009 #3
    Ohhh ok... can you kinda explain that? Isn't 1/(3t) basically (3t)^(-1) and then you can use the power rule?
     
  5. Sep 23, 2009 #4
    ohhhh i have to use quotient rule?!
     
  6. Sep 23, 2009 #5
    ahhh yay i got it thanks!!! picking up calc againafter the summer sucks...
     
  7. Sep 23, 2009 #6

    Dick

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    Ok. d/dt (3t)^(-1)=(-1)*(3t)^(-2)*(d/dt (3t)). The last part comes from the chain rule. I think that's what your are forgetting. That's -3/(3t)^2=(-1)/(3t^2).
     
  8. Sep 23, 2009 #7

    Office_Shredder

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    Yes, but then you need to use the chain rule
    [tex] \frac{d}{dt} (3t)^{-1} = -1*(3t)^{-2}*\frac{d}{dt}(3t) = -1*(3t)^{-2}*3[/tex]
     
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