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Math Help - Find vectors T, N, and B at the given point

  1. #1
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    Find vectors T, N, and B at the given point

    Find the vectors T, N, and B at the given point.

    r(t) = {cos(t), sin(t), ln(cos(t))}, (1,0,0)
    I got T(t)= r'(t)/|r'(t)|= (-sint, cost, -sint/cost)/square root(1+tan^2t) = (-sint,cost,-tant)=(0,1,0) because t=0.

    then for N(t)= T'(t)/|T'(t)|=(-cost,-sint,-sec^2)/ sqaure root(1+sec^4t)=(-cost/sqaure root 2,-sint/sqaure root 2, sec^2t/sqaure root 2)

    I am just wondering if my math is correct or if I went wrong somewhere.

    B(t)= T(t)xN(t)= -1/sqaure root2[(sect+sinttant)i+(sint(sec^2t + 1)j -k]
    =(-1/sqaure root2,0,1/sqaure root 2)

    Did I solve this correctly?
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  2. #2
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    Looks like you found the correct answer, only

    \mathbf{N}=\frac{\mathbf{T'}}{|\mathbf{T'}|}=\frac  {1}{\sqrt{1+\sec^4 t}}\left<-\cos t,-\sin t,-\sec^2 t\right>.

    Therefore,

    \mathbf{B}=\mathbf{T}\times\mathbf{N}=\frac{|\cos t|}{\sqrt{1+\sec^4 t}}<br />
\left|\begin{array}{ccc}<br />
\mathbf{i} & \mathbf{j} & \mathbf{k} \\<br />
-\sin t & \cos t & -\tan t \\<br />
-\cos t & -\sin t & -\sec^2 t<br />
\end{array}\right|.
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  3. #3
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    if t=0 wouldn't 1/sqaure root(1+sec^4t)= 1/sqaure root 2?

    And where did you get |cost| from?
    Last edited by mr fantastic; October 17th 2009 at 06:28 PM. Reason: Merged posts
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  4. #4
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    Quote Originally Posted by andreaspohle View Post
    if t=0 wouldn't 1/sqaure root(1+sec^4t)= 1/sqaure root 2?
    Correct.

    And where did you get |cost| from?
    It is the value of |\mathbf{r'}(t)|^{-1} as shown by

    \frac{1}{\sqrt{1+\tan^2 t}}=\frac{1}{\sqrt{\sec^2 t}}=\frac{1}{|\sec t|}=|\cos t|.
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