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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

    Hi,
    find vectors T, N, and B at the given point
    r(t)= <cost, sint, lncost>, (1,0,0)

    I found T(t) = <-sint/tant, cost,tant, -1>
    Is this correct?
    Finding N would be messy 'quotient rule' or could I multiply T(t) by tant to remove it from the denominator?
    I don't know what to do with the point (1,0,0)

    Thanks.
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  2. #2
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    help?

    please
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by khuezy View Post
    Hi,
    find vectors T, N, and B at the given point
    r(t)= <cost, sint, lncost>, (1,0,0)

    I found T(t) = <-sint/tant, cost,tant, -1>
    Is this correct?
    Finding N would be messy 'quotient rule' or could I multiply T(t) by tant to remove it from the denominator?
    I don't know what to do with the point (1,0,0)

    Thanks.
    Recall the definitions of \bold T(t),~\bold N(t),\textit{ and }\bold B(t):

    \bold T(t)=\frac{\bold r'(t)}{||\bold r'(t)||}

    We need to find \bold r'(t)\textit{ and }||\bold r'(t)||

    \bold r'(t)=\left<-\sin t,~\cos t,~-\tan t\right>

    ||\bold r'(t)||=\sqrt{\sin^2t+\cos^2t+\tan^2t}=\sqrt{1+\ta  n^2t}=\sqrt{\sec^2t}=\sec t=\frac{1}{\cos t}

    Therefore, \bold T(t)=\left<-\tfrac{1}{2}\sin(2t),~\cos^2t,~-\sin t\right>

    What do we do with the point?

    After a little examination, this point corresponds to the value of t=0 [To see why this is, compare the x,y,z coordinates to the corresponding counterpart defined in the vector \bold r(t); i.e. when is \cos t=1, \sin t=0 and \ln (\cos t)=0]

    Thus, \bold T(0)=\left<0,~1,~0\right>

    -------------------------------------------------------------------------

    \bold N(t)=\frac{\bold T'(t)}{||\bold T'(t)||}

    We need to find \bold T'(t)\textit{ and }||\bold T'(t)||

    \bold T'(t)=\left<-\cos (2t),~-\sin(2t),~-\cos t\right>

    ||\bold T'(t)||=\sqrt{\sin^2(2t)+\cos^2(2t)+\cos^2t}=\sqrt  {1+\cos^2t}

    Therefore, \bold N(t)=\left<-\frac{\cos(2t)}{\sqrt{1+\cos^2t}},~-\frac{\sin(2t)}{\sqrt{1+\cos^2t}},~-\frac{\cos t}{\sqrt{1+\cos^2t}}\right>

    Thus, \bold N(0)=\left<-\frac{1}{\sqrt{2}},~0,~-\frac{1}{\sqrt{2}}\right>

    -------------------------------------------------------------------------

    Since I found \bold T(t) and \bold N(t) for you, I would like you to find \bold B(t), given that

    \bold B(t)=\bold T(t)\times \bold N(t)

    -------------------------------------------------------------------------

    Does this make sense?

    --Chris
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