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Math Help - Space curve and unit tangent vector.

  1. #1
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    Space curve and unit tangent vector.

    Consider the space curve given by the equation

    r(t)=t^2i +[sin(2t)-2tcos(2t)]j+[cos(2t)+2tsin(2t)]k


    (a) Find the unit tangent vector T(t) and the equation of the tangent line at t=Pi.


    Everytime I attempt this problem I end up with the most despicable derivatives. I have tried to use trigonometric identities to clean up the problem I'm trying to solve a bit but it still leads to unatural results (a huge repetitive derivative computation). Is there some trigonometric identity that I'm missing out on or a trick that could produce results that make sense?
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  2. #2
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    Because T(t) = \frac{{r'(t)}}{{\left\| {r'(t)} \right\|}} just find r'(\pi ) and proceed.
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  3. #3
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    Yeah I know, but when I try to compute the unit tangent vector r'(t)/[r'(t)] the result I get for the derivative and when I try to make this derivative the magnitude is beyond disgusting. Is there some trigonometric identity involved or a trick? The most obvious one is cos^2(2t)+sin^2(2t)=1.
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  4. #4
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    Look \begin{gathered}<br />
  r'(t) = \left( {2t} \right)i + \left[ {4t\sin (2t)} \right]j + \left[ {4t\cos (2t)} \right]k \hfill \\<br />
  r'(\pi ) = \left( {2\pi } \right)i + \left( {4\pi } \right)k \hfill \\ <br />
\end{gathered}

    What is hard about finding T(\pi ) = \frac{{r'(\pi )}}<br />
{{\left\| {r'(\pi )} \right\|}}?
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  5. #5
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    What you posted isn't hard. Maybe the mistake I was making was that I didn't substitute the Pi into the differentiated equation. Thanks.
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