# Thread: Space curve and unit tangent vector.

1. ## 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?

2. Because $T(t) = \frac{{r'(t)}}{{\left\| {r'(t)} \right\|}}$ just find $r'(\pi )$ and proceed.

3. 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.

4. Look $\begin{gathered}
r'(t) = \left( {2t} \right)i + \left[ {4t\sin (2t)} \right]j + \left[ {4t\cos (2t)} \right]k \hfill \\
r'(\pi ) = \left( {2\pi } \right)i + \left( {4\pi } \right)k \hfill \\
\end{gathered}$

What is hard about finding $T(\pi ) = \frac{{r'(\pi )}}
{{\left\| {r'(\pi )} \right\|}}$
?

5. 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.