Given that
[tex]\vec r(t) = \left\langle 4t e^{-t}, 8 \arctan(t), 8 e^t \right\rangle[/tex]
we first differentiate with respect to t to get the tangent vector, [tex]\vec T(t)[/tex] :
[tex]\vec T(t) = \dfrac{d\vec r}{dt} = \left\langle (4-4t) e^{-t}, \dfrac8{1+t^2}, 8e^t \right\rangle[/tex]
At t = 0, the tangent vector is
[tex]\vec T(0) = \left\langle 4, 8, 8 \right\rangle[/tex]
To get the unit tangent vector, multiply this by 1/(norm of tangent vector) :
[tex]\|\vec T(0)\| = \sqrt{4^2 + 8^2 + 8^2} = \sqrt{144} = 12[/tex]
Then the unit tangent vector is
[tex]\dfrac{\vec T(0)}{\|\vec T(0)\|} = \left\langle \dfrac4{12}, \dfrac8{12}, \dfrac8{12}\right\rangle = \boxed{\left\langle\dfrac13, \dfrac23, \dfrac23\right\rangle}[/tex]
