need some clarification on this because my book isn't always clear with the notation

let $\displaystyle \vec{c}(t) = <e^t, e^{-t}, cos (t) > $

the particle will follow this path and flies off at $\displaystyle t = t_{0} = 1$ find where it is a $\displaystyle t_{1} = 2$

$\displaystyle \vec{c}'(t) = <e^{t} , -e^{-t}, -sin (t) >$

so $\displaystyle \vec{l}(2) = <e, \frac{1}{e}, cos (1) > + (2-1)<e, - \frac{1}{e}, - sin(1) >$

$\displaystyle = <e, \frac{1}{e}, cos (1) > + <e, - \frac{1}{e}, - sin(1) > $

correct?