Answer
$$\left\{ {t:0 \leqslant t \leqslant 2} \right\}{\text{ }}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \sqrt {4 - {t^2}} {\bf{i}} + \sqrt t {\bf{j}} - \frac{2}{{\sqrt {1 + t} }}{\bf{k}} \cr
& {\bf{r}}\left( t \right) = \left\langle {\sqrt {4 - {t^2}} ,\sqrt t , - \frac{2}{{\sqrt {1 + t} }}} \right\rangle \cr
& {\text{the vector value is of the form }}{\bf{r}}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \cr
& {\text{The domain of }}{\bf{r}}{\text{ is the largest set of values of }}t{\text{ on which all of }} \cr
& f\left( t \right){\text{, }}g\left( t \right){\text{ }}{\text{, and }}h\left( t \right){\text{ are defined}}{\text{. then }} \cr
& f\left( t \right) = \sqrt {4 - {t^2}} {\text{, }}g\left( t \right) = \sqrt t {\text{ and }}g\left( t \right) = - \frac{2}{{\sqrt {1 + t} }} \cr
& {\text{for }}f\left( t \right) = \sqrt {4 - {t^2}} {\text{ the function is defined for }}4 - {t^2} \geqslant 0,{\text{ then}} \cr
& 4 - {t^2} \geqslant 0 \cr
& {t^2} \leqslant 4 \cr
& \left| t \right| \leqslant 2 \cr
& {\text{the domain of }}f\left( t \right){\text{ is }}\left[ { - 2,2} \right] \cr
& {\text{for }}g\left( t \right) = {e^{\sqrt t }}{\text{ the function is defined for }}t \geqslant 0,{\text{ then}} \cr
& {\text{the domain of }}g\left( t \right){\text{ is }}\left[ {0,\infty } \right) \cr
& {\text{for }}h\left( t \right) = - \frac{2}{{\sqrt {1 + t} }}{\text{ the function is defined for }}1 + t > 0,{\text{ then}} \cr
& {\text{ }}t < - 1 \cr
& {\text{the domain of }}h\left( t \right){\text{ is }}\left( { - \infty , - 1} \right) \cr
& {\text{The domain of }}{\bf{r}}\left( t \right){\text{ is the intersection of the domains of }}f\left( t \right){\text{, }}g\left( t \right){\text{ and }}h\left( t \right) \cr
& \left[ { - 2,2} \right] \cap \left[ {0,\infty } \right) \cap \left( { - \infty , - 1} \right) = \left[ {0,2} \right] \cr
& {\text{The domain is}}:{\text{ }}\left\{ {t:0 \leqslant t \leqslant 2} \right\}{\text{ }} \cr} $$
