If $\mathbf{r}(t) =\left(2t+2\right) \mathbf{i}+\sqrt{t}\mathbf{j}-\ln \left(t+3\right) \mathbf{k}$, then the components of $\mathbf{r=r}(t)$ are $$\begin{equation*} {x(t) =2t+2\qquad y(t) =\sqrt{t}\qquad z(t) =- \ln \left( t+3\right)} \end{equation*}$$

Since no domain is specified, it is assumed that the domain consists of all real numbers $t$ for which each of the component functions is defined. Since $x=x(t)$ is defined for all real numbers, $y=y(t)$ is defined for all real numbers $t\geq 0,$ and $z=z(t)$ is defined for all real numbers $t\,{>}\!\,-\!3$, the domain of the vector function $\mathbf{r=r}(t)$ is the intersection of these three sets, namely, the set of nonnegative real numbers, {$t|t\geq 0$}.