Finding the Domain of a Vector Function

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, {\(t|t\geq 0\)}.