## Calculus (3rd Edition)

The domain of $r(t)$ is $$D=\{t\in R; t\neq 0, -1\}.$$
Given $$r(t)=e^ti+\frac{1}{t}j+(t+1)^{-3}k$$ We know that $1/t$ is not defined for $t=0$ and $(t+1)^{-3}$ is not defined for $t=-1$ (because we can not divide by $0$). Thus, the domain of $r(t)$ is $$D=\{t\in R; t\neq 0, -1\}.$$