Answer
(a) $ - 1 < t < 0$ ${\ \ }$ or ${\ \ }$ $0 < t \le 1$
(b) $0 < t \le 2$
Work Step by Step
a) Notice that ${t^{ - 1}}$ and ${\left( {t + 1} \right)^{ - 1}}$ are undefined at $t=0$ and $t=-1$, respectively. Since ${\sin ^{ - 1}}t$ is defined for $ - 1 \le t \le 1$, so, the domain of ${{\bf{r}}_1}\left( t \right)$ is $ - 1 < t < 0$ or $0 < t \le 1$.
(b) We must have $8 - {t^3} \ge 0$ and $t \ge 0$. Since $\ln t$ is defined for $0 < t < \infty $, so, the domain of ${{\bf{r}}_2}\left( t \right)$ is $0 < t \le 2$.
