Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 853: 1

$t\in(-1,3)$

Here, $t+1\gt0$, since $ln(x)$ is undefined when $x\leq0$ This means that $t \gt-1$ Also, $9-t^2\gt0$, since $\sqrt(x)$ is undefined when $x\leq0$ This means that $t^2\leq 9 \implies t \leq\pm3$ or, $-3\leq t\leq3$ Since $t$ should be greater than $-1$ and must be less than or equal to $3$, we can conclude that $t\in ]-1,3]$.