Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 895: 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 $t^2\lt 9 \implies t \lt\pm3$ or, $-3\lt t\lt3$ Since $t$ should be greater than $-1$ and must be less than $3$, we can conclude that $t\in(-1,3)$.