Given the vector function r(t) = <t2-t, sqrt(t+8), ln(t)/sin(t)>. (a) Find the domain of r(t).

For a rational function, the denominator cannot be zero.

$$t - 1 \neq 0 \implies t \neq 1$$

So, the domain for $f_1(t)$ is $D_1 = (-\infty, 1) \cup (1, \infty)$.

For a square root function, the expression inside the square root must be non-negative.

$$t + 8 \ge 0 \implies t \ge -8$$

So, the domain for $f_2(t)$ is $D_2 = [-8, \infty)$.

For the natural logarithm $\ln t$ to be defined, its argument must be positive.

$$t > 0$$

Also, the denominator cannot be zero.

$$\ln t \neq 0$$

This implies $t \neq e^0$, so $t \neq 1$.
Combining these conditions, the domain for $f_3(t)$ is $D_3 = (0, 1) \cup (1, \infty)$.

The domain of $\vec{r}(t)$ is $D = D_1 \cap D_2 \cap D_3$.

$$D_1 = (-\infty, 1) \cup (1, \infty)$$ $$D_2 = [-8, \infty)$$ $$D_3 = (0, 1) \cup (1, \infty)$$

We need $t$ to satisfy all conditions: $t \neq 1$, $t \ge -8$, and $t > 0$.
Combining $t \ge -8$ and $t > 0$, we get $t > 0$.
So, the overall conditions are $t > 0$ and $t \neq 1$.
This can be written in interval notation as $(0, 1) \cup (1, \infty)$.

The domain of $\vec{r}(t)$ is: