To find the domain of the vector function $\mathbf{r}(t) = \left( \frac{t-1}{t+1} \right) \mathbf{i} + \sin(t) \mathbf{j} + \ln(25 - t^2) \mathbf{k}$, we need to identify the values of $t$ for which all three components of the vector are defined.

1. First component: $\frac{t-1}{t+1}$

   • This is a rational function, so we need to exclude any values of $t$ that make the denominator zero.
   • Set $t+1 = 0$, which gives $t = -1$.
   • Therefore, the first component is defined for all $t \neq -1$.

2. Second component: $\sin(t)$

   • The sine function is defined for all real numbers, so there are no restrictions on $t$ from this component.

3. Third component: $\ln(25 - t^2)$

   • The natural logarithm function $\ln(x)$ is only defined when $x > 0$.
   • Therefore, $25 - t^2 > 0$, which simplifies to: $25 > t^2 \quad \text{or} \quad -5 < t < 5$
   • This means the third component is defined when $t \in (-5, 5)$.

Combining the restrictions:

• From the first component: $t \neq -1$
• From the third component: $t \in (-5, 5)$

Thus, the domain of $\mathbf{r}(t)$ is $(-5, -1) \cup (-1, 5)$.

Final answer in interval notation:

$(-5, -1) \cup (-1, 5)$

Would you like further clarification on any part of this solution?

Here are 5 related questions:

1. How do you find the domain of a rational function?
2. What are the conditions for the domain of a logarithmic function?
3. Can you explain how to find the domain of a vector function in general?
4. How does the sine function affect the domain of a vector function?
5. What would happen if we had an exponential function in the vector function?

Tip: Always check for restrictions from all components when finding the domain of a vector function.