To find the domain of the vector function $\mathbf{r}(t) = \left\langle 36 - t^2, e^{-2t}, \ln(t + 4) \right\rangle$, we need to determine the set of all $t$ values for which each component of the vector function is defined.

Step-by-Step Analysis:

1. First component: $36 - t^2$

   • This is a polynomial expression and is defined for all real numbers $t$.
   • Domain: $(-\infty, \infty)$.

2. Second component: $e^{-2t}$

   • The exponential function is defined for all real numbers $t$.
   • Domain: $(-\infty, \infty)$.

3. Third component: $\ln(t + 4)$

   • The natural logarithm function $\ln(x)$ is defined only for $x > 0$.
   • Thus, $t + 4 > 0$, which simplifies to $t > -4$.
   • Domain: $(-4, \infty)$.

Combined Domain:

The overall domain of the vector function $\mathbf{r}(t)$ is the intersection of the domains of the individual components.

• The first and second components are defined for all $t$, which is $(-\infty, \infty)$.
• The third component is defined for $t > -4$, which is $(-4, \infty)$.

Therefore, the domain of the vector function $\mathbf{r}(t)$ is the intersection of $(-\infty, \infty)$ and $(-4, \infty)$, which is:

$(-4, \infty)$

This is the domain in interval notation.

Would you like more details on any part of the solution or have any questions?

Here are some related questions to further explore this topic:

1. What is the range of the function $\ln(t + 4)$?
2. How do you find the derivative of the vector function $\mathbf{r}(t)$?
3. What is the significance of the domain in vector functions?
4. How would the domain change if the logarithmic term were $\ln(t - 4)$ instead?
5. What is the behavior of the vector function as $t$ approaches $-4$?

Tip: When working with logarithmic functions, always ensure the argument inside the logarithm is positive to determine the domain.