The problem asks to determine the domain $D$ of the vector function:

$\mathbf{R}(t) = 3t^2 \mathbf{i} - e^{2t} \mathbf{j} + \ln(t+1) \mathbf{k}$

To find the domain:

We analyze the individual components:

1. $3t^2 \mathbf{i}$: This is a polynomial, so it's defined for all real numbers $t \in \mathbb{R}$.

2. $e^{2t} \mathbf{j}$: The exponential function is also defined for all real numbers $t \in \mathbb{R}$.

3. $\ln(t+1) \mathbf{k}$: The natural logarithm $\ln(t+1)$ is only defined when $t + 1 > 0$, i.e., $t > -1$.

Conclusion:

The domain of the vector function $\mathbf{R}(t)$ is $t > -1$, or $D = (-1, \infty)$.

Thus, the correct answer is: $D = (-1, \infty)$

Would you like more details on this explanation? Here are some follow-up questions you might consider:

1. How do we find the domain of a logarithmic function in general?
2. What is the domain of the exponential function, and why?
3. How do you compute the domain for vector-valued functions?
4. Can you explain how polynomial functions impact the domain?
5. What happens to the domain when multiple functions are combined?

Tip: When finding the domain of a vector function, ensure that all components are defined for the same range of values of $t$.