To find the domain of the vector-valued function $\vec{r}(t) = \langle -4\sin(t), -5\cos(t), \ln(t+2) \rangle,$ we need to examine the domain of each component function separately.

1. For $-4 \sin(t)$ and $-5 \cos(t)$:

   • Both $\sin(t)$ and $\cos(t)$ are defined for all real values of $t$.
   • Therefore, there are no restrictions on $t$ from these two components.

2. For $\ln(t + 2)$:

   • The natural logarithm function, $\ln(x)$, is only defined when its argument is positive, i.e., $x > 0$.
   • Therefore, we require $t + 2 > 0$.
   • Solving this inequality gives $t > -2$.

Conclusion: The domain of $\vec{r}(t)$ is all values of $t$ such that $t > -2$. Thus, we can write the domain as: $\{ t \mid t > -2 \}.$

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Related Questions

1. What is the domain of $\ln(t - 1)$?
2. How would you find the domain of $\frac{\sin(t)}{\ln(t+2)}$?
3. What is the range of $\vec{r}(t) = \langle -4\sin(t), -5\cos(t), \ln(t+2) \rangle$?
4. How do you evaluate $\vec{r}(t)$ at a specific $t$ value, such as $t = 0$?
5. How would the domain change if $\ln(t+2)$ was replaced with $\ln(t^2 + 1)$?

Tip

Remember that the domain of a vector-valued function is the intersection of the domains of each component.