1. **State the problem:** Find the domain of the vector function $$\mathbf{r}(t) = \langle \ln(t + 2), \frac{t}{\sqrt{36 - t^2}}, 2^t \rangle$$ in interval notation. 2. **Analyze each component separately:** - For $$\ln(t + 2)$$, the argument must be positive: $$t + 2 > 0 \implies t > -2$$ - For $$\frac{t}{\sqrt{36 - t^2}}$$, the denominator $$\sqrt{36 - t^2}$$ must be positive (not zero or negative): $$36 - t^2 > 0 \implies -6 < t < 6$$ - For $$2^t$$, the domain is all real numbers, so no restriction here. 3. **Combine the domain restrictions:** - From $$\ln(t + 2)$$: $$t > -2$$ - From denominator: $$-6 < t < 6$$ The combined domain is the intersection: $$(-2, 6)$$ 4. **Final answer:** The domain of $$\mathbf{r}(t)$$ is $$\boxed{(-2, 6)}$$. This means $$t$$ can be any real number strictly greater than $$-2$$ and less than $$6$$.