Find the domain of the following vector-valued function. $$ r(t)=\sqrt{81-t^{2}} i+\sqrt{t} j+\frac{8}{\sqrt{6+t}} k $$ Select the correct choice below and fill in any answer boxes within your choice. $$ \begin{array}{ll} ext { A. }\{t: t \leq \square ext { or } t\square\} & ext { B. }\{t: \square

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Find the domain of the following vector-valued function. $$ r(t)=\sqrt{81-t^{2}} i+\sqrt{t} j+\frac{8}{\sqrt{6+t}} k $$ Select the correct choice below and fill in any answer boxes within your choice. $$ \begin{array}{ll} ext { A. }\{t: t \leq \square ext { or } t>\square\} & ext { B. }\{t: \square<\square ext { or } t \geq \square\} \ ext { E. }\{t: t<\square ext { or } t>\square\} & ext { D. }\{t: t \leq \square ext { or } t \geq \square\} \ ext { G. }\{t: \square \leq t<\square\} & ext { F. }\{t: \square

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To determine the domain of the vector-valued function:

$$
r(t) = \sqrt{81 - t^2}\, \mathbf{i} + \sqrt{t}\, \mathbf{j} + \frac{8}{\sqrt{6 + t}}\, \mathbf{k}
$$

we need to ensure that each component of the function is well-defined. This involves checking the conditions under which the square roots are real and the denominators are non-zero.

1. **First Component: $$\sqrt{81 - t^2}$$**
   $$
   81 - t^2 \geq 0 \implies t^2 \leq 81 \implies -9 \leq t \leq 9
   $$

2. **Second Component: $$\sqrt{t}$$**
   $$
   t \geq 0
   $$

3. **Third Component: $$\frac{8}{\sqrt{6 + t}}$$**
   $$
   6 + t > 0 \implies t > -6
   $$
   Additionally, the denominator $$\sqrt{6 + t}$$ cannot be zero, so:
   $$
   6 + t \neq 0 \implies t \neq -6
   $$

Combining all these conditions, we need $$t$$ to satisfy:
$$
0 \leq t \leq 9
$$
This is because:
- $$t \geq 0$$ ensures that $$\sqrt{t}$$ is defined.
- $$t \leq 9$$ ensures that $$\sqrt{81 - t^2}$$ is defined.
- $$t > -6$$ is automatically satisfied since $$t \geq 0$$.

Thus, the domain of $$r(t)$$ is all real numbers $$t$$ such that $$0 \leq t \leq 9$$.

**Answer:**

G. $$\{ t :\ 0 \leq t \leq 9 \}$$
The domain of the function is all real numbers $$ t $$ where $$ 0 \leq t \leq 9 $$.

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