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Chapter 11: Problem 57

Find the domain of the following vector-valued functions. $$\mathbf{r}(t)=\sqrt{t+2} \mathbf{i}+\sqrt{2-t} \mathbf{j}$$

Short Answer

Expert verified
Answer: The domain of the vector-valued function is \(-2\leq t\leq 2\).

Step by step solution

01

Identify the square root expressions

In this vector-valued function, $$\mathbf{r}(t)=\sqrt{t+2} \mathbf{i}+\sqrt{2-t} \mathbf{j}$$ the square root expressions are \(\sqrt{t+2}\) and \(\sqrt{2-t}\).
02

Find the domain for each square root expression separately

We need to ensure that the expressions inside the square roots are non-negative, i.e., \(t+2 \geq 0\) and \(2-t \geq 0\). For \(t+2 \geq 0\), we have \(t\geq-2\). And for \(2-t \geq 0\), we have \(t\leq2\).
03

Find the intersection of the individual domains

Since both conditions need to be met (as both coordinate functions need to be real), the domain of the vector-valued function is the intersection of the two individual domains, i.e., \(-2\leq t\leq 2\).
04

Write the final answer

The domain of the vector-valued function $$\mathbf{r}(t)=\sqrt{t+2} \mathbf{i}+\sqrt{2-t} \mathbf{j}$$ is \(\boxed{-2\leq t\leq 2}\).

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