Question

Find the domain of the vector-valued function. \(\mathbf{r}(t)=\mathbf{F}(t)+\mathbf{G}(t)\) where \(\mathbf{F}(t)=\cos t \mathbf{i}-\sin t \mathbf{j}+\sqrt{t} \mathbf{k}, \quad \mathbf{G}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}\)

Short answer

The domain of the vector-valued function \( \mathbf{r}(t) \) is \( t \geq 0 \).

Step-by-step solution

Analyze function components

Vector-valued function \(\mathbf{r}(t)\) is the vector sum of two separate vector functions, \( \mathbf{F}(t) \) and \( \mathbf{G}(t) \). While function \( \mathbf{G}(t)=\cos t \mathbf{i}+\sin t \mathbf{j} \) contains components that take on all real values of t (since both sine and cosine functions are defined for all real numbers), function \( \mathbf{F}(t)=\cos t \mathbf{i}-\sin t \mathbf{j}+\sqrt{t} \mathbf{k} \) has a square root function, which is only defined for \( t \geq 0 \) .

Determine the restricted domain

Since a domain constitutes the set of all possible t-values, we need to consider all restrictions. In this case, the square root term \( \sqrt{t} \) in \(\mathbf{F}(t) = \cos t \mathbf{i}-\sin t \mathbf{j}+\sqrt{t} \mathbf{k}\) implies that \( t \geq 0 \). This is because square root of a negative number is not defined in the real number system.

State the domain

Considering both vector functions making up the vector-valued function \( \mathbf{r}(t) \), the greatest restriction is given by \( t \geq 0 \) due to the square root term in \( \mathbf{F}(t) \). Thus, the domain of \( \mathbf{r}(t) \) is \( t \geq 0 \). This indicates that all non-negative real numbers comprise the domain of the vector-valued function.