Find the domain of the vector-valued function given below. $\mathbf { r } ( t ) = \mathbf { F } ( t ) \times \mathbf { G } ( t )$ where
$\begin{array} { l }
\mathbf { F } ( t ) = t ^ { 3 } \mathbf { i } + \frac { 1 } { t + 4 } \mathbf { j } + ( t - 4 ) \mathbf { k } \
\mathbf { G } ( t ) = \sqrt { 1 - t ^ { 2 } } \mathbf { i } + t \mathbf { j } + ( t - 1 ) \mathbf { k }
\end{array}$
A) $( - \infty , \infty )$
B) $[ - 1,1 ]$
C) $[ - 4 , \infty )$
D) $[ 0,1 ]$
E) $[ 1 , \infty )$

Question

Find the domain of the vector-valued function given below. $\mathbf { r } ( t ) = \mathbf { F } ( t ) \times \mathbf { G } ( t )$ where 
$\begin{array} { l } 
\mathbf { F } ( t ) = t ^ { 3 } \mathbf { i } + \frac { 1 } { t + 4 } \mathbf { j } + ( t - 4 ) \mathbf { k } \
\mathbf { G } ( t ) = \sqrt { 1 - t ^ { 2 } } \mathbf { i } + t \mathbf { j } + ( t - 1 ) \mathbf { k }
\end{array}$
A) $( - \infty , \infty )$
B) $[ - 1,1 ]$
C) $[ - 4 , \infty )$
D) $[ 0,1 ]$
E) $[ 1 , \infty )$

Quizplus · Accepted Answer

The domain of $\mathbf{r}(t)$ is determined by the restrictions imposed by both $\mathbf{F}(t)$ and $\mathbf{G}(t)$. For $\mathbf{F}(t)$, the only restriction is that $t \neq -4$ to avoid division by zero. For $\mathbf{G}(t)$, $\sqrt{1 - t^2}$ requires $1 - t^2 \geq 0$, which simplifies to $-1 \leq t \leq 1$. Since the domain of $\mathbf{G}(t)$ is more restrictive, it determines the overall domain of $\mathbf{r}(t)$, which is $[-1, 1]$.

Find the Domain of the Vector-Valued Function Given Below r(t)=F(t)×G(t)\mathbf { r } ( t ) = \mathbf { F } ( t ) \times \mathbf { G } ( t )r(t)=F(t)×G(t)

Find the Domain of the Vector-Valued Function Given Below $\mathbf { r } ( t ) = \mathbf { F } ( t ) \times \mathbf { G } ( t )$