Find the domain of the vector function $\mathbf { r } ( t ) = \left\langle 8 \sqrt { t } , \frac { 1 } { t - 9 } , \ln t \right\rangle$.
A) $( 0,9 ) \cup ( 9 , \infty )$
B) $[ 0,8 ) \cup ( 8,9 ) \cup ( 9 , \infty )$
C) $( 0,8 ) \cup ( 8 , \infty )$
D) $( 0,8 ) \cup ( 8,9 ) \cup ( 9 , \infty )$

Question

Quizplus · Accepted Answer

The domain of the vector function $\mathbf{r}(t)$ is determined by the domains of its component functions. The first component, $8\sqrt{t}$, is defined for $t \geq 0$. The second component, $\frac{1}{t-9}$, is defined for all $t$ except $t = 9$, since division by zero is undefined. The third component, $\ln(t)$, is defined for $t > 0$. Combining these conditions, the overall domain is $t > 0$ and $t \neq 9$, which is $(0,9) \cup (9, \infty)$.

Find the Domain of the Vector Function r(t)=⟨8t,1t−9,ln⁡t⟩\mathbf { r } ( t ) = \left\langle 8 \sqrt { t } , \frac { 1 } { t - 9 } , \ln t \right\rangler(t)=⟨8t​,t−91​,lnt⟩

Find the Domain of the Vector Function $\mathbf { r } ( t ) = \left\langle 8 \sqrt { t } , \frac { 1 } { t - 9 } , \ln t \right\rangle$