Find the domain of the vector function $\mathbf { r } ( t ) = \left\langle 9 \sqrt { t } , \frac { 1 } { t - 3 } , \ln t \right\rangle$ .
A) $[ 0,3 ) \cup ( 3,9 ) \cup ( 9 , \infty )$
B) $( 0,3 ) \cup ( 3 , \infty )$
C) $( 0,3 ) \cup ( 3,9 ) \cup ( 9 , \infty )$
D) $( 0,9 ) \cup ( 9 , \infty )$

Question

Quizplus · Accepted Answer

The domain is determined by the constraints: $t > 0$ for $\ln t$, and $t \neq 3$ for $\frac{1}{t-3}$. Thus, the domain is $(0, 3) \cup (3, \infty)$.

Find the Domain of the Vector Function r(t)=⟨9t,1t−3,ln⁡t⟩\mathbf { r } ( t ) = \left\langle 9 \sqrt { t } , \frac { 1 } { t - 3 } , \ln t \right\rangler(t)=⟨9t​,t−31​,lnt⟩

Find the Domain of the Vector Function $\mathbf { r } ( t ) = \left\langle 9 \sqrt { t } , \frac { 1 } { t - 3 } , \ln t \right\rangle$