Let $\mathbf { r } ( t ) = \left\langle \sqrt { 6 - t } , \frac { \left( e ^ { t } - 2 \right) } { t } , \ln ( t + 1 ) \right\rangle$ . Find the domain of $r$ .
A) $( - 2,6 ]$
B) $( - 2,0 ) \cup ( 0,6 ]$
C) $( 6 , \infty ]$
D) $( - \infty , - 2 )$
E) $[ 6,0 ) \cup ( 0,2 )$

Question

Quizplus · Accepted Answer

The domain is determined by the conditions $6 - t \geq 0$, $t \neq 0$, and $t + 1 > 0$, which results in $(-2, 0) \cup (0, 6]$.

Let r(t)=⟨6−t,(et−2)t,ln⁡(t+1)⟩\mathbf { r } ( t ) = \left\langle \sqrt { 6 - t } , \frac { \left( e ^ { t } - 2 \right) } { t } , \ln ( t + 1 ) \right\rangler(t)=⟨6−t​,t(et−2)​,ln(t+1)⟩

Let $\mathbf { r } ( t ) = \left\langle \sqrt { 6 - t } , \frac { \left( e ^ { t } - 2 \right) } { t } , \ln ( t + 1 ) \right\rangle$