Answer
a) $t \in (-1,0) \cup (0,2]$
b) $\lim\limits_{t \to 0} r(t)=\lt \sqrt 2, 1, 0 \gt$
c) $r'(t) =\lt \dfrac{-1}{2\sqrt {2-t}},\dfrac{te^t-e^t+1}{t^2}, \dfrac{1}{t+1} \gt$
Work Step by Step
a) Domain of the $x$ component is given as:
$\sqrt{2-t}$;
domain is $t \in (-\infty, 2]$
Domain of the $y$ component is given as:
$\dfrac{e^t-1}{t}$;
domain is $t \in (-\infty,0) \cup (0, \infty)$
Domain of the $z$ component is:
$ln(1+t)$
domain is $t \in (-1, \infty)$
Hence, we have $t \in (-1,0) \cup (0,2]$
b) Here, we have $\lim\limits_{t \to 0}\sqrt {2-t}=\sqrt 2$
and $\lim\limits_{t \to 0} \dfrac{e^t}{t}=1$; $ \lim\limits_{t \to 0} \ln(1+t)=\ln (1)=0$
Hence, we have $\lim\limits_{t \to 0} r(t)=\lt \sqrt 2, 1, 0 \gt$
c) Take the derivative of each component.
Thus, we have
$r'(t) =\lt \dfrac{-1}{2\sqrt {2-t}},\dfrac{te^t-e^t+1}{t^2}, \dfrac{1}{t+1} \gt$
