Question

Find the domain of the vector functions, r(t), listed below.
using interval notation.

a) r(t)=⟨ln(2t),√(t+7),1/(√(14−t))⟩
b) r(t)=⟨√(t−1),sin(1t),t^2⟩
c) r(t)=⟨ e^(−1t),t/(√(t^2−1),t^(1/3)⟩

Answer 1

The domain of the vector function is the intersection of the domains of its component functions.

a) $r(t)=⟨ln(2t),\sqrt{t+7},\frac{1}{\sqrt{14−t}}⟩.$

$ln(2t)$ exists for $2t>0$, hence $t>0$ or $t \in (0, \infty);$

$\sqrt{t+7}$ exists for $t+7 \geq 0$, $t \geq -7$ or $t \in [-7, \infty);$

$\frac{1}{\sqrt{14−t}}$ exists when $14-t>0$, or $t<14,$ $t \in (-\infty, 14).$

$(0, \infty)\bigcap [-7, \infty) \bigcap (-\infty, 14)=(0,14).$

Answer. $(0,14).$

b) $r(t)=⟨\sqrt{t−1},sin(1t),t^2⟩.$

$\sqrt{t−1}$ exists when $t-1 \geq0,$ $t \geq 1$ or $t \in [1, \infty);$

$sin(1t)$ exists for $t \in \R;$

$t^2$ exists for $t \in \R.$

$[1, \infty) \cap \R \cap \R=[1, \infty).$

Answer. $[1, \infty).$

c) $r(t)=⟨ e^{−1t},\frac{t}{\sqrt{t^2−1}},t^{1/3}⟩.$

$e^{−1t}$ exists for $t \in \R;$

$\frac{t}{\sqrt{t^2−1}}$ exists when $t^2-1>0,$ $t^2>1,$ $t \in (-\infty,-1) \cup (1, \infty).$

$t^{1/3}$ exists for $t \in \R.$

$((-\infty,-1) \cup (1, \infty)) \cap \R \cap \R=(-\infty,-1) \cup (1, \infty).$

Answer. $(-\infty,-1) \cup (1, \infty).$