Solved: "Find the Domain of the vector-valued function R..."

Clifton Sanchez

Answered question

2021-12-02

Find the Domain of the vector-valued function R
Let

\vec{R}

be a vector-valued function defined by

\vec{R} (t) = {\begin{cases} << \sqrt{4 - t} \frac{\sin 3 t}{t}, I n (t + 1) >> & , i f t \neq 0 \\ << 2, 3, 0 >> & , i f t = 0 \end{cases}

Give dom

\vec{R}

Determine whether

\vec{R}

is a continuous at

t = 0

Answer & Explanation

Drood1980

Beginner2021-12-03Added 16 answers

$\vec{R} (t) = {\begin{cases} << \sqrt{4 - t} \frac{\sin 3 t}{t}, I n (t + 1) >> & , i f t \neq 0 \\ << 2, 3, 0 >> & , i f t = 0 \end{cases}$
Domain of $\sqrt{4 - t}$
$\Rightarrow 4 - t \geq 0 \Rightarrow t \leq 4 \Rightarrow t ϵ (- \infty, 4]$
Domain of $\frac{lim 3 t}{t} \Rightarrow t \neq 0 \Rightarrow t ϵ (- \infty, 0) \cup (-, \infty)$
Domain of $\leq g (t + 1) \Rightarrow t + 1 > 0$
$\Rightarrow t ≻ 1 \Rightarrow t ϵ (- 1, \infty)$
Domain of $\sqrt{4 - t}, \frac{lim 3 t}{t} \neq \log (t + 1)$
$= (- \infty, 4] \cap {(- \infty, 0) \cap \infty)} \cap (- 1, + \infty)$
$= (- 1, 0) \cap (0, 4]$
but the $\vec{R} (t)$ is defined at $t = 0$
$∴$ dom $\vec{R} (t) = (- 1, 4]$
For contimuity of $\vec{R}$ at $t = 0$ we have to check the contimuityof cmponent of $\vec{R}$ .
$lim_{t \Rightarrow 0} \vec{R} (t) = lim_{t \Rightarrow 0} ⟨ \sqrt{4 - t}, \frac{\sin 3 t}{t}, \log (t + 1) ⟩$
$= ⟨ lim_{t \Rightarrow 0} \sqrt{4 - t}, lim_{t \Rightarrow 0} \frac{\sin 3 t}{t}, lim_{t \Rightarrow 0} \log (t + 1) ⟩$
$= ⟨ \sqrt{4 - 0}, 3 lim_{t \Rightarrow 0} \frac{\sin 3 t}{3 t,}, \log (0 + 1) ⟩$
$= ⟨ \sqrt{4}, 3 \times 1, \log 1 ⟩ {∴ lim_{t \Rightarrow} \frac{\sin x}{x} = 1}$
$= ⟨ 2, 3, 0 ⟩$
$= \vec{R} (0)$
$\Rightarrow lim_{t \Rightarrow 0} \vec{R} (t) <$

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