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jippie771h

Answered question

2021-12-04

Working with vector-valued functions For each vector-valuedfunction r, carry out the following steps.
a. Evaluate

lim_{t \Rightarrow 0} r (t) and lim_{t \Rightarrow \infty} r (t)

, if each exists.
b. Find

r^{'} (t)

and evaluate

r^{'} (0)

.
c. Find

r t)

.
d. Evaluate

\int r (t) d t .

r (t) = ⟨ t + 1, t^{2} - 3 ⟩

Answer & Explanation

Dona Hall

Beginner2021-12-05Added 15 answers

Tofind:
a) $lim_{t \Rightarrow 0} r (\begin{matrix} t \end{matrix}) a n d lim_{t \Rightarrow \infty} r (\begin{matrix} t \end{matrix})$
b) r'(t) and evaluate r'(0).
c) r''(t)
d) $\int r (t) d t$
d) fr(t)dr
Given vector valued function is $r (t) = < t + 1, t^{2} - 3 >$
a) To find: $lim_{t \Rightarrow 0} r (\begin{matrix} t \end{matrix}) a n d lim_{t \Rightarrow \infty} r (\begin{matrix} t \end{matrix})$
$lim_{t \Rightarrow 0} r (\begin{matrix} t \end{matrix}) = lim_{t \Rightarrow 0} =< 1, - 3 >$
and
$lim_{t \Rightarrow \infty} r (\begin{matrix} t \end{matrix}) = lim_{t \Rightarrow \infty} =$ does not exist
b) To find r'(t) and evaluate r'(0):
$r^{'} (t) = < \frac{d}{d t} (t + 1), \frac{d}{d t} (t^{2} - 3) \geq < 1, 2 t >$
Substitute t=0 in r'(t),
$\Rightarrow r^{'} (0) = < 1, 0 >$
c) To find: r''(t)
$r^{″} (t) = \frac{d}{d t} (r^{'} (t)) = < \frac{d}{d t} (1), \frac{d}{d t} (2 t) \geq < 0, 2 >$
d) To find $\int r (t) d t :$
$\int r (t) d t = \int d t = \int (t + 1) d t, \int (t^{2} - 3) d t$
$= < \frac{t^{2}}{2} + t + C, \frac{t^{3}}{3} - 3 t + D >$
Therefore,
a) $lim_{t \Rightarrow 0} r (\begin{matrix} t \end{matrix}) =< 1, - 3 >$
b) $r^{'} (t) = < 1, 2 t >$
$r^{'} (0) = < 1, 0 >$
c) $r^{″} (t) = < 0, 2 >$
d) $\int r (t) d t = < \frac{t^{2}}{2} + t + C, \frac{t^{3}}{3} - 3 t + D >$

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