Step by step solution to "Find T(t), N(t), aT, and aN at the..."

socorraob

Answered question

2021-12-24

Find

T (t), N (t), a_{T}

, and

a_{N}

at the given time t for the curve r(t).

r (t) = t^{2} i + 2 t j, t = 1

Answer & Explanation

Elois Puryear

Beginner2021-12-25Added 30 answers

Consider $r (t) = t^{2} i + 2 t j$
$T (t) = \frac{r^{'} (t)}{| | r^{'} (t) | |}$
Here, $r^{'} (t) = t^{2} i + 2 t j$
S Thus, $T (t) = \frac{r^{'} (t)}{| | r^{'} (t) | |}$
$= \frac{2 t i + 2 j}{| | 2 t i + 2 j | |}$
$= \frac{2 t i + 2 j}{\sqrt{4 t^{2} + 4}}$
$= \frac{2 t i + 2 j}{2 \sqrt{t^{2} + 1}}$
$= \frac{t i + j}{\sqrt{t^{2} + 1}}$
Therefore, $T (1) = \frac{i + j}{\sqrt{2}}$ Substitute t=1 in T(t)

Paineow

Beginner2021-12-26Added 30 answers

N (t) = \frac{T^{'} (t)}{| | T^{'} (t) | |}

First find

T^{'} (t)

and

| | T^{'} (t) | |

to find N(t) as follows:

T (t) = \frac{t i + j}{\sqrt{t^{2} + 1}}

T^{'} (t) = \frac{i \times \frac{1}{\sqrt{t^{2} + 1}} - (t i + j) (- \frac{1}{2}) \frac{1}{{(1 + t^{2})}^{\frac{3}{2}}}}{(t^{2} + 1)}

T^{'} (t) = \frac{- j t + i}{{(1 + t^{2})}^{\frac{3}{2}}}

And

| | T^{'} (t) | | = \sqrt{\frac{{(- t)}^{2}}{{(1 + t^{2})}^{3}} + \frac{1}{{(1 + t^{3})}^{3}}}

| | T^{'} (t) | | = \sqrt{\frac{1 + {(t)}^{2}}{{(1 + t^{2})}^{3}}}

= \frac{1}{1 + t}

Therefore,

N (t) = \frac{T^{'} (t)}{| | T^{'} (t) | |}

N (t) = \frac{T^{'} (t)}{| | T^{'} (t) | |}

= \frac{i - j t}{{(1 + t^{2})}^{\frac{3}{2}}} \times (1 + t)

= \frac{i - j t}{\sqrt{1 + t^{2}}}

Substitute

t = 1

N (t)

as follows:

N (1) = \frac{i - j}{\sqrt{2}}

user_27qwe

Skilled2021-12-30Added 375 answers

Find $a_{T}, a_{N}$ at t=1 as follows
$\begin{matrix} a_{T} (1) = \frac{r^{'} (1) \times r^{″} (1)}{| | r^{'} (1) | |} \\ = \frac{(2 i + 2 j) \times (2 i)}{2 \sqrt{2}} \\ = \frac{4}{2 \sqrt{2}} \\ = \sqrt{2} \\ a_{N} (1) = \frac{| | r^{'} (1) \times r^{″} (1) | |}{| | r^{'} (1) | |} \\ = \frac{| | (2 i + 2 j) \times (2 i) | |}{2 \sqrt{2}} \\ = \frac{| | 4 k | |}{2 \sqrt{2}} \\ = \frac{4}{2 \sqrt{2}} \\ = \sqrt{2} \end{matrix}$

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