Find the vectors T, N, and B at the given point. r(t) =<t^2, \frac{2}{3}t^3 , t> and point <4,-\frac{16}{3},-2>

Question

Find the vectors T, N, and B at the given point.

r (t) =< t^{2}, \frac{2}{3} t^{3}, t >

and point

< 4, - \frac{16}{3}, - 2 >

Answer 1

smallq9

Answered 2021-06-02 Author has 106 answers

The solution to the vector is below:

This is helpful 14

Show answer

Answer 2

Given $R (t) =< t^{2}, \frac{2}{3} t^{3}, t >$ and point $< 4, - \frac{16}{3}, - 2 >$

The point $< 4, - \frac{16}{3}, - 2 >$ occursat t=-2

Find the derivative of the vector,

$R^{'} (t) =< 2 t, 2 t^{2}, 1 >$

$| R^{'} (t) | = \sqrt{(2 t)^{2} + (2 t^{2})^{2} + 1^{2}}$

$= \sqrt{4 t^{2} + 4 t^{4} + 1}$

$= \sqrt{(2 t^{2} + 1)^{2}}$

$= 2 t^{2} + 1$

Tangent vectors:

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

$= \frac{1}{2 t^{2} + 1} < 2 t, 2 t^{2}, 1 >$

$T (- 2) = \frac{1}{2 (- 2)^{2} + 1} < 2 (- 2), 2 (- 2)^{2}, 1 >$

$=< - \frac{4}{9}, \frac{8}{9}, \frac{1}{9} >$

$T^{'} (t) =< \frac{(2 t^{2} + 1) 2 - 2 t (4 t)}{(2 t^{2} + 1)^{2}}, \frac{(2 t^{2} + 1) 4 t - (2 t^{2}) (4 t)}{(2 t^{2} + 1)^{2}}, - \frac{4 t}{(2 t^{2} + 1)^{2}} >$

$=< \frac{4 t^{2} + 2 - 8 t^{2}}{(2 t^{2} + 1)^{2}}, \frac{8 t^{3} + 4 t - 8 t^{3}}{(2 t^{2} + 1)^{2}}, - \frac{4 t}{(2 t^{2} + 1)^{2})} >$

$=< \frac{2 - 4 t^{2}}{(2 t^{2} + 1)^{2}}, \frac{4 t}{(2 t^{2} + 1)^{2}}, - \frac{4 t}{(2 t^{2} + 1)^{2}} >$

$| T^{'} (t) | = \sqrt{\frac{(2 - 4 t^{2})^{2} + (4 t)^{2} + (- 4 t)^{2}}{(2 t^{2} + 1)^{4}}}$

$= \frac{1}{(2 t^{2} + 1)^{2}} \sqrt{4 - 16 t^{2} + 16 t^{4} + 16 t^{2} + 16 t^{2}}$

$= \frac{1}{(2 t^{2} + 1)^{2}} \sqrt{16 t^{4} + 16 t^{2} + 4}$

$= \frac{2 (2 t^{2} + 1)}{(2 t^{2} + 1)^{2}}$

$= \frac{2}{2 t^{2} + 1}$

The normal vectors.

$N (t) = \frac{T^{'} (t)}{| T^{'} (t) |} = Not exactly what you’re looking for? Ask My QuestionThis is helpful 88Show answerExpert Community at Your Service Live experts 24/7 Questions are typically answered in as fast
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Find the vectors T, N, and B at the given point. r(t) =<t^2, \frac{2}{3}t^3 , t> and point <4,-\frac{16}{3},-2>

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