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Q. 28

Expert-verified

Calculus

Found in: Page 880

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

For each of the vector-valued functions, find the unit tangent vector.
$r (t) = (\sin 2 t, \cos 2 t, t)$

$\frac{1}{\sqrt{5}} ⟨ 2 \cos 2 t, - 2 \sin 2 t, 1 ⟩$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1. Given Information

$Consider r (t) = ⟨ \sin 2 t, \cos 2 t, t ⟩ The objective is to find the unit tangent vector to r (t) The unit tangent vector The unit tangent vector of r (t) denoted by T (t) is defined as T (t) = \frac{r^{'} (t)}{∥r^{'} (t)∥} Consider r (t) = ⟨ \sin 2 t, \cos 2 t, t ⟩ First we compute r^{'} (t) : \begin{array}{l} r^{'} (t) = \frac{d}{d t} ⟨ \sin 2 t, \cos 2 t, t ⟩ \\ = <\frac{d}{d t} (\sin 2 t), \frac{d}{d t} (\cos 2 t), \frac{d}{d t} (t)> \\ = ⟨ 2 \cos 2 t, - 2 \sin 2 t, 1 ⟩ \end{array}$

Step2. Continue

$\begin{array}{l} ∥r^{'} (t)∥ = ∥ ⟨ 2 \cos 2 t, - 2 \sin 2 t, 1 ⟩ ∥ \\ = \sqrt{(2 \cos 2 t)^{2} + (- 2 \sin 2 t)^{2} + (1)^{2}} \\ = \sqrt{4 \cos^{2} 2 t + 4 \sin^{2} 2 t + 1} \\ = \sqrt{4 (\cos^{2} 2 t + \sin^{2} 2 t) + 1} \end{array} = \sqrt{5} \sin c e \sin^{2} 2 t + \cos^{2} 2 t = 1 The unit tangent vector to r (t) is \begin{array}{l} T (t) = \frac{r^{'} (t)}{∥r^{''} (t)∥} \\ = \frac{⟨ 2 \cos 2 t, - 2 \sin 2 t, 1 ⟩}{\sqrt{5}} \\ Thus the unit tangent vector to the given vector - valued function is \\ \frac{1}{\sqrt{5}} ⟨ 2 \cos 2 t, - 2 \sin 2 t, 1 ⟩ \end{array}$

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