Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises: 23

Answer

$\textbf{T}(t) = \langle 0, \frac{-sin(2t)}{\sqrt {1 + 3cos^2(2t)}}, \frac{2cos(2t)}{\sqrt {1 + 3cos^2(2t)}}\rangle $

Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$ $\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ $\textbf{r}(t) = \langle 8, cos(2t), 2sin(2t)\rangle$ $\textbf{r}'(t) = \langle 0, -2sin(2t), 4cos(2t)\rangle$ $|\textbf{r}'(t)| = \sqrt {(0)^2 + (-2sin(2t))^2 + (4cos(2t))^2} = \sqrt {4sin^2(2t) + 16cos^2(2t)}$ $\textbf{T}(t) = \frac{\textbf{r}'(t)}{|\textbf{r}'(t)|} = \frac{\langle 0, -2sin(2t), 4cos(2t)\rangle}{\sqrt {4sin^2(2t) + 16cos^2(2t)}} = \langle 0, \frac{-sin(2t)}{\sqrt {sin^2(2t) + 4cos^2(2t)}}, \frac{2cos(2t)}{\sqrt {sin^2(2t) + 4cos^2(2t)}}\rangle = \langle 0, \frac{-sin(2t)}{\sqrt {1 + 3cos^2(2t)}}, \frac{2cos(2t)}{\sqrt {1 + 3cos^2(2t)}}\rangle $
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