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: 25

Answer

$\textbf{T}(t)= \langle \frac{t^2}{\sqrt {t^4 + 4}}, 0, \frac{-2}{\sqrt {t^4 + 4}}\rangle$

Work Step by Step

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