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: 21

Answer

$\textbf{T}(t) = \langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3}\rangle$

Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$ $\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ $\textbf{r}(t) = \langle 2t,2t,t\rangle$ $\textbf{r}'(t) = \langle 2,2,1\rangle$ $|\textbf{r}'(t)| = \sqrt {(2)^2 + (2)^2 + (1)^2} = \sqrt {9} = 3$ $\textbf{T}(t) = \frac{\textbf{r}'(t)}{|\textbf{r}'(t)|} = \frac{\langle 2,2,1\rangle}{3} = \langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3}\rangle$
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