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: 28

Answer

$\langle \frac{1}{\sqrt 2}, 0,\frac{-1}{\sqrt 2}\rangle$

Work Step by Step

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