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: 26

Answer

$\textbf{T}(t) = \frac{\langle e^{2t}, 2e^{2t}, -3e^{-3t}\rangle}{\sqrt {5e^{2t}+9e^{-6t}}}$

Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$ $\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ $\textbf{r}(t) = \langle e^{2t}, 2e^{2t}, 2e^{-3t}\rangle$ $\textbf{r}'(t) = \langle 2e^{2t}, 4e^{2t}, -6e^{-3t}\rangle$ $|\textbf{r}'(t)| = \sqrt {(2e^{2t})^2 + (4e^{2t})^2 + (-6e^{-3t})^2} = \sqrt {20e^{2t}+36e^{-6t}} = 2\sqrt {5e^{2t}+9e^{-6t}}$ $\textbf{T}(t) = \frac{\textbf{r}'(t)}{|\textbf{r}'(t)|} = \frac{\langle 2e^{2t}, 4e^{2t}, -6e^{-3t}\rangle}{2\sqrt {5e^{2t}+9e^{-6t}}} = \frac{\langle e^{2t}, 2e^{2t}, -3e^{-3t}\rangle}{\sqrt {5e^{2t}+9e^{-6t}}}$
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