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: 29

Answer

$\textbf{T}(1) = \langle\frac{2}{\sqrt 5},0,\frac{-1}{\sqrt 5}\rangle$

Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$ $\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ $\textbf{r}(t) = \langle 6t,6,3/t\rangle$ $\textbf{r}'(t) = \langle 6,0,-3/t^2\rangle$ $|\textbf{r}'(t)| = \sqrt {(6)^2 + (0)^2 + (-3/t^2)^2} = \sqrt {36 +9/t^4}$ $\textbf{T}(t) = \frac{\textbf{r}'(t)}{|\textbf{r}'(t)|} = \frac{\langle 6,0,-3/t^2\rangle}{ \sqrt {36 +9/t^4}}$ $\textbf{T}(1) = \frac{\langle 6,0,-3/(1)^2\rangle}{ \sqrt {36 +9/(1)^4}} = \langle\frac{2}{\sqrt 5},0,\frac{-1}{\sqrt 5}\rangle$
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