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: 17

Answer

$\textbf{r}'(\pi/2) = \langle 1, 0, 0\rangle$

Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$ $\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ $\textbf{r}(t) = \langle t,cos(2t),2sin(t)\rangle$ $\textbf{r}'(t) = \langle 1, -2sin(2t), 2cos(t)\rangle$ $\textbf{r}'(\pi/2) = \langle 1, -2sin(2(\pi/2)), 2cos(\pi/2)\rangle = \langle 1, 0, 0\rangle$
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