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: 9

Answer

$\textbf{r}'(t) = \langle 6t^2, \frac{3}{\sqrt t}, \frac{-3}{t^2}\rangle$

Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$ $\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ $\textbf{r}(t) = \langle 2t^3, 6\sqrt t, \frac{3}{t}\rangle$ $\textbf{r}'(t) = \langle 6t^2, \frac{3}{\sqrt t}, \frac{-3}{t^2}\rangle$
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