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: 14

Answer

$\textbf{r}'(t) = \langle \frac{-1}{(t+1)^2}, \frac{1}{1+t^2}, \frac{1}{t+1}\rangle$

Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$ $\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ $\textbf{r}(t) = \langle (t+1)^{-1}, tan^{-1} t, ln(t+1)\rangle$ Chain Rule: $\textbf{r}'(t) = \langle \frac{-1}{(t+1)^2}, \frac{1}{1+t^2}, \frac{1}{t+1}\rangle$
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