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: 42

Answer

$\textbf{r}''(t) = \langle 396t^{10}-2, 56t^6 + 6t, 20t^{-6}\rangle$ $\textbf{r}'''(t) = \langle 3960t^9, 336t^5+6, -120t^{-7}\rangle$

Work Step by Step

To find the derivative, simply find the derivative of each component. $\textbf{r}(t) = \langle 3t^{12}-t^2, t^8+t^3, t^{-4}-2\rangle$ $\textbf{r}'(t) = \langle 36t^{11}-2t, 8t^7+3t^2, -4t^{-5}\rangle$ $\textbf{r}''(t) = \langle 396t^{10}-2, 56t^6 + 6t, 20t^{-6}\rangle$ $\textbf{r}'''(t) = \langle 3960t^9, 336t^5+6, -120t^{-7}\rangle$
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