Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises: 10

Answer

$\mathbf{r}'(t)=\langle-e^{-t}, 1-3t^2, \frac{1}{t}\rangle$

Work Step by Step

1) Differentiate $\mathbf{i}$ component of $\mathbf{r}(t)$ using chain rule. --> $\frac{d(e^{-t}\mathbf{i})}{dt}=(e^{-t})(-1)\mathbf{i}=-e^{-t}\mathbf{i}$ 2) Differentiate $\mathbf{j}$ component of $\mathbf{r}(t)$ using power rule --> $\frac{d(t-t^3\mathbf{j})}{dt}=(1-3t^2)\mathbf{j}$ 3) Differentiate $\mathbf{k}$ component of $\mathbf{r}(t)$ --> $\frac{d(ln(t)\mathbf{k})}{dt}=\frac{1}{t}\mathbf{k}$ 4) Therefore, $\mathbf{r}'(t)=-e^{-t}\mathbf{i}+(1-3t^2)\mathbf{j}+\frac{1}{t}\mathbf{k}=\langle-e^{-t}, 1-3t^2, \frac{1}{t}\rangle$
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