Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 13


$$\mathbf{r}=\left[(t+1)^{3 / 2}-1\right] \mathbf{i}+\left(1-e^{-t}\right) \mathbf{j}+[1+\ln (t+1)] \mathbf{k}$$

Work Step by Step

Since \begin{align*} \mathbf{r}&=\int\left[\left(\frac{3}{2}(t+1)^{1 / 2}\right) \mathbf{i}+e^{-t} \mathbf{j}+\left(\frac{1}{t+1}\right) \mathbf{k}\right] d t\\ &=(t+1)^{3 / 2} \mathbf{i}-e^{-t} \mathbf{j}+\ln (t+1) \mathbf{k}+\mathbf{C} \end{align*} Since \begin{align*} \mathbf{r}(0)&=(0+1)^{3 / 2} \mathbf{i}-e^{-0} \mathbf{j}+\ln (0+1) \mathbf{k}+\mathbf{C}\\ \mathbf{k}&=(0+1)^{3 / 2} \mathbf{i}-e^{-0} \mathbf{j}+\ln (0+1) \mathbf{k}+\mathbf{C}\\ \mathbf{C}&=-\mathbf{i}+\mathbf{j}+\mathbf{k} \end{align*} Hence $$\mathbf{r}=\left[(t+1)^{3 / 2}-1\right] \mathbf{i}+\left(1-e^{-t}\right) \mathbf{j}+[1+\ln (t+1)] \mathbf{k}$$
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