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: 11


$$\mathbf{r}=\left(-\frac{t^{2}}{2}+1\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+2\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+3\right) \mathbf{k}$$

Work Step by Step

Given \begin{aligned} \frac{d \mathbf{r}}{d t} &=-t \mathbf{i}-t \mathbf{j}-t \mathbf{k} \\ \mathbf{r}(0) &=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k} \end{aligned} Then \begin{align*} \mathbf{r}&=\int(-t \mathbf{i}-t \mathbf{j}-t \mathbf{k}) d t\\ &=-\frac{t^{2}}{2} \mathbf{i}-\frac{t^{2}}{2} \mathbf{j}-\frac{t^{2}}{2} \mathbf{k}+\mathbf{C} \end{align*} Since $\mathbf{r}(0) =\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}$, then \begin{align*} \mathbf{r}(0)&=0 \mathbf{i}-0 \mathbf{j}-0 \mathbf{k}+\mathbf{C}\\ &=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k} \\ \mathbf{C}&=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k} \end{align*} Hence $$\mathbf{r}=\left(-\frac{t^{2}}{2}+1\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+2\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+3\right) \mathbf{k}$$
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