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: 16


$$ \mathbf{r}=\left(-\frac{t^{2}}{2}+10\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{k}$$

Work Step by Step

Since \begin{align*} \frac{d \mathbf{r}}{d t}&=\int-(\mathbf{i}+\mathbf{j}+\mathbf{k}) d t\\ &=-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k})+\mathbf{C}_{1} \end{align*} and $$ \frac{d \mathbf{r}}{d t}(0)=\mathbf{0} \Rightarrow-(0 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k})+\mathbf{C}_{1}=\mathbf{0} \Rightarrow \mathbf{C}_{1}=\mathbf{0} $$ Then $$ \frac{d \mathbf{r}}{d t}=-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k}) $$ and \begin{align*} \mathbf{r}&=\int-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k}) d t\\ &=-\left(\frac{t^{2}}{2} \mathbf{i}+\frac{t}{2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k}\right)+\mathbf{C}_{2} \end{align*} $$ \mathbf{r}(0)=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k}$$ Then $$\mathbf{C}_{2}=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k}$$ and $$ \mathbf{r}=\left(-\frac{t^{2}}{2}+10\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{k}$$
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