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: 15


$$\mathbf{r}=8 t \mathbf{i}+8 t \mathbf{j}+\left(100-16 t^{2}\right) \mathbf{k}$$

Work Step by Step

Since $$\frac{d \mathbf{r}}{d t}=\int(-32 \mathbf{k}) d t=-32 t \mathbf{k}+\mathbf{C}_{1}$$ and $$ \frac{d \mathbf{r}}{d t}(0)=8 \mathbf{i}+8 \mathbf{j} \Rightarrow-32(0) \mathbf{k}+\mathbf{C}_{1}=8 \mathbf{i}+8 \mathbf{j} \Rightarrow \mathbf{C}_{1}=8 \mathbf{i}+8 \mathbf{j} $$ Then $$ \frac{d \mathbf{r}}{d t}=8 \mathbf{i}+8 \mathbf{j}-32 t \mathbf{k}$$ and \begin{align*} \mathbf{r}&=\int(8 \mathbf{i}+8 \mathbf{j}-32 t \mathbf{k}) d t\\ &=8 t \mathbf{i}+8 t \mathbf{j}-16 t^{2} \mathbf{k}+\mathbf{C}_{2} \end{align*} $$ \mathbf{r}(0)=100 \mathbf{k} \Rightarrow 8(0) \mathbf{i}+8(0) \mathbf{j}-16(0) \mathbf{j}-16(0)^{2} \mathbf{k}+\mathbf{C}_{2}=100 \mathbf{k}$$ Then $ \mathbf{C}_{2}=100 \mathbf{k}$$$\mathbf{r}=8 t \mathbf{i}+8 t \mathbf{j}+\left(100-16 t^{2}\right) \mathbf{k}$$
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