## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 12

#### Answer

$$\mathbf{r}=90 t^{2} \mathbf{i}+\left(90 t^{2}-\frac{16}{3} t^{3}+100\right) \mathbf{j}$$

#### Work Step by Step

Given \begin{aligned} \frac{d \mathbf{r}}{d t} &=\int (180 t) \mathbf{i}+\left(180 t-16 t^{2}\right) \mathbf{j} \\ \mathbf{r}(0) &=100 \mathbf{j} \end{aligned} Then \begin{align*} \mathbf{r}&=\int\left[(180 t) \mathbf{i}+\left(180 t-16 t^{2}\right) \mathbf{j}\right] d t\\ &=90 t^{2} \mathbf{i}+\left(90 t^{2}-\frac{16}{3} t^{3}\right) \mathbf{j}+\mathbf{C} \end{align*} Since $\mathbf{r}(0) =100 \mathbf{j}$, then \begin{align*} \mathbf{r}(0)&=90(0)^{2} \mathbf{i}+\left[90(0)^{2}-\frac{16}{3}(0)^{3}\right] \mathbf{j}+\mathbf{C}\\ 100 \mathbf{j}&=\mathbf{C} \end{align*} Hence $$\mathbf{r}=90 t^{2} \mathbf{i}+\left(90 t^{2}-\frac{16}{3} t^{3}+100\right) \mathbf{j}$$

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