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: 14


$$\mathbf{r}=\left(\frac{t^{4}}{4}+2 t^{2}+1\right) \mathbf{i}+\left(\frac{t^{2}}{2}+1\right) \mathbf{j}+\frac{2 t^{3}}{3} \mathbf{k} $$

Work Step by Step

Since \begin{align*} \mathbf{r}&=\int\left[\left(t^{3}+4 t\right) \mathbf{i}+t \mathbf{j}+2 t^{2} \mathbf{k}\right] d t\\ &=\left(\frac{t^{4}}{4}+2 t^{2}\right) \mathbf{i}+\frac{t^{2}}{2} \mathbf{j}+\frac{2 t^{3}}{3} \mathbf{k}+\mathbf{C} \end{align*} \begin{align*} \mathbf{r}(0)&=\left(\frac{0^{4}}{4}+2(0)^{2}\right) \mathbf{i}+\frac{0^{2}}{2} \mathbf{j}+\frac{2(0)^{3}}{3} \mathbf{k}+\mathbf{C}\\ \mathbf{i}+\mathbf{j}&=\mathbf{C}=\mathbf{i}+\mathbf{j} \end{align*} Hence $$\mathbf{r}=\left(\frac{t^{4}}{4}+2 t^{2}+1\right) \mathbf{i}+\left(\frac{t^{2}}{2}+1\right) \mathbf{j}+\frac{2 t^{3}}{3} \mathbf{k} $$
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