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 13.3 Arc Length in Space - Exercises 13.3 - Page 759: 6


\begin{align*} \mathrm{T} &=\frac{6}{7} \mathrm{i}-\frac{2}{7} \mathrm{j}-\frac{3}{7} \mathrm{k}\\ \text{Length } &=49\end{align*}

Work Step by Step

Since $\mathbf{r}=6 t^{3} \mathbf{i}-2 t^{3} \mathbf{j}-3 t^{3} \mathbf{k}$ Then $ \mathbf{v}=18 t^{2} \mathbf{i}-6 t^{2} \mathbf{j}-9 t^{2} \mathbf{k} $ and $$|\mathbf{v}|=\sqrt{\left(18 t^{2}\right)^{2}+\left(-6 t^{2}\right)^{2}+\left(-9 t^{2}\right)^{2}}=\sqrt{441 t^{4}}=21 t^{2}$$ Hence \begin{align*} \mathrm{T}&=\frac{\mathrm{v}}{\mathrm{|v|}}\\ &=\frac{18 t^{2}}{21 t^{2}} \mathrm{i}-\frac{66^{2}}{21 t^{2}} \mathrm{j}-\frac{9^{2} t^{2}}{21 t^{2}} \mathrm{k}\\ &=\frac{6}{7} \mathrm{i}-\frac{2}{7} \mathrm{j}-\frac{3}{7} \mathrm{k}\\ \text{Length }&=\int_{1}^{2} 2 \mathrm{l} t^{2} d t\\ &=\left[7 t^{3}\right]_{1}^{2}\\ &=49\end{align*}
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