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: 4


\begin{align*} \mathbf{T}&=\frac{1}{\sqrt{3}} \mathbf{i}-\frac{1}{\sqrt{3}} \mathbf{j}+\frac{1}{\sqrt{3}} \mathbf{k}\\ \text{Length }& =3 \sqrt{3} \end{align*}

Work Step by Step

Since $\mathbf{r}=(2+t) \mathbf{i}-(t+1) \mathbf{j}+t \mathbf{k} $ Then $\mathbf{v}=\mathbf{i}-\mathbf{j}+\mathbf{k} \Rightarrow|\mathbf{v}|=\sqrt{1^{2}+(-1)^{2}+1^{2}}=\sqrt{3}$ Hence \begin{align*} \mathbf{T}&=\frac{\mathbf{v}}{|\mathbf{v}|}\\ &=\frac{1}{\sqrt{3}} \mathbf{i}-\frac{1}{\sqrt{3}} \mathbf{j}+\frac{1}{\sqrt{3}} \mathbf{k}\\ \text{Length }&=\int_{0}^{3} \sqrt{3} d t\\ &=[\sqrt{3} t]_{0}^{3}\\ &=3 \sqrt{3} \end{align*}
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