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: 3


\begin{align*} \mathbf{T} &=\frac{1}{\sqrt{1+t}} \mathbf{i}+\frac{\sqrt{t}}{\sqrt{1+t}} \mathbf{k}\\ \text{Length }& =\frac{52}{3} \end{align*}

Work Step by Step

Since $\mathbf{r}=t \mathbf{i}+\frac{2}{3} t^{3 / 2} \mathbf{k}$ Then $\mathbf{v}=\mathbf{i}+t^{1 / 2} \mathbf{k} \Rightarrow|\mathbf{v}|=\sqrt{1^{2}+\left(t^{1 / 2}\right)^{2}}=\sqrt{1+t}$ Hence \begin{align*} \mathbf{T}&=\frac{\mathbf{v}}{|\mathbf{v}|}\\ &=\frac{1}{\sqrt{1+t}} \mathbf{i}+\frac{\sqrt{t}}{\sqrt{1+t}} \mathbf{k}\\ \text{Length }&=\int_{0}^{8} \sqrt{1+t} d t\\ &=\left[\frac{2}{3}(1+t)^{3 / 2}\right]_{0}^{8}\\ &=\frac{52}{3} \end{align*}
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