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: 8


\begin{align*} \mathbf{T} &=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}\\ \text{Length }& =1\end{align*}

Work Step by Step

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