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 760: 15


$$\sqrt{2}+\ln (1+\sqrt{2}) $$

Work Step by Step

Since $$\mathbf{r}=(\sqrt{2} t) \mathbf{i}+(\sqrt{2} t) \mathbf{j}+\left(1-t^{2}\right) \mathbf{k} $$ Then \begin{align*} \mathbf{v}&=\sqrt{2} \mathbf{i}+\sqrt{2} \mathbf{j}-2 t \mathbf{k} \\|\mathbf{v}|&=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}+(-2 t)^{2}}\\ &=\sqrt{4^{2}+4 t}=2 \sqrt{1+t^{2}} \end{align*} Hence \begin{align*} \text{Length}&=\int_{0}^{1} 2 \sqrt{1+t^{2}} d t\\ &=\left[2\left(\frac{t}{2} \sqrt{1+t^{2}}+\frac{1}{2} \ln (t+\sqrt{1+t^{2}})\right)\right]_{0}^{1}\\ &=\sqrt{2}+\ln (1+\sqrt{2}) \end{align*}
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