Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 12 - Vector-Valued Functions - 12.5 Curvature - Exercises Set 12.5 - Page 880: 49

Answer

$k(\theta) \propto 1 / r$

Work Step by Step

For a curve in polar coordinates, $f(\theta)=r,$ the curvature is: \[ \frac{|2\left(\frac{d r}{d \theta}\right)^{2}+ r^{2}-r \frac{d^{2} r}{d \theta^{2}} 1}{\left|r^{2}+\left(\frac{d r}{d \theta}\right)^{2}\right|^{3 / 2}} \] Since: \[ \begin{array}{c} e^{a \theta}=r(\theta) \\ \frac{d y}{d \theta}=a e^{a \theta}=a r(\theta) \\ \frac{d^{2} r}{d \theta^{2}}=\frac{d}{d \theta}(a r(\theta))=a r^{\prime}(\theta)=a^{2} r(\theta) \end{array} \] So: \[ \begin{aligned} &\frac{\left|2(a r)^{2}+r^{2}-r\left(a^{2} r\right)\right|}{\left|(a r)^{2}+r^{2}\right|^{3 / 2}}= k(\theta) \\ &=\frac{\left(2 a^{2}-a^{2}+1\right)r^{2}}{r^{3}\left(a^{2}+1\right)^{3 / 2}} \\ \frac{1}{r} \frac{a^{2}+1}{\left(a^{2}+1\right)^{3 / 2}} =\frac{1}{\sqrt{a^{2}+1}}(1 / r) \end{aligned} \] $k(\theta) \propto 1 / r$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.