Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.4 Curvature - Exercises - Page 734: 13


$\dfrac{\sqrt {\pi^2+5}}{(1+\pi^2)^{3/2}}$

Work Step by Step

The curvature $\kappa$ for a plane curve system is: $\kappa (t)= \dfrac{||r'(t) \times r''(t)||}{||r'(t)||^3}$ We have: $r'(t) =\lt -\sin t, \cos t, 2t \gt$ and $r''(t) =\lt -\cos t ,-\sin t, 2 \gt$ Thus, $\kappa(t) = \dfrac{||2(\cos t+t\sin t)i-2(t \cos t -\sin t )j+k|}{||\lt -\sin t, \cos t, 2t \gt||^3} \\=\dfrac{\sqrt {4(\cos t+t\sin t)^2+4(t \cos t -\sin t )^2+1}}{[\sin^2t+\cos^2 t+4t^2]^{3/2}}$ Now, we will compute $\kappa (t)$ at $t=\dfrac{\pi}{2}$ Therefore, $\kappa(t)= \dfrac{\sqrt {4 (\dfrac{\pi^2}{4})+4(1)+1}}{(1+\pi^2)^{3/2}}=\dfrac{\sqrt {\pi^2+5}}{(1+\pi^2)^{3/2}}$
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