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: 11


$\dfrac{2 \sqrt {74}}{27}$

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: $r'(t) =\lt -t^{-2}, -2t^{-3}, 2t \gt$ and $r''(t) =\lt 2t^{-3}, 6t^{-4}, 2 \gt$ Thus, $\kappa(t) = \dfrac{||16t^{-3}i+6t^{-2}j-2t^{-6}k ||}{||\lt -t^{-2}, -2t^{-3}, 2t \gt||^3} \\=\dfrac{\sqrt {256t^{-6}+36t^{-4}+4t^{-12}}}{[(-t^{-4})^2+4t^{-6}+4t^2]^{3/2}}$ Now, we will compute $\kappa (t)$ at $t=1$ Therefore, $\kappa(t)= \dfrac{\sqrt {296}}{(9)^{3/2}}=\dfrac{2 \sqrt {74}}{27}$
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