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: 10


$\kappa(t)= \dfrac{2t^{-3}}{(t^{-4}+1)^{3/2}}$

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: $r'(t) =\lt -t^{-2}, 0, 1 \gt$ and $r''(t) =\lt 2t^{-3}, 0, 0 \gt$ Thus, $\kappa(t) = \dfrac{||2t^{-3} j ||}{||\lt -t^{-2}, 0, 1 \gt||^3} \\=\dfrac{2t^{-3}}{[(-t^{-2})^2+0^2+1^2]^{3/2}}$ Therefore, $\kappa(t)= \dfrac{2t^{-3}}{(t^{-4}+1)^{3/2}}$
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