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: 12



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: $r'(t) =\lt -1, e^{t-4}, 8-2t \gt$ and $r''(t) =\lt 0, e^{t-4}, -2 \gt$ Thus, $\kappa(t) = \dfrac{||(2te^{t-4}-10e^{t-4})i-2j-e^{t-4}k||}{||\lt -1, e^{t-4}, 8-2t \gt||^3} \\=\dfrac{\sqrt {(2te^{t-4}-10e^{t-4})^2+4+e^{2t-8}}}{[1+e^{2t-8}+(8-2t)^2]^{3/2}}$ Now, we will compute $\kappa (t)$ at $t=4$ Therefore, $\kappa(t)= \dfrac{\sqrt {(-2)^2+4+1}}{(1+1)^{3/2}}=\dfrac{3}{2^{3/2}}$
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