Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Section 13.3 - Arc Length and Curvature - 13.3 Exercise - Page 868: 25



Work Step by Step

Given: $r(t)=\lt t,t^2, t^3 \gt$ To calculate the curvature of the curve we will have to use Theorem 10 such as: $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}$ Thus, $r'(t)=\lt 1,2t, 3t^2 \gt$ and $r''(t)=\lt 0,2,6t \gt$ Now, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{|\sqrt {36t^4+36t^2+4}|}{[1+4t^2+9t^4]^{3/2}}$ At point $(1,1,1)$ $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{|\sqrt {36t^4+36t^2+4}|}{[1+4t^2+9t^4]^{3/2}}=\dfrac{|\sqrt {36+36+4}|}{[1+4+9]^{3/2}}=\dfrac{\sqrt{76}}{[14]^{3/2}}$ Hence, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}|_{(1,1,1)}=\dfrac{\sqrt{76}}{14^{3/2}}$
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