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: 21



Work Step by Step

Given: $r(t)=t^3j+t^2k$ 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, $r'(t)=3t^2j+2tk$ and $r''(t)=6tj+2k$ and $|r'(t)|=\sqrt {(3t^2)^2+(2t)^2}=\sqrt {9t^4+4t^2}$ Now, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{| (3t^2j+2tk)\times(6tj+2k)|}{|\sqrt {9t^4+4t^2}|^3}$ $=\dfrac{6t^2}{|\sqrt {9t^4+4t^2}|^3}$ $=\dfrac{6t^2}{(9t^4+4t^2)^{3/2}}$
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