Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.3 Arc Length and Curvature - 13.3 Exercises - Page 908: 22

Answer

$\dfrac{\sqrt{4t^2e^{2t}-8te^{2t}+5e^{2t}+4}}{[1+4t^2+e^{2t}]^{3/2}}$

Work Step by Step

As we are given that $r(t)=ti+t^2j+e^tk$ Write Theorem 10. $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}$ Find. $r'(t)=1i+2tj+e^tk$ and $r''(t)=2j+e^tk$ or, $|r'(t)|=\sqrt {(1)^2+(2t)^2+(e^t)^2}=\sqrt {1+4t^2+e^{2t}}$ Thus, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{| (1i+2tj+e^tk)\times(2j+e^tk)|}{|\sqrt {1+4t^2+e^{2t}}|^3}$ or, $=\dfrac{|(2te^t-2e^t)-(e^t-0)+(2-0)|}{|\sqrt {1+4t^2+e^{2t}}|^3}$ $=\dfrac{|2e^t(t-1)-e^t+2|}{|\sqrt {1+4t^2+e^{2t}}|^3}$ Hence, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{\sqrt{4t^2e^{2t}-8te^{2t}+5e^{2t}+4}}{[1+4t^2+e^{2t}]^{3/2}}$
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