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: 32


$y=\pm 2 (1+b^2)^{3/2}x^2+bx$

Work Step by Step

Write formula 11. $\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$ Let us consider $f(x)=ax^2+bx$ . This is the general equation of a parabola. Now, $f'(x)=2ax+b$ and $f''(x)=2a$ Thus,$\kappa(x)=\dfrac{|2a|}{[1+(2ax+b)^2]^{3/2}}$ or, $\kappa(0)=\dfrac{|2a|}{[1+b^2]^{3/2}}$[ $ \kappa(0)$ is the curvature at origin.] or, $ 4=\dfrac{|2a|}{[1+b^2]^{3/2}}$ or, $a=\pm 2 (1+b^2)^{3/2}$ Therefore, the equation of parabola is $y=\pm 2 (1+b^2)^{3/2}x^2+bx$
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