Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.4 - Curvature and Normal Vectors of a Curve - Exercises 13.4 - Page 765: 17


$y=ax^2$ shows the maximum curvature at $(0,0)$ and has no minimum curvature.

Work Step by Step

We have: $f(x)=ax^2$ This implies that $\dfrac{d f(x)}{dx}=2ax$ and $\dfrac{d^2 f(x)}{dx^2}=2a$ We know that $\kappa (x)=\dfrac{[\dfrac{d^2 f(x)}{dx^2}]}{[1+(\dfrac{d f(x)}{dx}^2)]^{3/2}}$ So, $\kappa (x)=\dfrac{[2a]}{[1+4a^2x^2]^{3/2}}$ and $\dfrac{d \kappa}{dx}=-\dfrac{[24a^3x]}{[1+4a^2x^2]^{5/2}}$ and,$ \dfrac{d^2 \kappa}{dx^2}=\dfrac{d}{dx}[-\dfrac{[24a^3x]}{[1+4a^2x^2]^{5/2}}]=\dfrac{-24a^3[(1+4a^2x^2)^{5/2}-20a^2x^2 (4a^2x^2+1)^{3/2}]}{(1+4a^2x^2)^5}$ Set $x=0$, so $ \dfrac{d^2 \kappa}{dx^2}=-24a^3 \lt 0$ This means that the maximum occurs at $x=0, y=0$. Therefore, $y=ax^2$ shows the most curvature at $(0,0)$ and has no minimum curvature.
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