Curvature of a Semi Cubicle Parabola
The curvature of a curve\[y=f(x)\]
is given by \[\kappa = \frac{\frac{d^2y}{dx^2}}{(1+ (\frac{dy}{dx})^2)^{\frac{3}{2}}}\]
.Sup[pose a semi cubical parabola has equation
\[y^2=a^2x^3\]
then \[y=a x^{\frac{3}{2}}\]
and \[\frac{dy}{dx}=\frac{3a}{2} x^{\frac{1}{2}}\]
and \[\frac{d^2 y}{dx^2}= \frac{3a}{4}x^{ - \frac{1}{2}}\]
.Then
\[\kappa = \frac{\frac{3a}{4}x^{ - \frac{1}{2}}}{(1+(\frac{3a}{2} x^{\frac{1}2})^2)^{\frac{3}{2}}}= \frac{6a}{\sqrt{x} (4+9a^2x)^{\frac{3}{2}}}\]
.The degree of the denominator is 6 and the degree of the denominator is 4, so as
\[x \rightarrow \infty\]
the curvature tends to zero.