## 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.