## Curvature Example

A curve
$y=f(x)$
has curvature given by
$\kappa = \frac{\frac{d^2 y}{dx^2}}{(1+(\frac{dy}{dx})^2)^{\frac{3}{2}}}$
.
Example:
$y=e^x$
.
$\frac{d^2y}{dx^2}=\frac{dy}{dx}=e^e$
so
$\kappa = \frac{e^x}{(1+e^{2x})^{\frac{3}{2}}}$
.
As
$x \rightarrow \infty$
the denominator increases much faster than the numerator, so the curvature tends to zero.
$\frac{d^2 \kappa}{dx^2}=\frac{(1+e^{2x})^{\frac{3}{2}}e^x- e^x(\frac{3}{2}2(1+e^{2x})^{\frac{1}{2}}}{(1+e^{2x})^3}= \frac{e^{3x}-2e^x}{(1+e^{2x})^{\frac{5}{2}}}$

$\frac{d^2 \kappa}{dx^2}=0 \rightarrow e^{3x}-2e^x=0 \rightarrow e^x(e^{2x}-2)=0 \rightarrow x = \frac{ln2}{2}$

The curvature is always positive and tends to 0, so the curvature just calculated must be a maximum.