## Proof of Curvature Formula

The curvature formula for a curve
$y=f(x)$
is
$\frac{d \theta}{ds} = \kappa = \frac{\frac{d^2y}{dx^2}}{(1+ \frac{dy}{dx})^2)^\frac{3}{2}}$
, where
$\theta$
is the angle subtended by an arc of length
$s$
. To prove it note that
$tan \theta \frac{dy}{dx} \rightarrow \frac{d}{dx}= \frac{d^2y}{dx^2} \rightarrow sec^2 \theta \frac{d \theta}{dx}= \frac{d^2y}{dx^2}$

But
$sec^2 \theta = 1+tan^2 \theta = 1+ (\frac{dy}{dx})^2$
so
$(1+ (\frac{dy}{dx})^2) \frac{d \theta}{dx}= \frac{d^2y}{dx^2}$
(1)
Now use
$\frac{d \theta}{dx} =\frac{d \theta}{ds} \frac{ds}{dx}$
and note that
$ds= ((dx)^2+(dy)^2)^{\frac{1}{2}} \rightarrow \frac{ds}{dx} =(1+ (\frac{dy}{dx}))^{\frac{1}{2}}$
so
$\frac{d \theta}{dx}= (1+ (\frac{dy}{dx}))^{\frac{1}{2}}\frac{d \theta }{ds}$
.
Then (1) becomes
$(1+ (\frac{dy}{dx})^2)^{\frac{3}{2}} \frac{d \theta}{ds}= \frac{d^2y}{dx^2} \rightarrow \frac{d \theta}{ds} = \frac{\frac{d^2y}{dx^2}}{(1+ \frac{dy}{dx})^2)^\frac{3}{2}}$