\[y=f(x)\]
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
\[ (1+ (\frac{dy}{dx})^2)^{\frac{3}{2}} \frac{d \theta}{ds}= \frac{d^2y}{dx^2} \rightarrow \frac{d \theta}{ds} = \frac{\frac{d^2y}{dx^2}}{(1+ \frac{dy}{dx})^2)^\frac{3}{2}} \]