# Intrinsic equation / curvature question...?

## Prove if $| \psi |$ is small then $\kappa \approx \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \left(1 - \frac{3}{2} {\tan}^{2} \psi\right)$

Jul 29, 2018

For $y = y \left(x\right)$, curvature $\kappa$ is:

$\kappa : = \frac{\mathrm{dp} s i}{\mathrm{ds}} = \frac{y ' '}{1 + {\left(y '\right)}^{2}} ^ \left(\frac{3}{2}\right)$

$= y ' ' {\left(1 + {\left(y '\right)}^{2}\right)}^{- \frac{3}{2}}$

$= y ' ' \left(1 - \frac{3}{2} {\left(y '\right)}^{2} + \mathbb{O} \left({\left({y}^{'}\right)}^{3}\right)\right)$

Because:

• $y ' = \tan \psi$

Then:

• $\tan \psi \approx \psi \text{ for " abspsi " << "1 color(blue)(implies) abs(y') " << " 1 " if " abspsi "<< } 1$

So ignoring $\mathbb{O} \left({\left({y}^{'}\right)}^{3}\right)$ and higher:

• $\kappa \approx y ' ' \left(1 - \frac{3}{2} {\tan}^{2} \psi\right) \text{ for " abspsi " << } 1$