# Curvatures, Curvatures, Everywhere Curvatures

Fig. 1

In his book “What is Mathematics”, Courant wrote:

Here is my proof:

$\alpha$ is the slope-angle tells us that

$\tan{(\alpha)} = f'(x)$

which implies

$\frac{d\tan(\alpha)}{dx}=\frac{d\tan(\alpha)}{d \alpha}\frac{d\alpha}{dx}$

$=\sec^2(\alpha)\frac{d \alpha}{dx}=(1+\tan^2(\alpha))\frac{d \alpha}{dx}=(1+(f'(x))^2)\frac{d\alpha}{dx}=f''(x),$

i.e.,

$\frac{d\alpha}{dx}=\frac{1}{1+(f'(x))^2} f''(x)$.

We also know that

$s=\int^{x}_{}\sqrt{1+(f'(x))^2}\;dx$.

Therefore,

$\frac{ds}{dx} = \sqrt{1+(f'(x))^2}$.

Hence by definition, the curvature

$\kappa= h'(s)= \frac{d \alpha}{ds}= \frac{d \alpha}{dx} \frac{d x}{ds}= \frac{d\alpha}{dx}\frac{1}{\frac{ds}{dx}}=\frac{f''(x)}{1+(f'(x))^2}\frac{1}{\sqrt{1+(f'(x))^2}}$

$=\frac{f''(x)}{(1+(f'(x))^2)^{1+\frac{1}{2}}}$

$= \frac{f''(x)}{(1+(f'(x))^2)^{\frac{3}{2}}}$

I am now suggesting an exercise :

Calculate the curvature of figure in Fig. 1 at a point of your choice.