Daily Archives: May 20, 2017

Curvatures, Curvatures, Everywhere Curvatures

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Fig. 1

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

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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.