Let $P_0$ be a point on a curve C with positive curvature κ. Define the radius of curvature at $P_0$

Given:

$P_0$be a point on a curve C with positive curvature κ

We have to define the radius of curvature at$P_0$.

Let $r\left(t\right)=⟨t,\alpha \sin \beta t,\alpha \cos \beta t⟩$and t=0

$r\left(t\right)=⟨t,\alpha \sin \beta t,\alpha \cos \beta t⟩$

at $t=0,r\left(0\right)=⟨0,0,\alpha ⟩$

Differentiate r(t) with respect to t,

$r^{\mathrm{\prime }}\left(t\right)=⟨1,\alpha \beta \cos \beta t,-\alpha \beta \sin \beta t⟩$

and,

localid="1649998970850" $$\begin{gathered} ∥r^{\mathrm{\prime }}\left(t\right)∥=\sqrt{1+\left(\alpha \beta \cos \beta t^2\right)+(-\alpha \beta \sin \beta t{)}^2} \\ =\sqrt{1+{\alpha }^2{\beta }^2} \end{gathered}$$

Now,

localid="1649999060548"

Since, $∥r^{\mathrm{\prime }}\left(t\right)∥=\sqrt{1+{\alpha }^2{\beta }^2}$then to have $∥r^{\mathrm{\prime }}\left(0\right)∥=\sqrt{1+{\alpha }^2{\beta }^2}$

Since, $∥T^{\mathrm{\prime }}\left(t\right)∥=\frac{\alpha {\beta }^2}{\sqrt{1+\alpha {\beta }^2}}$which is a constant.

Then,

$∥T^{\mathrm{\prime }}\left(0\right)∥=\frac{\alpha {\beta }^2}{\sqrt{1+{\alpha }^2\beta }}$

The curvature k of c at a point on the curve is given by :

$k=\frac{∥T^{\mathrm{\prime }}\left(0\right)∥}{∥r^{\mathrm{\prime }}\left(0\right)∥}$

localid="1649999096384" $$\begin{gathered} =\frac{\alpha {\beta }^2}{\sqrt{1+{\alpha }^2{\beta }^2}\sqrt{1+{\alpha }^2{\beta }^2}} \\ =\frac{\alpha {\beta }^2}{1+{\alpha }^2{\beta }^2} \end{gathered}$$

Therefore, the radius of the curvature is :

localid="1649999122418" $$\begin{gathered} \rho =\frac{1}{k} \\ \rho =\frac{1+{\alpha }^2{\beta }^2}{\alpha {\beta }^2} \end{gathered}$$