Radius of curvature formula derivation

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SUMMARY

The radius of curvature for a curve defined by the function y=f(x) is derived using the formula \(\frac{f''(x)}{(1+f'(x)^2)^{3/2}}\). The discussion emphasizes understanding curvature in the context of vector functions, specifically using the formula \(\kappa=\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}\). The derivation involves simplifying the vector representation of the curve, where \(\mathbf{r} = xi + f(x)j\), and computing the cross product of the first and second derivatives of the vector function.

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chandran
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for a curve defined by y=f(x) the radius of curvature is defined as
[f""(x)/(1+f"(x))] power 3/2. I need a good neat & understandable derivation for that. can anybody show a web.
 
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Differential Geometry of Curves

I don't like the format of the forum reply, so click on the following link to view your derivative: DGC.
 
Well first understand that curvature for a vector function is given by:

\kappa=\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}

Now, let r = xi + f(x)j and simplify. To prove the first formula \kappa, use the following fact and compute r' x r''. The answer should be clear from there.

\kappa=\frac{d\mathbf{T}}{ds}

\mathbf{r}'=\frac{ds}{dt}\mathbf{T}
 

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