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```				   can anyone give me the proof of the formula of radius of curvature of a curve?
```

6 years ago

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```										hey Sudhindra....
To understand the radius of curvature proof. you need to understand it like wise as:
1.  The radius of curvature is the radius of the "osculating circle,"
i.e., the circle that is tangent to the curve at that point.
Clearly the circle itself is its own osculating circle everywhere,
and the radius is R, so that the radius of curvature is 1/K = R.

2.  Another "cheat" is to use the polar equation for the radius of
curvature.  If the curve in polar coordinates is given by
r = r(@), then the radius of curvature is

[r^2 + (dr/d@)^2]^(3/2)
1/K =  --------------------------------.
r^2 + 2(dr/d@)^2 - r*(d^2r/d@^2)

This makes things trivial for a circle, because r(@) = R and
all the derivatives of r with respect to @ vanish, so we obtain

1/K = (R^2)^(3/2) / R^2 = R^3/R^2 = R.

3.  Now let's do this in x-y coordinates, using the straightforward
expression for the top half of the circle.  It's helpful to
define the variable u = R^2 - x^2.

y = sqrt(R^2 - x^2) = sqrt(u)

We will need the first and second derivatives of y with respect
to x.

du/dx = -2x

dy/dx = [1/(2sqrt(u))]*(du/dx) = -2x/[2sqrt(u)] = -x/sqrt(u).

d^2y/dx^2 = [sqrt(u)*(-1) - (-x)(dy/dx)]/u
= [-sqrt(u) - x^2/sqrt(u)]/u

So we plug this into the formula for the curvature K:

d^2y/dx^2           [-sqrt(u) - x^2/sqrt(u)]/u
K = ---------------------  =  --------------------------
[1 + (dy/dx)^2]^(3/2)         [1 + x^2/u]^(3/2)

-u^(-1/2) * [u + x^2] * u^-1     multiply terms in
= -----------------------------     brackets by sqrt(u).
u^(-3/2) * [u + x^2]^(3/2)

u^0
= ----------------                  lots of cancellation!
(u + x^2)^(1/2)

= 1/(R^2 - x^2 + x^2)^(1/2)

= 1/R

and again we arrive at the expected result that the radius of
curvature is 1/K = R for the top semicircle. You can argue the
result for the bottom semicircle (y = -sqrt(R^2 - x^2)) either
by symmetry or by grinding through the derivatives again.
```
5 years ago

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