The curvature of a given curve at a particular point is the curvature of the approximating circle at that point.

The curvature depends on the radius - the smaller the radius, the greater the curvature (approaching a point at the extreme) and the larger the radius, the smaller the curvature. (A very large approximating circle means the curve is almost a straight line at that point.)

The radius of curvature R is simply the reciprocal of the curvature, K. That is,

`R = 1/K`

So we'll proceed to find the curvature first, then the radius will just be the reciprocal of that curvature.

Let P and `P_1` be 2 points on a curve, "very close" together, as shown.

radius of curvature

`Delta s` is the length of the arc `PP_1`.

`Delta theta` is the angle turned by the tangent line as it moves from P to `P_1`.

The curvature of the arc from P to `P_1` is given by

`(Delta theta)/(Delta s)`

Now, the curvature K at point P is given by:

`K=lim_(Delta theta->0)(Delta theta)/(Delta s) = (d theta)/(ds)`

We now need to find `(d theta)/(ds)` and we use the Chain Rule: ` (d theta)/(ds) = (d theta)/(dx)dx/(ds)`

Note that `tan\ theta=dy/dx`, so `theta=arctan(dy/dx)`

Returning to our formula, ` (d theta)/(dx)= d/(dx)arctan ((dy)/(dx))`

In the Differentiation of Transcendental Functions chapter we'll learn the derivative of `y=arctan\ u`, where `u=f(x)`, is given by `(dy)/(dx)=((du)/(dx))/(1+u^2)`

With `u= arctan ((dy)/(dx))` we differentiate as follows:

` (d theta)/(dx)= d/(dx)arctan ((dy)/(dx))`



We also need `dx/(ds)`, which is given by:


Putting it all together gives us the formula for curvature, `K`:

`K= (d theta)/(ds)=((d^2y)/(dx^2))/[1+(dy/dx)^2]^(3/2)`

Now the radius of curvature is just the reciprocal of this expression, that is:

`R= [1+(dy/dx)^2]^(3text(/)2)/((d^2y)/(dx^2))`

Of course, the radius needs to be positive, so we take the absolute value of the denominator (bottom) of the fraction.

`R= [1+(dy/dx)^2]^(3text(/)2)/(|(d^2y)/(dx^2)|)`