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.
`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:
Of course, the radius needs to be positive, so we take the absolute value of the denominator (bottom) of the fraction.
Please support IntMath!