# Numerical calculation about curve length

1. Sep 2, 2007

### xyz3003

I think I have returned all my math back to teachers without any refund.

y=f(x);
h=xb-xa, which is very small.

My Q is to calculate curve length rather than area numerically.
But let me use area as example to show you what i want.

to calculate area between xa to xb, we have 2 ways:
1) area=(f(xa)+f(xb))*h/2; (trapezoid?)
2) area=(f(xa)+4*f(xm)+f(xb))*h/6; here xm=(xa+xb)/2; (parabola?)
As my test, second one is much better than first.

for curve length:
1) len=square root( (f(xb)-f(xa))*(f(xb)-f(xa)) + h*h);
actually, it is distance from (xa, f(xa)) to (xb, f(xb)).

do you know second way to calculate curve length as in area sample above, simple, easy-to-use and better?

any links or explainations are highly appreciated.

thanks.

.
.

2. Sep 2, 2007

### arildno

Well, in regions where the curvature is slowly varying, you might interpolate with circular arc segments instead.

Just a suggestion..

3. Sep 3, 2007

### AlephZero

Start from an integral representing the length of the curve

$$\int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$$

or $$\int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right)}\,dt$$ for a curve defined by parametric equations.

Evaluate the integrals with your favorite numerical formula.

4. Sep 5, 2007

### xyz3003

I just want to know the "favorite numerical formula", if which is better than summing line distances from one point to another.

5. Sep 6, 2007

### HallsofIvy

As I said before, in most cases, Simpson's rule is most efficient.

6. Sep 6, 2007

### JonF

I like the trapezoidal rule because it does a good enough job for most things and is much less tedious than simpons.