# Length of f()

1. Jan 11, 2005

### Alkatran

Are there any formulas for finding the length of the line traced out by some function f()?

IE: If f(x) = cx + b where c and b are constants
The length from 0 to x is:
l(x) = sqr(x^2 + (cx)^2)

But I don't know what to do for any polynomial above a line.
I can make a summation for it, but don't know how to simplify.

l(x) = lim[t->infinity](sum[n = 0 to t](sqr( 1/x^2 + (f(x*n/t)-f(x*(n+1)/t)))^2)))

Basically the sum of arbitrarily small linear approximations.

I figure the length of sin() and cos() are related to pi somehow...

2. Jan 11, 2005

### quasar987

As written in my calculus textbook:

Theorem: Let f be afunction s.t. f ' is continuous on [a,b]. The lenght L of the curve joining the points R(a,f(a)) and S(b,f(b)) is given by

$$L=\int_a^b\sqrt{1+(f '(x))^2}dx$$

3. Jan 11, 2005

### dextercioby

For a polynomial y=y(x) of degree larger or equal to three,the integral cannot be computed exactly.U'd be dealing with so-called "LEGENDRE ELLIPTIC INTEGRALS".
Only numerical methids would work.

Daniel.

4. Jan 11, 2005

### Galileo

This is exactly how the length is defined if you take the limiting process.
I'll use slightly different notation then you did.

A small line segment of width $\Delta x$ can be approximated by:
$$\sqrt{(\Delta x)^2+(\Delta y)}$$

by cutting up the interval [a,b] into n subintervals of width $\Delta x$, you can approximate the length by:

$$L \approx \sum_{i=1}^n\sqrt{(\Delta x_i)^2+(\Delta y_i)}=\sum_{i=1}^n\sqrt{1+(\frac{\Delta y_i}{\Delta x_i})^2}\Delta x_i$$

The approximation gets better of n gets larger.
The length of the curve is defined by:

$$L=\lim_{n \to \infty}\sum_{i=1}^n\sqrt{1+(\frac{\Delta y_i}{\Delta x_i})^2}\Delta x_i=\int_a^b \sqrt{1+y'(x)^2}dx$$

The square root often makes it very difficult or impossible to evaluate explicitly. We'll have to resort to approximating the length. With Simpson's rule for example.

Last edited: Jan 11, 2005