Length of f()

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Alkatran

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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...
 

quasar987

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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

[tex]L=\int_a^b\sqrt{1+(f '(x))^2}dx[/tex]
 

dextercioby

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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.
 

Galileo

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Alkatran said:
Basically the sum of arbitrarily small linear approximations.
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 [itex]\Delta x[/itex] can be approximated by:
[tex]\sqrt{(\Delta x)^2+(\Delta y)}[/tex]

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

[tex]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[/tex]

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

[tex]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[/tex]

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.
 
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