3
$\begingroup$

I know the originals formula for arc-length is:

$$\int_{a}^b \sqrt{1+{f'(x)}^2}$$

However most of the formulas don't have closed formed solutions, and are unsolvable in terms of this equation.

So far, I tried taking $$\lim_{n\rightarrow\infty}\sum_{i=1}^n \sqrt{1+\left(z\left(a+\frac{a-b}{n}\right)i\right)^2}\left(\frac{a-b}{n}\right)$$

Where $z(x)$ is the equation of the derivative of $f(x)$, $i$ is the number of line segments made around the intervals of the equation, and n is the sub-intervals. The only problem is my Desmos online graphing calculator has trouble computing the arclength as n goes to infinity.

I would like to know some of the techniques for calculating the arc length numerically in a easier way, from the range of $[a,b]$,and maybe an example.

$\endgroup$
  • 5
    $\begingroup$ Oftentimes it is much easier to find a parametrization $f:\mathbb{R} \rightarrow \mathbb{R}^2$ than it is to find an explicit formula $f:\mathbb{R} \rightarrow \mathbb{R}$ for a given curve. In such a case, the arclength formula is instead $\displaystyle \int_{t = t_1}^{t = t_2} \sqrt{\Big(\frac{dx}{dt}\Big)^2 + \Big( \frac{dy}{dt} \Big)^2 }$. $\endgroup$ – Kaj Hansen Jun 30 '14 at 21:13
  • 1
    $\begingroup$ Well if you cant find the parameter of something like $x^x+x^2$, and if it doesn't have a indefinite integral, than then the only was is to "numerically" compute it, not analytically. $\endgroup$ – Arbuja Jun 30 '14 at 22:02
  • 1
    $\begingroup$ Unfortunately, you are correct. $\endgroup$ – Kaj Hansen Jun 30 '14 at 22:30
  • $\begingroup$ See math.stackexchange.com/a/61796/589. $\endgroup$ – lhf Jul 1 '14 at 0:29
2
$\begingroup$

Assuming $f(x)$ is single-valued and differentiable on $(a,b)$, and writing $g(x) \equiv f^\prime(x)$ to avoid confusion, you have $s$ as an integral: $$ s = \int_a^b \sqrt{1+g(x)^2} \, dx $$

The form you gave, while correct in principle, does this integration by a sum of $n$ rectangles; the error term goes down as $1/n$ and this is much too slow. For most will behaved functions $g(x)$, a reasonable numerical integration scheme is Simpson's rule, which will exhibit an error of order $1/n^4$ so it is feasible to get decent accuracy.

Simpson's rule consists of using an even $n$ and weighting the even points at double strength compared to the odd points, and weighting the two endpoint values at half strength.

See Numerical Recipes if you need a more robost method for poorly behaved functions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.