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Finding the arc length of a function on an interval is another application of the definite integral. Everyone knows the distance formula, sqrt((x2-x1)^2 + (y2-y1)^2), but that only works for straight lines formed from connecting two points. The equation for finding arc length is simple, but taking the integral of the equation may be extremely difficult. In short, get a graphing calculator that can calculate integrals, or get math software (Derive, Mathematica, etc.) for the same purpose.

Enough rambling. The equation to find arc length is Sab sqrt(1 + (dy/dx)^2)dx. If you find it difficult or impossible to find the integral in this manner, try Scd sqrt(1 + (dx/dy)^2)dy. Both of these equations assume that their functions are differentiable on the intervals [a,b] and [c,d], respectively.

Literally, the length of an arc. Mathematically, the length of the curve produced by some function f(x) between two points. Let's say f(x) looks something like this:

```          _
|      / \
|     /   \       /
|    /     \     /
|   /       \   /
|            \_/
_|_____________________
|   a              b
|
```

Now, you need to find the length of the arc between x=a and x=b. (Don't ask why you need to find it. Math doesn't care about why. You just do.) Well, you could produce a pretty good guess by breaking up the arc into line segments and measuring each of them separately:

```          _
|      / \
|     /  |\       /
|    /|  | \     /|
|   / |  |  \   / |
|  |  |  |  |\_/  |
_|__|__|__|__|__|__|___
| a                b
|
```

This, of course, is only an approximation. But you can make the approximation better and better by making your line segments shorter and shorter. The easy way to measure the line segments would be with a flexible tape measure, but that's not very mathematical, is it? So let's use right triangles and pull the Pythagorean Theorem out of our dusty old geometry textbooks.

```          _
|      /|\
|     /_| \       /
|    /| |  \     /
|   / | |   \   /
|     | |    \_/
_|_____|_|_____________
|  a a   a         b
|     i   i+1
```

We'll call the horizontal distance between those two vertical lines Δx (delta-x), because d stands for "distance" and we like using Greek to make our proofs sound more mathematical. The vertical line at x=ai strikes the curve at f(x)=f(ai), so the vertical distance (Δyi) between the two points is f(ai+1)-f(ai). So far so good?

Now Pythagoras comes in, and we have the length of the line segment equal to the square root of Δx2+Δyi2. Ugly, but useful. We'll clean it up later.

If you break up the curve using, oh, about n vertical lines of equal Δ (er, distance apart), you can now write the length of the entire curve as the sum of all the line segments. Each line segment's length equals the square root of Δx2+Δyi2, where i is between 1 and n, so we can add up all the segments and write it out using even more Greek, in the form of sigma notation:

```  _n_      _________
\       /  2    2
/__   \/ Δx +Δy
i=1            i
```

But that's ugly. Too many letters, for one thing. So let's work on getting rid of those "y"s, because they have funny tails and we don't like things like that sticking out of our mathematical formulas.

We said already that

```  Δyi = f(ai+1)-f(ai)
```

That looks familiar to you, so you go rummaging through the earlier chapters in your textbooks and find... yes! The Mean Value Theorem, an intuitively obvious bit of math which is actually useful for once! That theorem tells us there is some value for x, call it xi between ai and ai+1, such that

```  f'(xi)(ai+1-ai) = f(ai+1)-f(ai) = f'(xi) Δx
```

Well, there you go. We now have

```  Δyi = f'(xi) Δx
```

And substituting that into our messy y-infested formula for the arc length, we have

```  _n_      ________________
\       /  2           2
/__   \/ Δx +(f'(x )Δx)
i=1               i
```

or, after cleaning up under the square root:

```  _n_      ___________
\       /         2
/__   \/ 1 +f'(x )   Δx
i=1             i
```

But that's still not precise enough for your math professor, because he wants Δx to be infinitely small. So you let n increase to infinity using something he showed you months ago called a limit. Pretending for a moment that % means infinity, we have

```      _n_     ____________
lim  \      /          2
n->% /__  \/ 1 + f'(x )   Δx
i=1             i
```

But wait! This is the definition of an integral! (Well, at least it is when the limit is finite and the function is continuous. But then, you wouldn't want to try and find the length of an infinite or broken-in-half curve anyhow.) Rejoicing and huzzahs abound as you simplify one more time:

```           __________
|\b    /         2
|     / 1 + f'(x)  dx
\| a \/
```

This tidy formula you can now plug into your advanced calculator or Mathematica program or (if you're really geeky) do it in your own head to find the exact length of the arc, down to the last micrometer. Or whatever. It may take ten times as long as the tape measure, but at least you can go to bed knowing with confidence that your answer was precise.

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