## Analytically and numerically computed arc length

on 10 Jul 2012

### Teja Muppirala (view profile)

Hi,

I'm trying to compute the length of a curve defined in parametric form:

```t = linspace(0,pi); % Actually t could go from 0 to any angle lower than 2*pi
```
```r = 1 ./ ( 1 - t / (2*pi) );
```
```x = r.*cos(t); y = r.*sin(t);
```
```dx = diff(x); dy = diff(y);
```
```l = sum( sqrt(dx.^2 + dy.^2) ); % Arc length. Linear aprox.
```

This way the length is equal to 4.4725.

If I do the calculations analytically, I find the length is:

```l = -2*pi*log( 1 - angle/(2*pi) ); % Being the initial point angle = 0
```

using angle = pi the result is 4.3552.

What's the reason of this difference?

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### Teja Muppirala (view profile)

on 10 Jul 2012

I'm not sure how you are analytically calculating path length. It seems from your expression that you just integrated r(t) straight up, which is not the correct way to do it. You need to use the formula for path length.

For example, it is given for cylindrical coordinates here:

http://tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx (Google: arc length cylindrical coordinates)

That equation is difficult to integrate by hand, but it can be done symbolically in MATLAB very easily.

```syms r t
r = 1 ./ ( 1 - t / (2*pi) );
ds_dt = sqrt(r^2 + diff(r)^2);
path_length = int(ds_dt,0,pi)
subs(path_length)
```

You get a long analytical expression for path length, which turns out to be the same answer as you got numerically, 4.47.

### Carlos (view profile)

on 10 Jul 2012

You are right. I made a mistake and I came here too soon.

Thanks. Should I delete that question?

Walter Roberson

### Walter Roberson (view profile)

on 10 Jul 2012

No, it's okay, it will help other people in future.

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