Analytically and numerically computed arc length
Asked by Carlos on 10 Jul 2012
Latest activity Commented on by Walter Roberson on 10 Jul 2012
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?
Thanks in advance.
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2 Answers
Answer by Teja Muppirala on 10 Jul 2012
Accepted answer
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.
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Answer by Carlos on 10 Jul 2012
You are right. I made a mistake and I came here too soon.
Thanks. Should I delete that question?
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