## Finding the length of curve using MATLAB

As per integral calculus, the length of a continuous and differentiable curve f(x) from x=a to x=b is given by

S=$\int_a^b \sqrt{(1+(dy/dx)^2} dx$

Now how do we find the length of a curve in MATLAB.

Let us do this via an example. Assume one asked you to find the length of $x^2*sin(x)$ from Π to 2Π. In the book, How People Learn, the authors mention that learning a concept in multiple contexts prolongs retention. Although it may not be the context that the authors of the book are talking about, let us find the length of the curve multiple ways within MATLAB. Try the program for functions and limits of your own choice to evaluate the difference.

METHOD 1: Use the formula S=$\int_a^b \sqrt{(1+(dy/dx)^2} dx$ by using the diff and int function of MATLAB

METHOD 2: Generate several points between a and b, and join straight lines between consecutive data points. Add the length of these straight lines to find the length of the curve.

METHOD 3. Find the derivative dy/dx numerically using forward divided difference scheme, and then use trapezoidal rule (trapz command in MATLAB) for discrete data with unequal segments to find the length of the curve.

1. Why does METHOD 3 giving inaccurate results? Can you make them better by using better approximations of derivative like central divided difference scheme?
2. Redo the problem with f(x)=$x^{\frac{3}{2}}$ with a=1 and b=4 as the exact length can be found for such a function.

% Simulation : Find length of a given Curve
% Language : Matlab 2007a
% Authors : Autar Kaw
% Last Revised : June 14 2008
% Abstract: We are finding the length of the curve by three different ways
% 1. Using the formula from calculus
% 2. Breaking the curve into bunch of small straight lines
% 3. Finding dy/dx of the formula numerically to use discrete function
% integration
clc
clear all

disp(‘We are finding the length of the curve by three different ways’)
disp(‘1. Using the formula from calculus’)
disp(‘2. Breaking the curve into bunch of small straight lines’)
disp(‘3. Finding dy/dx of the formula numerically to use discrete function integration’)

%INPUTS – this is where you will change input data if you are doing
% a different problem
syms x;
% Define the function
curve=x^2*sin(x)
% lower limit
a=pi
% b=upper limit
b=2*pi
% n = number of straight lines used to approximate f(x) for METHOD 2
n=100
%p = number of discrete data points where dy/dx is calculated for METHOD 3
p=100

% OUTPUTS
% METHOD 1. Using the calculus formula
% S=int(sqrt(1+dy/dx^2),a,b)
% finding dy/dx
poly_dif=diff(curve,x,1);
% applying the formula
integrand=sqrt(1+poly_dif^2);
leng_exact=double(int(integrand,x,a,b));
fprintf (‘\nExact length =%g’,leng_exact)
%***********************************************************************

% METHOD 2. Breaking the curve as if it is made of small length
% straight lines
% Generating n x-points from a to b

xi= a:(b-a)/n:b;
% generating the y-values of the function
yi=subs(curve,x,xi);
% assuming that between consecutive data points, the
% curve can be approximated by linear splines.
leng_straight=0;
m=length(xi);
% there are m-1 splines for m points
for i=1:1:m-1
dx=xi(i+1)-xi(i);
dy= yi(i+1)-yi(i);
leneach=sqrt(dx^2+dy^2);
leng_straight=leng_straight+leneach;
end
fprintf (‘\n\nBreaking the line into short lengths =%g’,leng_straight)

% METHOD 3. Same as METHOD1, but calculating dy/dx
% numerically and integrating using trapz
xi=a:(b-a)/p:b;
% generating the dy/dx-values
m=length(xi);
for i=1:1:m-1
numer=yi(i+1)-yi(i);
den=xi(i+1)-xi(i);
dydxv(i)=numer/den;
end
% derivative at last point using Backward divided difference formula
% is same as Forward divided difference formula
dydxv(m)=dydxv(m-1);
integrandi=sqrt(1+dydxv.*dydxv);
length_fdd=trapz(xi,integrandi);
disp(‘ ‘)
disp(‘ ‘)
disp (‘Using numerical value of dy/dx coupled’)
disp (‘with discrete integration’)
fprintf (‘ =%g’,length_fdd)

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu

## A simple MATLAB program to show that High order interpolation is a bad idea

In a previous post, we talked about that higher order interpolation is a bad idea.

In this post I am showing you a MATLAB program that will allow you to experiment by changing the number of data points you choose, that is, the value of n (see the input highlighted in red in the code – this is the only line you want to change) and see for yourself why high order interpolation is a bad idea. Just, cut and paste the code below (or download it from http://www.eng.usf.edu/~kaw/download/runge.m) in the MATLAB editor and run it.

### % Simulation : Higher Order Interpolation is a Bad Idea

% Language : Matlab r12
% Authors : Autar Kaw
% Last Revised : June 10 2008
% Abstract: In 1901, Carl Runge published his work on dangers of high order
% interpolation. He took a simple looking function f(x)=1/(1+25x^2) on
% the interval [-1,1]. He took points equidistantly spaced in [-1,1]
% and interpolated the points with polynomials. He found that as he
% took more points, the polynomials and the original curve differed
% even more considerably. Try n=5 and n=25
clc
clear all
clf

disp(‘In 1901, Carl Runge published his work on dangers of high order’)
disp(‘interpolation. He took a simple looking function f(x)=1/(1+25x^2) on’)
disp(‘the interval [-1,1]. He took points equidistantly spaced in [-1,1]’)
disp(‘and interpolated the points with a polynomial. He found that as he’)
disp(‘took more points, the polynomials and the original curve differed’)
disp(‘even more considerably. Try n=5 and n=15’)

%
% INPUT:
% Enter the following
% n= number of equidisant x points from -1 to +1
n=15;

% SOLUTION
disp(‘ ‘)
disp(‘SOLUTION’)
disp(‘Check out the plots to appreciate: High order interpolation is a bad idea’)
fprintf(‘\nNumber of data points used =%g’,n)
% h = equidisant spacing between points
h=2.0/(n-1);
syms xx
% generating n data points equally spaced along the x-axis
% First data point
x(1)=-1;
y(1)=subs(1/(1+25*xx^2),xx,-1);
% Other data points
for i=2:1:n
x(i)=x(i-1)+h;
y(i)=subs(1/(1+25*xx^2),xx,x(i));
end

% Generating the (n-1)th order polynomial from the n data points
p=polyfit(x,y,n-1);

% Generating the points on the polynomial for plotting
xpoly=-1:0.01:1;
ypoly=polyval(p,xpoly);

% Generating the points on the function itself for plotting
xfun=-1:0.01:1;
yfun=subs(1/(1+25*xx^2),xx,xfun);

% The classic plot
% Plotting the points
plot(x,y,’o’,’MarkerSize’,10)
hold on
% Plotting the polynomial curve
plot(xpoly,ypoly,’LineWidth’,3,’Color’,’Blue’)
hold on
% Plotting the origianl function
plot(xfun,yfun,’LineWidth’,3,’Color’,’Red’)
hold off
xlabel(‘x’)
ylabel(‘y’)
title(‘Runges Phenomena Revisited’)
legend(‘Data points’,’Polynomial Interpolant’,’Original Function’)
%***********************************************************************
disp(‘ ‘)
disp(‘ ‘)
disp(‘What you will find is that the polynomials diverge for’)
disp(‘0.726<|x|<1. If you started to choose same number of points ‘)
disp(‘but more of them close to -1 and +1, you would avoid such divergence. ‘)
disp(‘ ‘)
disp(‘However, there is no general rule to pick points for a general ‘)
disp(‘function so that this divergence is avoided; but some rules do exist for ‘)
disp(‘certian types of functions.’)

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu