Finding the perimeter of a 2D ellipse using composite trapezoidal integration

I am trying to find the perimeter of a 2D ellipse given by the parametric equations: x = 4 * cos(t), y = 3 * sin(t), for 0 <= t <= 2*pi. I'm using a provided trapezoid function, called trap given below:

% trap function

% a = lower limit of integral, b = upper limit of integral, n = # of subintervals

function g=trap(a,b,n)

xd=linspace(a,b,n+1);

h=xd(2)-xd(1);

sum=0.5*f(xd(1));

for j=2:n

sum=sum+f(xd(j));

end

g=h*(sum+0.5*f(xd(n+1)));

end

I also have a function for the f(x) version of the ellipse function: (x/4)^2 + (y/3)^2 = 1

function y=f(x)

y=3*sqrt(1-(x^2/16));

end

However, this does not seem to be working. My values turn out to be complex since you cannot find the square root of a negative number. I know I'm doing something wrong, but cannot seem to wrap my head around it. My main question is how do I get the equation of the ellipse into y = f(x) form. I don't think I'm doing it correctly. I have a = 0 and b = 2*pi since that's the range we are integrating over. However, I'm also not sure about that. Any help would be greatly appreciated.

Answer 1

[Torsten]

The arclength of a curve in parametrized form C = ((f1(t),f2(t)) ,a<=t<=b) is given as

integral_{a}^{b} sqrt(f1'(t)^2+f2'(t)^2) dt.

This is used in the code below.

If you want to use your trap function instead of trapz, you can implement it as

perimeter = trap(f,0,2*pi,100) 

and

function g = trap(f,a,b,n)
...
end

  syms t
  x = 4*cos(t);
  y = 3*sin(t);
  dfx = diff(x,t);
  dfy = diff(y,t);
  f = sqrt(dfx^2+dfy^2);
  f = simplify(f);
  f = matlabFunction(f);
  T = linspace(0,2*pi,100);
  F = f(T);
  perimeter = trapz(T,F)

[Lorson Blair]

Thank you so much. This works.