finding lengths of an ellipse

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PB on 18 Apr 2019
Commented: David Wilson on 18 Apr 2019
I have to find the lengths of an ellipse which is given by y = cost and x = 5 sint. Evaluate the length from t1=0 to the following points,
t2=[0.5,1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6]. Have to use one of the methods, trapezoidal, midpoint or simpson's to two digit accuracy. How do i write a script from 0 to 0.5, 0 to 1, 0 to 1.5 etc ( from 0 to each point) to get 12 values for the lengths.
PB on 18 Apr 2019
yes the parametric equations given this way and we have to MATLAB code for it.

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Accepted Answer

David Wilson
David Wilson on 18 Apr 2019
If you need a little more help, then it pays to plot the ellipse and look at the arc lengths.
t = linspace(0,2*pi)';
x = @(t) 5*sin(t); y = @(t) cos(t);
plot(x(t),y(t),'r-', ...
x(0), y(0), 'bs', ...
x(t2), y(t2), 'kh');
for i=1:length(t2)
text(x(t2(i)), y(t2(i)),sprintf('%2.1f',t2(i)))
grid on; axis equal;
xlabel('x'); ylabel('y')
It's important to have the axis equal in order to roughly approximate the arc lengths. You can see that from t=0 to t=0.5, the arc length is a little over 2.5. That makes a good check.
Now I've used the formlar above for the arc length, but I've had to convert your sine to my cosine etc. I;'ve also used integral, your (homework?) requested you use a simpler Simpson etc, which I'll leave you to implement. At least you have a tentative solution.
>> f = @(t) sqrt(25*sin(pi/2-t).^2 + cos(pi/2-t).^2);
>> arcLengths = arrayfun(@(t2) integral(f,0,t2), t2)
arcLengths =
As a check, you can see that my guess of 2.5 is not too far off.
David Wilson
David Wilson on 18 Apr 2019
Following my way, you don't need to adjust the stepsize since integral will (magically) take care of it using its adaptive step size ability. I didn't bother add any tolerance to integral, since 2 decimal places is pretty lax.
HOWEVER, you were asked to use a simple integrator without error control, so you will have to choose a suitable stepsize when using Simpson's etc. If you are going to use a straight-forward approach, I'd subdivied the interval (in t) to say about 100 steps as a first approximation.

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