Solving for arc length of an ellipse

[redeemer90]

Homework Statement

The task is to solve for the arc length of an ellipse numerically. a & b are given for an ellipse centered at the origin and a value for x is given.

Homework Equations

Equation of ellipse is given to be
[tex]x^{2}/a^{2} + y^{2}/b^{2} = 1[/tex]
and the equation to solve for the arc length is given as
[tex]a \int^{\theta}_{0}\sqrt{1-k^{2} sin^{2}t} dt[/tex]
Assuming a is the major axis

The Attempt at a Solution

The additional condition is that
[tex]-a \leq x\leq a[/tex], so [tex]\theta[/tex] can be [tex]\ge 0.5 \pi[/tex]
Since [tex]\int^{\theta}_{0}\sqrt{1-k^{2} sin^{2}t} dt[/tex] does not seem to work when [tex]{\theta} \ge 0.5 \pi[/tex]

The only solution I can think of is as follows

1. If x < 0
2. pb4 = quarter the perimeter of the ellipse
3. Set x= -x (reflect about the x axis)
4. ptemp = arc length for the positive x
5. The final answer would be p=pb4+(pb4-ptemp)

This would mean evaluating equation [tex]\int^{\theta}_{0}\sqrt{1-k^{2} sin^{2}t} dt[/tex] twice.
Is there a better solution to this problem?

Thanks,

- Sid

 

[mighty2000]

Dear Sid,

I would recommend using numerical integration methods to solve for the arc length of an ellipse. This would involve breaking down the integral into smaller intervals and using numerical techniques such as the trapezoidal rule or Simpson's rule to approximate the integral. This method would be more accurate and efficient than evaluating the integral twice.

Another approach would be to use the parametric equations of an ellipse, which can be written as x = a cos t and y = b sin t. Using these equations, we can derive the equation for arc length as S = \int^{\theta}_{0}\sqrt{a^{2}sin^{2}t + b^{2}cos^{2}t} dt. This integral can then be solved numerically using the techniques mentioned above.

I hope this helps. Good luck with your calculations!