User:Egm6341.s10.Team4.roni/HW5/5.10

= Problem 10: Application of Engineering Orbital Mech. =

Given
Refer Lecture slide 30-4 for problem statement

Polar form relative to focus
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 * [[Image:Ellipse Polar.svg|200px]]
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Eq (3) from Page [[media:Egm6341.s10.mtg30.djvu|30-3]]

$$\displaystyle \ \ r(\theta) = \frac{a(1-e^2)}{1-e \, cos(\theta)} $$

In our problem, a=1

where, eccentricity $$\displaystyle e = \sin(\frac{\pi}{12}) $$

First Method to compute the Arc Length for an elliptical curve is using Eq (4) from Page [[media:Egm6341.s10.mtg30.djvu|30-3]] Since for the Ellipse $$\displaystyle \theta$$ range is from 0 to 2 $$\displaystyle\pi $$

Circumference $$\displaystyle I( \theta) = \int_0^{2\pi} dl$$

Where Eq (1) $$\displaystyle dl= \sqrt{ {r}^{2}+ {\left(  \frac{dr}{d \theta}\right)}^{2}} d\theta $$.

The second method uses the elliptic integral of the second kind which is defined as:

$$E(e^2) = \int_0^{\pi/2}\sqrt {1-e^2 \sin^2\theta}\ d\theta\ $$    Eq(6)    in page     [[media:Egm6341.s10.mtg30.djvu|30-4]]

The circumference of an ellipse is $$\displaystyle C = 4 a E(e^2)$$    Eq(5)   in page   [[media:Egm6341.s10.mtg30.djvu|30-4]],

Giving the integral:


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$$\displaystyle I= 4 E(e^2) = 4\int_0^{\pi/2}\sqrt {1-e^2 \sin^2\theta}\ d\theta $$

Find
Find Arc length of ellipse using the two Integrals and the integration methods below.

compute $$\displaystyle I_n$$ to the error $$\displaystyle 10^{-10}$$ ,

compute time using tic/toc matlab command and error estimate

1) Composit Trapozoidal rule

2) Romberg Table

3) Clencurt

Solution
The Integration result are summarized in the following table:



Note that checking both integral calculations against Ramanujan approximation 's:
 * $$C \approx \pi \left[3(a+b) - \sqrt{(3a+b)(a+3b)}\right]= \pi(3(a+b)-\sqrt{10ab+3(a^2+b^2)})$$

where, a = longest radius, b= shortest radius in ellipse

the results of our integration method and Ramanujan approximation were same. $$\displaystyle (c \simeq 6.176601987)$$

Discussion: One can clearly see that for these integrals the Trapezoidal method gives superior performance This is due to the fact that the Integrand is a periodic function. Clencurt has slightly better performance than Romberg. First method Integral gives lower quality on all methods and need more CPU time for similar performance.

(1) Composite Trapezoidal rule

 Matlab Code: 

 subfunction ff5_10(x) 

(2) Romberg Table

 Matlab Code: 

(3) ClenCurt Integration)

 Matlab Code: 

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