Approximating the Circumference of an Ellipse with Parametric Equations

In summary, an ellipse has an equation which can be written parametrically as:x = a cos(t)y = b sin(t)It can be proved that the circumference of this ellipse is given by the integral:\int^{2\pi}_0 \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} \ \ dt \In summary, an ellipse has an equation which can be written parametrically, and it can be proved that the circumference is given by the integral: 2*pi*r(1+0.5*c^2)
  • #1
Libertine
16
0
An ellipse has an equation which can be written parametrically as:
x = a cos(t)
y = b sin(t)

It can be proved that the circumference of this ellipse is given by the integral:
[tex]\int^{2\pi}_0 \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} \ \ dt[/tex]

Prove that, if [tex]a=r(1+c)[/tex] and [tex]b=r(1-c)[/tex], where c is a positive number small enough for powers higher than [tex]c^2[/tex] to be neglected, then this circumference is approximately:
[tex]2 \pi r (1+\frac{1}{4}c^2)[/tex]

So I substituted in the expressions for a and b:
[tex]$ \int^{2\pi}_0 \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} \ \ dt \\
=\int^{2\pi}_0 \sqrt{(r(1+c))^2 \sin^2 t + (r(1-c))^2 \cos^2 t} \ \ dt \\
=r \int^{2\pi}_0 \sqrt{\sin^2 t + 2c \sin^2 t + c^2 \sin^2 t + \cos^2 t - 2c \cos^2 t + c^2 \cos^2 t} \ \ dt \\
=r \int^{2\pi}_0 \sqrt{(\sin^2 t + \cos^2 t)+ c^2( \sin^2 t + \cos^2 t) + 2c(\sin^2 t - \cos^2 t)} \ \ dt \\
=r \int^{2\pi}_0 \sqrt{1 + c^2 + 2c(\sin^2 t - \cos^2 t)} \ \ dt \\ $
[/tex]
After this point, I seem to hit a brick wall and can't simplify it any further or factorise to get rid of that annoying square root. Any help appreciated.

(I'm assuming the tex won't come out all right first time, so I'll be trying to correct it for a little while)
 
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  • #2
Although I can't explicitly tell you how to help, perhaps approximating the square root function in the integral will get you somewhere. I see one or two term Taylor expansions justify equations in classes all the time as "approximations" :) ( http://mathworld.wolfram.com/SquareRoot.html )

Edit: It was worth a shot. Came out to 2*pi*r(1+0.5*c^2) by my calculations.

Edit 2: Whoops I said Newton's Method instead of Taylor series.
 
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  • #3
where c is a positive number small enough for powers higher than [itex]c^2[/itex] to be neglected,
That sounds like a command to use Taylor series, and neglect all the terms with powers higher than .



You won't be able to compute this integral directly -- it doesn't have an expression in terms of "elementary" functions.
 
  • #4
Considering the first 3 terms in the power series expansion of

[tex](1 + x)^{\frac{1}{2}},[/tex]

where

[tex]x = c^2 + 2c \left( \sin^2 t - \cos^2 t \right) = c^2 - 2c \cos 2t[/tex]

is small, seems to give the right answer.

Edit: I got the same result as vsage when I considered the first 2 terms of the power series expansion, but this does not include all terms of order c^2.

Regards,
George
 
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  • #5
Ok, thanks guys. I'll have a go doing that (although my Taylor Series expansion knowledge is sketchy at best).
 
  • #6
[tex](1 + x)^{\frac{1}{2}}= 1 + \frac{1}{2} x - \frac{1}{8}x^2 + \frac{1}{16} z^3+\cdots , |x|<1[/tex]
 
  • #7
You don't actually have to know the Taylor expansion: you only need a couple terms, so it's easy enough to compute. You simply need to differentiate with respect to c twice to get all the terms up to order c^2.
 
  • #8
Just for the curious: I did managed to get the answer in the end - it was necessary to use the third term of the expansion. Thanks.
 

What is the formula for finding the circumference of an ellipse?

The formula for finding the circumference of an ellipse is 2π√((a²+b²)/2), where a and b are the semi-major and semi-minor axes respectively.

How does the circumference of an ellipse compare to the circumference of a circle?

The circumference of an ellipse is always longer than the circumference of a circle with the same radius. This is because an ellipse has two different radii, the semi-major and semi-minor axes, while a circle only has one radius.

Can the circumference of an ellipse be measured directly?

No, the circumference of an ellipse cannot be measured directly. It can only be calculated using the formula or approximated using a string or flexible measuring tape.

Why is the calculation of the circumference of an ellipse important in real-world applications?

The calculation of the circumference of an ellipse is important in many real-world applications, such as engineering and architecture. It is also used in the fields of astronomy and physics to measure the orbits of planets and other celestial bodies.

Are there any other methods for finding the circumference of an ellipse besides the formula?

Yes, there are other methods for finding the circumference of an ellipse, such as using the Ramanujan approximation or numerical integration. However, the formula is the most commonly used and accurate method for calculating the circumference of an ellipse.

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