
#1
May2306, 03:15 PM

P: 17

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(1c)[/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(1c))^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) 


#2
May2306, 08:51 PM

P: n/a

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. 



#3
May2306, 10:00 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

You won't be able to compute this integral directly  it doesn't have an expression in terms of "elementary" functions. 



#4
May2306, 11:09 PM

Mentor
P: 6,039

Circumference of an ellipse
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 



#5
May2506, 09:27 AM

P: 17

Ok, thanks guys. I'll have a go doing that (although my Taylor Series expansion knowledge is sketchy at best).




#6
May2506, 02:25 PM

HW Helper
P: 1,007

[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
May2506, 09:33 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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
May2606, 07:06 AM

P: 17

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.



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