Hi, anyone able to help with this. show that the perimeter(arc length) of the ellipse (x^2/a^2)+(y^2/b^2)=1 where a>b>0, is given by the elliptic intergral l=4a(intergral between 0-PI/2 of)(sqrt(1-e^2*cos(t)^2) dt where e=sqrt(1-(b/a)^2) completely lost
we may assum the arc length is given by l=(intergral between t0-t1) sqrt((dx/dt)^2+(dy/dt)^2)dt
x = acos(t); y = bsin(t) is the parametric equation of the ellipse. dx/dt = -asin(t); dy/dt = bcos(t); Plug it into the formula for arc length. And due to symmetry, the perimeter of the ellipse will be 4 times the arc length from t = 0 to t = pi/2.
Thank you, I'll give that a go
You are welcome.
when I plug it in, should I be trying to integrate it, or turn it into the elliptic intergral?
(dx/dt)^2 + (dy/dt)^2 = a^2sin^2(t) + b^2cos^(t) Replace sin^2(t) by (1 - cos^2(t)) (dx/dt)^2 + (dy/dt)^2 = a^2(1- cos^2(t)) + b^2cos^(t) = a^2 - a^2cos^2(t) + b^2cos^2(t) = a^2 - cos^2(t)(a^2 - b^2) = a^2 - a^2cos^2(t)(1 - (b/a)^2) = a^2{ 1 - cos^2(t)(1 - (b/a)^2) } = a^2(1 - e^2cos^2(t)
Perimeter = \[ 4\int_0^{\pi/2}\sqrt{(dx/dt)^2 + (dy/dt)^2}dt = 4\int_0^{\pi/2}\sqrt{a^2(1-e^2\cos^2(t))}dt = \\ 4a\int_0^{\pi/2}\sqrt{(1-e^2\cos^2(t))}dt = ?\\ \]
okay, Thimk I might be able to get it now, so intergrate this to show ( x^2/a^2)+(y^2/b^2)=1
I think they are just asking you to show that the above integral represents the perimeter of the ellipse which we have done already. They are not expecting you to evaluate the integral and so you can ignore my question mark above.
ahh, okay, thanks
We started with the given equation of an ellipse in Cartesian coordinates. made the substitution and converted the equation into parametric form and said because of the symmetry of the ellipse, the perimeter is 4 times the arc length from t = 0 to t = pi/2 and plugged dx/dt and dy/dt into the arc length formula and did some algebra and reduced it to the form they want you to show represents the perimeter of an ellipse which we have done.
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