Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

we may assum the arc length is given by l=(intergral between t0-t1) sqrt((dx/dt)^2+(dy/dt)^2)dt

OpenStudy (aum):

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.

OpenStudy (anonymous):

Thank you, I'll give that a go

OpenStudy (aum):

You are welcome.

OpenStudy (anonymous):

when I plug it in, should I be trying to integrate it, or turn it into the elliptic intergral?

OpenStudy (aum):

(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)

OpenStudy (aum):

Perimeter = \[ 4\int_0^{\pi/2}\sqrt{(dx/dt)^2 + (dy/dt)^2}dt = 4\int_0^{\pi/2}\sqrt{a^2(1-e^2\cos^2(t))}dt = \\ 4a\int_0^{\pi/2}\sqrt{(1-e^2\cos^2(t))}dt = ?\\ \]

OpenStudy (anonymous):

okay, Thimk I might be able to get it now, so intergrate this to show ( x^2/a^2)+(y^2/b^2)=1

OpenStudy (aum):

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.

OpenStudy (anonymous):

ahh, okay, thanks

OpenStudy (aum):

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.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!