Rotating ellipse

Geometry Level 5

An ellipse E E is rotating about its center with a constant angular speed ω \omega in the counter clockwise direction. However, during its rotation, its eccentricity, e e , is varying with time so as to satisfy the following criteria.

  1. The foci of the ellipse E E trace out another fixed, concentric ellipse E E' whose eccentricity is 1 2 \displaystyle \frac{1}{2} .

  2. The foci of the ellipse E E' always lie on the ellipse E E .

Evaluate the root mean square eccentricity of the ellipse E E i.e., < e 2 > \displaystyle \sqrt{< e^2 > } .

Details and Assumptions:

  • Both the ellipses are concentric as mentioned before. Moreover, the length of semi-major axis of both the ellipses is the same equal to a = 5 a = 5 .

  • The conditions are satisfied at all instants of time and hence the eccentricity is continuous function of time except when their axes are parallel.

  • The average has to be taken over one complete rotation.

  • The root mean square of continuous function p ( x ) p(x) defined as [ a , b ] R \displaystyle [ a , b ] \to \mathbb{R} is given as < ( p ( x ) ) 2 > = a b ( p ( x ) ) 2 d x a b d x \displaystyle \sqrt{< (p(x))^2 > } = \sqrt{\frac{\int \limits_a^b (p(x))^2 \text{ d}x }{\int \limits_a^b \text{ d}x }}


The answer is 0.931.

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1 solution

Vishad Viplav
Dec 9, 2014

the path traced by foci can be taken as x=5cos(wt) and y=5 (3^0.5)/2sin(wt) the distance of the point on path at time t can be calculated as sqrt(x^2+y^2) which would be equal to a e where e is the eccentricity e^2=3/4+cos^2(wt)

please elaborate someone

Zerocool 141 - 4 years, 6 months ago

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