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Thread: Word problem:-

  1. #1
    Member great_math's Avatar
    Oct 2008

    Word problem:-

    There are 2 discs each of radius 'a' placed one above the other.

    The bottom disc is fixed and the above disc moves in a straight line with a velocity inversely proportional to the area exposed of the bottom disc. The constant of proportionality is found to be one.

    Find the time after which the bottom disc will be fully exposed, i.e., the moving disc on top falls off.

    Thank you.
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  2. #2
    Super Member
    Aug 2008
    Seems we'd have to first calculate the area remaining as a function of the distance of the second disc is away from the center of the first. Assume the lower disc is at the origin and we're moving the top one along the positive x axis. Then the equation for the top one is $\displaystyle (x-s)^2+y^2=a^2$. Now, consider the plot below and the football shaped intersection of the two discs. The area of that region as a function of s is:

    $\displaystyle f(s)=4\int_{s/2}^{a} \sqrt{a^2-x^2}dx$

    where I'm just calculating the area of the upper right sector of the football and multiplying by 4 right? So that the exposed area is just $\displaystyle A(s)=\pi a^2-f(s)$.

    You said: $\displaystyle \frac{ds}{dt}=\frac{1}{A(s)}$

    (I think anyway).


    $\displaystyle 4\int_{0}^{s}\int_{s/2}^a \sqrt{a^2-x^2}dx ds=t$

    It looks confusing but I think it's valid: do the second integral with respect to the "dummy" variable s, then substitute "s" for the upper limit and zero for the lower. It gets messy but I get the eclipse is over at $\displaystyle t=\pi a^3-2/3 a^3\approx 2.47 a^3$, but to be honest I'd be sweating if this was a test question (I might be wrong).
    Attached Thumbnails Attached Thumbnails Word problem:--twodiscs.jpg  
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