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jan 03 pure 3 past paper q watch

    • Thread Starter

    Hi ya im a little stuck on this q, and iv got my exam on tues!! :confused:
    any help is gratefully appreciated!!

    6) the surface of a circular pond of radius a is bein covered by weeds. The weeds are growing in a circular region whose centre is at the centre of the pond. At time t the region covered by the weeds has radious r and area A. An ecologist models the growth of the weeds by assumin that the rate of increase of the area covered is proportional to the area of the pond not yet covered.

    i) i get this one,, show that dA/dt = 2pi r (dr/dt)

    A = pi r^2
    so dt/d! = 2pi.r (dr/dt)

    i get stuck here onwards!!

    ii) hence show that the ecologists model leads to the differential equation

    2r(dr/dt)= k(a^2 - r^2)
    where k is a constant.

    iii) By solving the differential equation in part (ii), express r in terms of t, a and k, given that r = 0 when t=0.

    iv) will the weeds ever cover the whole pond?? justify ur answer.

    thank u!!

    Area of region not covered = X = pi a² - pi r² = pi (a² - r²)

    The ecologist's model suggests that dA/dt = kX. So:
    dA/dt = k pi (a² - r²)
    2 pi r (dr/dt) = k pi (a² - r²)
    2 r (dr/dt) = k (a² - r²)

    2r dr = k (a²-r²) dt
    ∫ 2r/(a²-r²) dr = ∫ k dt
    - ln|a² - r²| = kt + C --- [The LHS was done using the sub u=a²-r².]
    ln|a² - r²| = C - kt
    a² - r² = e^C . e^(-kt)
    r² = a² - Ae^(-kt) --- [where A=e^C]

    Since r=0 when t=0, we have:
    0 = a² - A
    A = a²

    r² = a² - a².e^(-kt)
    r = [a² - a².e^(-kt)]0.5

    As t becomes very large, a².e^(-kt) tends to zero. So, after a sufficiently long time, r tends to a.

    So yes.
    • Thread Starter

    thank u very much!! it kinda makes sense now!!! wish i found it as easy as u den i wud easily ace this exam!!
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Updated: June 3, 2005

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