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    Hey guys,

    Really stuck on this question....

    When a plague of locusts attack a wheat crop, the proportion of the crop destroyed after t hours is denoted by x. In a model, it is assumed that the rate at which the crop is destroyed is proportional to x(1-x).
    A plague of locusts is discovered in a wheat crop when 1/4 of the crop has been destroyed.
    Given that the rate of destruction at this instant is such that if it remained constant, the crop would be completley destroyed in a further 6 hours.

    i) show that dx/dt= 2/3x(1-x)
    ii) find the percentage of the crop destroyed 3 hours after the plague of locusts is discovered.

    Okay so dx/dt= kx(1-x)

    conditions, t=o when X=1/4
    and dx/dt=k, t=0?

    Please help me wtf am I doing!
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    For part i)

    When x=0.25, the rate of destruction is such that the remaining 0.75 of the crops would be destroyed in 6 hours, so dx/dt=0.75/6=3/24 where x=0.25

    As you rightly stated dx/dt=kx(1-x)

    So when x=0.25, kx(1-x)=3/24
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    (Original post by Dj.Clay)
    For part i)

    When x=0.25, the rate of destruction is such that the remaining 0.75 of the crops would be destroyed in 6 hours, so dx/dt=0.75/6=3/24 where x=0.25

    As you rightly stated dx/dt=kx(1-x)

    So when x=0.25, kx(1-x)=3/24
    but it asks for dx/dt=2/3x(x-1)
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    (Original post by J DOT A)
    but it asks for dx/dt=2/3x(x-1)
    Substitute 1/4 in to get k
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    If a quarter of the field has been destroyed, then \frac{3}{4} remains, that is then destroyed in 6 hours so the hourly rate \dfrac{dx}{dt} = \dfrac{3}{4} \div 6 = \dfrac{3}{24}

    @ x = 0.25
    \dfrac{dx}{dt} = kx(1-x) = k\frac{1}{4}(1-\frac{1}{4}) = k\frac{3}{16} = \frac{3}{24}

    Therefore:
    k = \dfrac{3}{24} \div \dfrac{3}{16} = \dfrac{2}{3}

\dfrac{dx}{dt} = \dfrac{2}{3}x(1-x)

    To integrate:
    \int \dfrac{1}{x(1-x)} dx = \int \frac{2}{3} dt

    Use partial fractions to simplify the integral:
    \dfrac{1}{x(1-x)} = \dfrac{A}{x} + \dfrac{B}{1-x}

1 = A(1-x) + B(x)

    @ x = 0
    1 = A + 0B

A = 1

    @ x = 1
    1= 0A + B

B = 1
    Therefore \dfrac{1}{x(1-x)} = \dfrac{1}{x}+\dfrac{1}{1-x}

    \int \dfrac{1}{x} + \dfrac{1}{1-x} dx = \int \frac{2}{3} dt

    And then integrate:
    \ln{x} -\ln{(1-x)} = \frac{2}{3}t + c

    Put in known values to find c:
    @ t = 0, x = 0.25
    \ln0.25 - \ln0.75 = 0 + c = c = -ln3

\ln{x} -\ln{(1-x)} = \frac{2}{3}t - \ln3

    And now put the value of t in to find x:
    @ t = 3
    \ln{x} -\ln{(1-x)} = \frac{2}{3}3 - \ln3 = 2 - \ln3

\ln{x} -\ln{(1-x)} = \ln(\dfrac{x}{1-x})

\ln(\dfrac{x}{1-x}) + \ln3 = 2 = \ln(\dfrac{3x}{1-x})

    Which gives:
    e^2 = \dfrac{3x}{1-x}

    And then rearrange to make x the subject:
    (1-x)e^2 = 3x

e^2-xe^2 = 3x

e^2 = 3x + xe^2

e^2 = x(3+e^2)

x = \dfrac{e^2}{3+e^2} = 0.711

    Which means 71% of the crop has been destroyed 3 hours after discovering the locusts.
 
 
 
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