C4 problem (1st Order differential equations)

The surface of a circular pond of radius a is being 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 radius r and area A. An ecologist models the growth of the weeds by assuming that the rate of increase of the area covered is proportional to the area of the pond not yet covered.

(i) Show that dA/dt = 2(pi)r*dr/dt

(ii) Hence show that the ecologist's 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 your answer.
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Questions (i) and (ii) are done. But no idea how to do (iii). Please Help :redface:

Reply 1

m:)ckel

INT [2r / (a^2-r^2)]dr = INT (k)dt

- INT [-2r / (a^2-r^2)]dr = INT (k)dt

-ln (a^2 - r^2) = kt + c

t=0, r=0 => -ln(a^2) = c

=> -ln(a^2-r^2) = kt - ln(a^2)
ln(a^2-r^2) - ln(a^2) = -kt
ln(1 - r^2/a^2) = -kt (by using lna - lnb = ln(a/b))

1 - r^2/a^2 = e^-kt

r^2 = a^2(1-e^-kt)

So 'r' equals the sqaure root of all that.

(iv) The pond will be covered when r=a
i.e. rt(1-e^-kt) = 1
=> e^-kt = 0

e^-kt will tend to 0 as 't' tends to infinity. So, theoretically, the weeds will never cover the pond.