A culture of bacteria is growing at a rate proportional to the number,

N, of bacteria present, so that

dN/dt= kN

where t is the time elapsed in hours and k is a constant. Initially, the culture contains 3000 bacteria, and after three hours of uninterrupted growth, it contains 9000 bacteria.

(i) Find k.

Thereafter (that is, for t > 3) the culture is carefully washed with a solution

which is harmful to the bacteria, and the bacteria are killed o by the solution

at a constant rate of 75 each minute. [Thus, the rate of change in the number

of bacteria is as described above, but diminished by 4500 per hour.]

(ii) Find the number of bacteria in the culture for any time t > 3.

(iii) After how much washing time (in hours and minutes) do the bacteria die out?

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For (i)

dN/dt=kN

1/N*dN/dt=k

int(1/N)*(dN/dt)*dt=int k dt

int(1/N)*dN=kt+C

ln ㅣNㅣ=kt+C

ㅣNㅣ=e^(kt+c)

ㅣNㅣ=e^c*e^kt

and N=ħe^c*e^kt

=Ae^kt, where A=ħe^c

∴N=Ae^kt

when N(0)=3000,

3000=A

so, N(t)=3000e^kt,

when N(3)=9000

9000=3000e^kt

3=e^3k

3k ln e= ln 3

3k= ln 3

k=(1/3) ln (3), which is 0.3662 in 4 decimal places....

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First, can u plz check whether this answer is right or wrong??

and..i have no idea how to approach to question (ii) and (iii)

Thank you for your kindness