# Math Help - can anyone plz help me out with this problem?? maybe this is a piece of cake 4 u

1. ## can anyone plz help me out with this problem?? maybe this is a piece of cake 4 u

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=±e^c*e^kt
=Ae^kt, where A=±e^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)

2. your value of $k$ is correct ...

$k = \frac{\ln{3}}{3}$

see new post below ...

3. let me try this again ... the washing will affect the rate of growth, so a new DE is required.

$\frac{dN}{dt} = kN - 4500$

$\frac{dN}{kN-4500} = dt$

$\frac{k}{kN-4500} \, dN = k \, dt$

$\ln|kN-4500| = kt + C$

$kN - 4500 = Ae^{kt}$

at $t = 3$ , $N = 9000$ , $k = \frac{\ln{3}}{3}$

$3000\ln(3) - 4500 = 3A$

$A = 1000\ln(3) - 1500$

$kN = 4500 - (1500 - 1000\ln{3})e^{kt}$

$N = \frac{3}{\ln{3}}[4500 - (1500 - 1000\ln{3})e^{kt}]$