Question

Two kinds of bacteria are found in a sample of tainted food. It is found that the populations size of type 1,\\(N_1\\) and of type 2, \\(N_2\\) satisfy the equations \\(\\frac{dN_1}{dt}=-k_1N_1, N_1(0)=N_{1,0}\\) and \\(\\frac{dN_2}{dt}=-k_2N_2, N_2(0)=N_{2,0}\\) . Then the population sizes equal \\(N_1=N_2\\) at the following time

Answer 1

$\frac{dN_1}{dt}=-k_1N_1, \quad N_1(0)= N_{1,0}$

$\frac{dN_2}{dt}=-k_2N_2, \quad N_2(0)= N_{2,0}$

We have the following solutions:

$N_1(t)=N_{1,0}e^{-k_1t}$ and $N_2(t)=N_{2,0}e^{-k_2t}$

Suppose that at the time $t=t_0$ the population sizes are equal:

$N_1(t_0)=N_2(t_0)$

$N_{1,0}e^{-k_1t_0} =N_{2,0}e^{-k_2t_0}$

$\frac{N_{1,0}}{N_{2,0}}=e^{(k_1-k_2)t_0}$

$(k_1-k_2)t_0=\ln(\frac{N_{1,0}}{N_{2,0}})$

$t_0=\frac{1}{k_1-k_2} \ln(\frac{N_{1,0}}{N_{2,0}})$

Answer: at the time $t_0=\frac{1}{k_1-k_2} \ln(\frac{N_{1,0}}{N_{2,0}})$