Bacteria in a Colony

Question

There are initially 1,000 bacteria in a colony. The rate at which the number of bacteria in the colony changes is given by dN/dt = kN where k is a constant. If the number of bacteria doubles every 21 minutes, which function models the number of bacteria, N(t), after t minutes?

N(t) = 1,000e^–0.033t

N(t) = 1,000e^0.033t

N(t) = 1,000(2)^–0.033t

N(t) = 1,000(2)^0.033t

Answer 1

Hi Lissie,

dN/dt = kN is a differential equation that can be solved by separating the variables.

dN/N = k*dt

∫dN/N = ∫k*dt

ln(N) = kt + c

N = e^{kt + c}

N = e^c*e^kt

e^c is just a constant, so we'll call it C.

N(t) = Ce^kt

The colony initially has 1000 bacteria, so N(0) = 1000

1000 = Ce^0

C = 1000

N(t) = 1000e^kt

The problem also states that the colony doubles every 21 minutes. So N(21) = 2000.

2000 = 1000e^k*21

2 = e^k*21

ln(2) = k*21

k = ln(2) / 21 = 0.033

So,

N(t) = 1000e^0.033*t

Looks like that second option is your answer. Hope this helps!