Question

A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich, who identified it in 1885. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 40 cells. (a) Find the relative growth rate. k = hr−1 (b) Find an expression for the number of cells after t hours. P(t) = (c) Find the number of cells after 7 hours. cells

Answer 1

Answer:

a) k=2.07944 (1/hour)

b) $P(t)=40e^{2.0794t}$

c) P(7)=83,886,080

Step-by-step explanation:

We know that the cells duplicates after 20 minutes (t=1/3 hours).

We can write a model of that as:

$\frac{dP}{dt}=kP\\\\\frac{dP}{P}=kdt\\\\\int \frac{dP}{P}=k\int dt\\\\ln(P)+C_1=kt\\\\P=Ce^{kt}\\\\\\P(0)=40=Ce^0=C\\\\C=40\\\\\\P(1/3)=80=40e^{k*(1/3)}\\\\e^{k*(1/3)}=80/40=2\\\\k/3=ln(2)\\\\k=3*ln(2)=2.07944$

a) k=2.0794 h^(-1)

b) $P(t)=40e^{2.0794t}$

c) $P(7)=40e^{2.0794*7}=40*e^{14.556}=40*2,097,152=83,886,080$