Question

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20minutes. The initial population of a culture is 55 cells.
(a) Find the relative growth rate. (Assume t is measured in hours.)
k = _____



(b) Find an expression for the number of cells after t hours.
P(t) =_____



(c) Find the number of cells after 8 hours.
_____cells

(d) Find the rate of growth after 8 hours. (Round your answer to three decimal places.)
billion cells per hour
______
(e) When will the population reach 20,000 cells? (Round your answer to two decimal places.)
______hr

Answer 1

relative growth rate = ln(2)-ln(1)/20
= .034657/min = 2.07944

number of cells = n(t) = n₀ * e^tλ

λ = ln(2)/20 * 60 = 2.07944 /hr

n(t) = 55 * e^2.07944*t

after 8 hours t = 8;

n = 55 * e^2.07944*8 = 16.777009 million cells

dn/dt = nλ;

n = 16.777009 million;

rate of growth =34886783.762 /hr

n = 20,000

n₀ = 55

ln(20,000/55) =λ*t

time taken to reach 20,000 cells = t =2.84 hours