Calculus Question

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 20 minutes. The initial population of a culture is 54 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

a) 1 hour = 60 mins / 20 mins = 3 --> let t be in hours

P(1/3) = 2(60) = 120

P(t) = 54e^kt

120 = 54e^k/3

20/9 = e^k/3

ln(20/9) = ln(e^k/3)

k/3 = ln(20/9)

k = 3ln(20/9) ≈ 2.39552

b) P(t) = 54e^2.39552t

c) P(8) = 54e^2.39552(8) = 11,357,500,000 (or 1.13575 x 10¹⁰) cells

d) dP/dt = kP(8) = 2.39552(11,357,500,000) = 2.721 x 10¹⁰ billion cells per hour

e) 20,000 = 54e^2.39552t

370.37037 = e^2.39552t

ln(370.37037) = ln(e^2.39552t)

ln(370.37037) = 2.39552t

t = 2.47 hours