Question

The growth rate of a particular bacteria is modeled by the differential equation dP/dt = k P. Suppose a population at of bacteria doubles in size every 11 hours. Initially, there are 200 bacteria cells. If we begin growing the bacteria for our experiment at 7: 00pm on September 4, when is the earliest the necessary 5,000,000 bacteria cells will be ready? a) September 07 at 12: 00pm b) September 07 at 9: 00pm c) September 08 at 8: 00am d) September 08 at 11: 00pm e) September 11 at 12: 00pm f) September 12 at 5: 00am

Answer 1

Solution Given DE and bacteria doubles for every 11hoursand P For the DE dP Separating variables and integrating: dP dt 200 dt P(0)-ceo-200-C(1) P(t) 200e we have B-2004 C-200 Given bacteria doubles for every 11 hours Ater 11 hours, P(11) 2(200) 400 Substi tuting: 400- 200e*11In 2k-In2s0.063 P(t) = 200e0053 For P(t) 5000000- 200e3 25000-eoos 0. 063 1n (25000) t = 160. 74hours ~ 6days 17hours t07 pm on sept4 then t-160.74hours-7pmon sept4)+ 6days 17hours -12pmof Septl1