Question

The growth rate of a particular bacteria is modeled by the dP differential equation 17 = k P. Suppose a population dt of bacteria triples in size every 11 hours. Initially, there are 100 bacteria cells. If we begin growing the bacteria for our experiment at 8:00am on January 4, when is the earliest the necessary 5,000,000 bacteria cells will be ready?

Answer 1

Given at kp (Using seperation of de kat Variables and Integrating ) - Skat => lng P = kt+C Given Initially Po= 100 at t= 0 hes lie 100 = kxo+c => c= ln 100 EneP= kt + ln 100 = ln tookt IP=100 ekt This is the differential eqa showing the growth P(t) = 100 ekt rate of bacteria for every 11 hrs; population triples i..., at t= 11 hes; P = 3% = 3x100 = 300 300 = 100 ekx!! => 3-ek en (3) = lenk- uk k = lry (3) = 0·0998738 P(t) = 100 69.09987387 = → We need time at which Plt) - 5000000€ 5000066 = 100 e 0.09987385 lo (50000) = 0.0998738 t =) t= ln (soouo) 0-0728738 t= 108.33 hrs It N 108 hrs . Started at 8:00 am on January 4 108 hrs ~ 4 days 12 hos hence population will be 5000000 at January [8] 8pm lime taken: - 4 days => Date 12hss Time