Question

3. The growth of a certain bacteria in a reactor is assumed to be governed by the logistic equation: d P dt where P is the population in millions and t is the time in days. Recall that M is the carrying capacity of the reactor and k is a constant that depends on the growth rate (a) Suppose that the carrying capacity of the reactor is 10 million bacteria, and that the peak growth rate is 3 million bacteria per day. Determine the constants k and M in the above equation. (b) Supposing the bioreactor has 250,000 bacteria in it to begin with, find the number of bacteria in the tank. how long will it take for the population to reach 50% of the carrying capacity? Suppose that bacteria are being harvested from the tank continuously. Let h be the rate at which the bacteria are harvested in millions per day. Write down the new differential equation governing the bacteria population. What is the ma of harvesting h that will not cause the population of bacteria to go extinct? (Below this rate there will always be a stable equilibrium point where P is positive). (c) ximum rate

The growth of a certain bacteria in a reactor...

Answer 1

.Pr _ MMP dP=k(t-0)-c By partial fractions dP- P M MP P(t) The carrying capacity of the growth rate M-10million and the peak or maximum growth rate is 10 million 0 \r\nd (dp dP dt 0 k(M -2P)-0 M-2P- 0 10 Carrying capacity is M -10million. So the population is P - 5 million. From the data the peak growth rate is 3 million per day. From equation (1), dP 3-k.5(10-5) k (5) 25 k 0.12 Therefore the value of k is 0.12 and M is 10 millioin \r\n\r\nThe new differential equation is, dP dt =0.12P(10-P)-h -0.12 (10P-p2)-h --0.12P2 +1.2P -h dP dt To find the equilibrium points. take 0 -0.12P2 +1.2P-h-0 0.12 0.12 P2-10P +8.33333h-0 If h-0 then it has two equilibrium points. If h is non-zero then the quadratic form P2-10P +8.33333hcan be expressed as, (P-a) So it has one equilibrium point with multiplicity 2. Thus 8.33333h - a2 8.33333h-52 Since P2 -10P+8.33333h- (P-5)-25+8.33333h 25 8.33333 Hence, h3 million per day is the maximum allowable harvesting rate. "",""error"":null,""modified"":1551835182,""source"":""automated""}"