Question

3. (17 points) The growth in a population of bacteria follows a logistic growth model given by the differential equation dP 0.05P - 0.00001p? dt with units of number of bacteria and hours. (a) (3 points) What is the carrying capacity of this population? (b) (9 points) Given an initial population of 1000 bacteria, how long will it take for the population to double? (c) (5 points) What is the rate of change (per hour) in the size of the bacteria popula- tion at the point in time in part (b)?

Answer 1

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Given dp de = 0.05 p - 0.0000122 P .de dt = 0.05p l 0.05%0.00001 de dt 0-05 P P 5000 de = op (1-2) Capacity k = 5000 In the solve ode with Plo) = 1000 dp = 0.05PL dt Pli P 5000 de 0.05 dt 5000 [P(5000 --P)] + p ] de..0.00001 at 5000 Р 5000-P + (h ) dp = 0.05 dr 5000 - P = 0.05 1 0 0 en p-In 5000-P PO): 1000 in 1000 - In 4000 = 0.05x0+ coln (14) from In. 5000 - P -0.05T tlu (You)

let PCH) = 2000 2000 ln 0.05 + + In. (Yu) 3000 In (213) – In (Yu) 0:05 19.61 kr © The so AT P= 200.0 de dt 0-05 X 2000 0.00001 (2000) 100 40 60