The number of individuals in a population of bacteria is described by the follow

Question

The number of individuals in a population of bacteria is described by the following differential equation: dP(t) =kP(t) (i) where P(t) is the population, t is the time in hours, and k is a constant that depends on the type of bacteria. (a) Show that the general solution of (1) is: P(t) = Ct' CD where C is mi arbitrary constant. (b) Assume that a particular population of bacteria has 500 individuals at time equal zero, and 1500 individuals after three hours. Find the particular solution that mathematically describes this situation (i.e., find the values of C and k in (2) And write the resulting equation).

Explanation / Answer

Solution:

1)

solution:

The differential equation that governs the exponential model is P'(t) = kP(t). The solutions to these equations are precisely the exponential functions P(t) = Cekt. For example, if you take C = 1 and k = ln(2), you have the exponential function with base 2: P(t) = etln(2) = 2t.

2) solution:

(a) We use the exponential growth model with n0 = 500 and r = 0.4 to get

n(t) = 500e 0.4t where t is measured in hours.

(b) Using the function in part (a), we find that the bacterium count after 1 hour is n(1) = 500e ^(0.4(1)

= 500e^ 0.4 746 Note that

n(1) 6= 500(1 + 0.4) = 700.

(c) Using the function in part (a), we find that the bacterium count after 10 hours is n(10) = 500e ^0.4(10) = 500^e 4 27, 300 Note that n(10) 6= 500(1 + 0.4)^10 14, 463.

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