A Colony Of Bacteria Is Growing In A Petri Dish Which Has A
A Colony Of Bacteria Is Growing In A Petri Dish Which Has A Maximum Ca
A colony of bacteria is growing in a petri dish which has a maximum capacity of 140mg. The mass of bacteria is increasing at a rate given by the logistic equation. Initially, there is 2mg of bacteria and the rate of increase is 1mg per day. The general form of the logistic differential equation is given as
y' = k y (1 - y/M)
where y(t) is the mass of bacteria at time t, M is the maximum capacity (140 mg), and k is a constant to be determined.
Given the initial conditions y(0) = 2 mg, and the initial rate y'(0) = 1 mg/day, we can find k and then write the explicit equation for the rate of change of mass.
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The logistic growth model is a classical mathematical representation used to describe populations or quantities that grow rapidly at first but slow as they approach a maximum carrying capacity. It is widely applied in biological studies, including bacterial growth, as it realistically portrays the limited resources and environmental constraints impacting growth dynamics.
In this context, the differential equation governing bacterial mass y(t) is given by:
y' = k y (1 - y / 140)
where y(t) is in milligrams (mg) at time t (days), M = 140 mg represents the maximum capacity, and k is the growth rate constant to be determined based on initial conditions.
Determining the value of k
At t = 0, the initial mass y(0) = 2 mg, and the rate of increase y'(0) = 1 mg/day. Substituting into the differential equation:
1 = k 2 (1 - 2/140)
Calculating the term in parentheses:
1 - 2/140 = 1 - 1/70 = 69/70
Thus:
1 = k 2 (69/70)
Solving for k:
k = 1 / [2 * (69/70)] = 1 / [(138/70)] = 70 / 138 = 35 / 69 ≈ 0.5072
Formulating the logistic differential equation
Therefore, the logistic equation describing the bacterial growth is:
m' = (35/69) m (1 - m / 140)
Finding when the mass reaches 70 mg
Our goal is to determine the time t when y(t) = 70 mg.
Given the logistic model, the solution can be expressed explicitly as:
y(t) = M / [1 + A * e-kMt
where A is a constant determined by initial conditions.
Calculating A using initial conditions
At t = 0, y(0) = 2 mg:
2 = 140 / [1 + A * e0] = 140 / (1 + A)
Rearranged:
1 + A = 140 / 2 = 70
A = 69
Solving for t when y(t) = 70 mg
Set y(t) = 70:
70 = 140 / (1 + 69 e-k 140 * t)
Cross-multiplied:
70 (1 + 69 e-k 140 t) = 140
Divide both sides by 70:
1 + 69 e-k 140 * t = 2
Subtract 1:
69 e-k 140 * t = 1
Express e-k 140 t:
e-k 140 t = 1 / 69
Take the natural logarithm:
-k 140 t = ln(1/69) = -ln(69)
Solve for t:
t = [ln(69)] / (k * 140)
Recall k ≈ 0.5072, thus:
t ≈ [4.234] / (0.5072 * 140) ≈ 4.234 / 70.998 ≈ 0.0596 days
Therefore, the bacteria reach 70 mg approximately after 0.06 days, essentially within the first few hours.
Mass in grams after 10 days when starting with 2 mg
Next, we determine the mass of bacteria in grams after 10 days, given initial conditions and the exponential model.
Using the explicit logistic solution:
y(t) = 140 / [1 + 69 e-k 140 * t]
At t = 10 days:
Calculate the exponential term:
-k 140 10 = -0.5072 140 10 = -0.5072 * 1400 ≈ -709.28
e-709.28 is effectively zero due to the large negative exponent.
Thus, y(10) ≈ 140 / (1 + 0) = 140 mg
Convert 140 mg to grams:
140 mg = 0.140 grams
Summary
- The logistic differential equation satisfied by the mass y is: m' = (35/69) m (1 - m / 140)
- The bacteria will reach 70 mg approximately at 0.06 days.
- The mass of bacteria after 10 days is approximately 0.140 grams.
References
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