Question

WRITE COMPLETE SOLUTION.

4. The growth rate of a certain colony of bacteria in a controlled room temperature is given by the expression

dP(t)/dt = -4.375x10^-5 P(t)^2 + 0.0525P(t)

Where:
P(t) = bacterial count
t      = time

Initial biomass count was performed by the researchers and found a total of 400 grams.
Determine:
a) The time when population reaches 500 grams.
b) The time when population reaches 850 grams.
c) The time when half of the maximum growth rate is reached.

Answer 1

First we have to find equation relating bacterial count [ P(t) ] and time (t) from the given equation

we need to perform integration

So, our final equation is :

............(i)

In the given question, the unit of time is not provided, so we are assuming unit of time as " minutes"

(a).

when P(t) = 500g

Putting P(t) = 500 in equation(i), we get

Taking (Ln) both sides, we get

0.0525t = ln(1.428)

0.0525t = 0.3562

t = 6.786 min

(b).

when P(t) = 850g

Putting P(t) = 850 in equation(i), we get

Taking (Ln) both sides, we get

0.0525t = ln(4.857)

0.0525t = 1.58

t = 30.10 min

(c)

time when maximum half the maximum growth rate is reached

To find the maximum growth rate, we need to equate growth rate equal to zero

It is given in the question

from this, P(t) = 1200g

half the maximum growth rate = 1200/2 = 600g

Putting P(t) = 600g in equation(i), we get

Taking (Ln) both sides, we get

0.0525t = ln(2)

0.0525t = 0.6931

t = 13.20 min