A bacterial population follows the law of exponential growth. If between noon and 2 p.m. the population triples, at what time should the population become 100 times what it was at noon? At 10 a.m. what percentage was present?

SOLUTION:

First, we denote

P as the population of bacteria at anytime

P_o as the original bacterial population

t = 0 (12 noon)

t = 2 (2 p.m.)

Let us determine the given and the required

GIVEN:

@12nn to 2p.m.; P= 3P_o

REQUIRED:

Using the formula of Applications of Ordinary First-Ordered Differential Equations under Exponential Growth or Decay

\frac{dP}{dt}=kP \\
\int \:\frac{dP}{P}=\int \:kdt\\
e^{ln\:P}\:=\:e^{kt\:+\:C}\\
P=\:Ce^{kt}\:\:\:\:\:(Eq.1)\\
@t=0; P=P_o\\
P_o=Ce^{kt}\\
P_o=Ce^{k\left(0\right)}\\
P_o = C

Substituting to Eq.1., we get

P=\:P_{o\:}e^{kt}\:\:\:\:\:\:\:(Eq.2)

Then from the given condition, from 12 noon to 2 p.m., the population triples (using Eq.2), we will solve for the value of k

@t= 2\:;\:P= 3P_o\\
P=\:P_{o\:}e^{kt}\\
3P_{o\:}=\:P_{o\:}e^{k\left(2\right)}\\
k=0.54931

We will then come up with the working equation (WE), this will help us solve the required problems

P_{\:}=\:P_{o\:}e^{\left(0.54931\right)t}

1.) what time should the population become 100 times

Using WE,

t=?\:\:;\:\:P=100P_o\\
P_{\:}=\:P_{o\:}e^{\left(0.54931\right)t}\\
100P_{o\:}=\:P_{o\:}e^{\left(0.54931\right)t}\\
t=8.38\: hrs.\\
t= 8:22:48\: p.m. \; or\:8:23\:p.m.

2.) at noon

P=P_o

3.) percentage at 10 a.m.

@10 a.m.\:\:;\:\:t=-2\\
P_{\:}=\:P_{o\:}e^{\left(0.549\right)\left(-2\right)}\\
P_{\:}=\:P_{o\:}\left(0.33333\right)\\
\%=\frac{P}{P_o}{(100)}=\frac{P_o\left(0.33333\right)}{P_o}{(100)}\\
\%=\:33.33\%