Question

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

Answer 1

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: \[\frac{dP}{dt}\alpha P\]
\[\Rightarrow \frac{dP}{dt} = aP\]
\[ \Rightarrow \frac{dP}{P} = adt\]
\[ \Rightarrow \log\left| P \right| = at + C . . . . . \left( 1 \right)\]
Now,
\[P = N\text{ at }t = 0 \]
\[\text{Putting }P = N\text{ and }t = 0\text{ in }\left( 1 \right), \text{ we get }\]
\[\log\left| N \right| = C \]
\[\text{Putting }C = \log\left| N \right|\text{ in }\left( 1 \right), \text{ we get }\]
\[\log\left| P \right| =\text{ at }+ \log\left| N \right|\]
\[ \Rightarrow \log\left| \frac{P}{N} \right| =\text{ at }. . . . . \left( 2 \right)\]
According to the question,
\[\log\left| \frac{3N}{N} \right| = 5a\]
\[ \Rightarrow a = \frac{1}{5}\log\left| 3 \right| = \frac{1}{5} \times 1 . 0986 = 0 . 21972\]
\[\text{ Putting }a = 0 . 21972\text{ in }\left( 2 \right),\text{ we get }\]
\[\log\left| \frac{P}{N} \right| = 0 . 21972t . . . . . \left( 3 \right) \]
\[ \Rightarrow e^{0 . 21972t} = \frac{P}{N} . . . . . \left( 4 \right)\]
\[\text{ Putting }t = 10\text{ in }\left( 4 \right)\text{ to find the bacteria after 10 hours, we get }\]
\[ e^{0 . 21972 \times 10} = \frac{P}{N}\]
\[ \Rightarrow e^{2 . 1972} = \frac{P}{N}\]
\[ \Rightarrow \frac{P}{N} = 9\]
\[ \Rightarrow P = 9N\]
To find the time taken when the number of bacteria becomes 10 times of the number of initial population, we have
\[P = 10N\]
\[ \therefore \log\left| \frac{10N}{N} \right| = \frac{1}{5}t\log 3\]
\[ \Rightarrow t = \frac{5 \log 10}{\log 3}\]