Question

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Answer 1

Let the count of bacteria at any instant t be y.

According to the question, the rate of growth of bacteria is proportional to number present.

dy dt ∝y dy dt =ky dy y =kdt

Integrate both sides.

∫ dy y =k ∫ dt logy=kt+C (1)

Let, initially y 0 be the number of bacteria at t=0 then,

log y 0 =C

Substitute the value of Cin equation (1).

logy=kt+log y 0 logy−log y 0 =kt kt=log y y 0 (2)

Given that the number of the bacteria increases by 10% in 2hours, then,

y=( 1+ 10 100 ) y 0 y= 110 100 y 0 y y 0 = 11 10

So,

k×2=log( 11 10 ) k= 1 2 log( 11 10 )

Substitute the value of k in equation (2).

t× 1 2 log( 11 10 )=log( y y 0 ) t= log( y y 0 ) 1 2 log( 11 10 )

Now, the time when number of bacteria increases from 100000to 200000.

y=2 y 0 at t= t 1

From equation (2), we get,

t 1 = 2log( 2 y 0 y 0 ) log( 11 10 ) t 1 = 2log2 log( 11 10 )

Thus, the time at which bacteria increases from 100000to 200000 is 2log2 log( 11 10 ) hours.