Question:

In a culture,the bacteria count is 1,00,000.The number is increased by 10%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?

Updated On: Dec 8, 2023
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
dydty∴\frac{dy}{dt}∝y
dydt=ky⇒\frac{dy}{dt}=ky(where k is a constant)
dyy=kdt⇒\frac{dy}{y}=kdt
Integrating both sides,we get:
dyy=kdt∫\frac{dy}{y}=k∫dt
logy=kt+C...(1)⇒logy=kt+C...(1)
Let y0y_0 be the number of bacteria at t=0.
logy0=C⇒logy_0=C
Substituting the value of C in equation(1),we get:
logy=kt+logy0logy=kt+logy_0
logylogy0=kt⇒logy-logy_0=kt
log(yy0)=kt⇒log(\frac{y}{y_0})=kt
kt=log(yy0)...(2)⇒kt=log(\frac{y}{y_0})...(2)
Also,it is given that the number of bacteria increased by 10%10\% in 2hours.
y=110100y0⇒y=\frac{110}{100y_0}
yy0=1110...(3)⇒\frac{y}{y_0}=\frac{11}{10}...(3)
Substituting this value in equation(2),we get:
k.2=log(1110)k.2=log(\frac{11}{10})
k=12log(1110)⇒k=\frac{1}{2}log(\frac{11}{10})
Therefore,equation(2) becomes:
12log(1110).t=log(yy0)\frac{1}{2}log(\frac{11}{10}).t=log(\frac{y}{y_0})
t=2log(yy0)log(1110)...(4)⇒t=\frac{2log(\frac{y}{y_0})}{log(\frac{11}{10})}...(4)
Now,let the time when the number of bacteria increases from 100000 to 200000 be t1t_1.
y=2y0 at t=t1⇒y=2y_0\space at\space t=t_1
From equation(4),we get:
t1=2log(yy0)log(1110)=2log2log(1110)t_1=\frac{2log(\frac{y}{y_0})}{log(\frac{11}{10})}=\frac{2log2}{log(\frac{11}{10})}
Hence,in 2log2log(1110)\frac{2log2}{log(\frac{11}{10})} hours the number of bacteria increases from 100000 to 200000.
Was this answer helpful?
1
0

Top Questions on Differential equations

View More Questions