In a culture,the bacteria count is 1,00,000. | vedclass

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(A) Let $y$ be the number of bacteria at any time $t$.Given that the rate of growth is proportional to the number present,we have $\frac{dy}{dt} = ky$

Vedclass · Accepted Answer

$\frac{2 \log 2}{\log (1.1)}$

Vedclass · Answer

(A) Let $y$ be the number of bacteria at any time $t$.Given that the rate of growth is proportional to the number present,we have $\frac{dy}{dt} = ky$,where $k$ is a constant.Separating variables and integrating,we get $\int \frac{dy}{y} = \int k dt$,which gives $\log y = kt + C$.At $t = 0$,let $y = y_0 = 1,00,000$. Thus,$C = \log y_0$.So,$\log y = kt + \log y_0$,or $\log \left(\frac{y}{y_0}\right) = kt$.Given that the number increases by $10 \%$ in $2$ hours,at $t = 2$,$y = y_0 + 0.1 y_0 = 1.1 y_0$.Substituting this into the equation: $\log(1.1) = k(2) \Rightarrow k = \frac{1}{2} \log(1.1)$.We want to find $t$ when $y = 2,00,000 = 2 y_0$.Substituting $y = 2 y_0$ and $k = \frac{1}{2} \log(1.1)$ into $\log \left(\frac{y}{y_0}\right) = kt$:$\log(2) = \left(\frac{1}{2} \log(1.1)\right) t$.Solving for $t$,we get $t = \frac{2 \log 2}{\log(1.1)}$ hours.

In a culture,the bacteria count is $1,00,000$. The number increases 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?

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