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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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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?

$\begin{array}{1 1}\large\frac{2.\log 2}{\log(\large\frac{11}{10})} \\ \large\frac{\log 2}{\log(\large\frac{11}{10})} \\\large\frac{2.\log 2}{\log(\large\frac{10}{11})} \\ \large\frac{\log 2}{\log(\large\frac{10}{11})}\end{array} $

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1 Answer

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  • $\large\frac{dy}{dt}$ is proportional to y where y is the number of bacteria at any instant $t$
Step 1:
From the given information we know that dy/dt is proportional to $y$
$\large\frac{dy}{dt}=$$ky$
on seperating the variables
we get $\large\frac{dy}{y}$$=k.dt$
Step 2:
Integrating on both sides
we get $\int \large\frac{dy}{y}$$=k \int dt$
$\log y=kt+c$ ----(1)
Let $y_0$ be the number of bacteria when $t=0$
Hence $\log y_0=C$
Substituting this value in equation (1) we get $\log y =kt+\log y_0$
$\log y- \log y_0=kt$
$\log(\large\frac{y}{y_0})$$=kt$ ---(2)
Step 3:
It is also given that the number of bacteria increases 10% in 2 hours
Hence $y=110\large\frac{y_0}{100}$
$\large\frac{y}{y_0}=\frac{11}{10}$
Substituting this in (2)
we get $2k=\log (\large\frac{ 11}{10})$
or $k=\large\frac{1}{2}$$\log \large\frac{11}{10}$
Therefore $\large\frac{1}{2} $$\log (\large\frac{11}{10})$$t = \log (\large\frac{ y}{y_0})$
$t=\large\frac{2\log (\Large\frac{y}{y_0})}{ \log(\Large\frac{11}{10})}$
Step 4:
Let the time when the number of bacteria increases from 1,00,000 to 2,00,000 be $t_1$
$y=2.y_0$ at $t=t_1$
Hence $t_1=\large\frac{2 \log (\large\frac{y}{y_0})}{\log (\large\frac{11}{10})} = \large\frac{2.\log 2}{\log(\large\frac{11}{10})}$
Hence in $\large\frac{2.\log 2}{\log(\large\frac{11}{10})}$ the number of bacteria increases from 1,00,000 to 2,00,000
answered Aug 14, 2013 by sreemathi.v
 

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