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# How many hours after the start will the instantaneous growth rate of the population be $200$ cells per hour?

Last updated date: 13th Sep 2024
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Hint: $\left( 1 \right)$ Here we have to find the solution in the form of $y\left( t \right)=a{{e}^{kt}}$
Where as $'a'$ is initial amount
$'t'$ is time
$'k'$ is decay factor
$\left( 2 \right)$ Then we have found $k$ using the information which is given here. After that consider the condition which is given in the question that is the rate of population be $200$ cells per hour and then solve it.

Complete step by step solution: As we know that,
A population of bacteria is $1100$ cells. It is increased to $1200$ cells after $2$ hour.
Let us assume that here the population is exponential. form.
$\because$ So we have to find the function in the form of exponential.
The exponential form will be,
$y\left( t \right)=a{{e}^{kt}}$
Where as, $a$ is the initial amount
$t$ is time
$k$ is decay factor.
Then,
First of all we have to find the value of $k$
We have,
Initial amount is $1100$ and it increase to $1200$ after $2$ hours.
So,
$1200=1100{{e}^{2k}}....$ [Using exponential form]
Solving for $'k':-$
$\dfrac{1200}{1100}={{e}^{2k}}$
$\therefore \ln \left[ \dfrac{12}{11} \right]=2k\ln e$
Then equation becomes,
$k=\dfrac{\ln \left( \dfrac{12}{11} \right)}{2}$
So now we have,
$y\left( t \right)=1100{{e}^{\dfrac{t}{2}\ln \left( \dfrac{12}{11} \right)}}$
Now, the condition given,
For population to be $200$ cells per how,
$\therefore 1300=1100{{e}^{\dfrac{t}{2}\ln \left( \dfrac{12}{11} \right)}}$
Now, the condition given,
For population to be $200$ cells per how,
$\therefore 1300=1100{{e}^{\dfrac{t}{2}\ln \left( \dfrac{12}{11} \right)}}$
The above equation can be written as:
$\ln \left( \dfrac{13}{11} \right)=\dfrac{t}{2}\ln \left( \dfrac{12}{11} \right)$
$\therefore t=2\dfrac{\ln \left( \dfrac{13}{11} \right)}{\ln \left( \dfrac{12}{11} \right)}$
By solving it we get
$t=3.84$
Substitute given value in $y$ we get.
$y=1100{{e}^{\dfrac{3.84}{2}\ln \left( \dfrac{12}{11} \right)}}$
$\therefore y=1300$
Hence,
$3.84$ hours after the start will the instantaneous growth rate of population be $200$ cells per hour.

(2) Also check if the numbers are negative or positive or not. Sometimes these are $\left( \pm \right)$ signs given, but we write the values in positive. So, it gets wrong. Then check the answer after solving.