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

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Last updated date: 16th Jun 2024
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Answer
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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.

Additional Information:
(1) First of all, see the question and check that what we have found or what the question shows us that we have to find or what the question shows us. If it is in the exponential form, then we will use the exponential form.
(2) Then also find all the things which are in exponential form. At last find the required answer to the question.

Note:
(1) First check all the possibilities, sometimes the equation is given but sometimes the question does not give an equation. Hence students get confused. So remember this the question shows some important words also so that we can get help for solving the problems.
(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.