A bacteria sample of certain number of bacteria is observed to grow exponentially in a given amount of time. Using exponential growth model, the rate of growth of this sample of bacteria is calculated.
The differential equation representing the growth of bacteria is given as: \[ \frac{dP}{dt} = kP, \] where \( P \) is the population of bacteria at any time \( t \). bf{Based on the above information, answer the following questions:} [(i)] Obtain the general solution of the given differential equation and express it as an exponential function of \( t \). [(ii)] If the population of bacteria is 1000 at \( t = 0 \), and 2000 at \( t = 1 \), find the value of \( k \).
Show Hint
For exponential growth or decay, always separate variables, integrate, and rewrite in exponential form.
(i) General Solution of the Differential Equation:
The provided differential equation is:
\[
\frac{dP}{dt} = kP.
\]
Variable separation yields:
\[
\frac{1}{P} \, dP = k \, dt.
\]
Integrating both sides:
\[
\int \frac{1}{P} \, dP = \int k \, dt.
\]
The integration results in:
\[
\ln |P| = kt + C,
\]
where \( C \) is the constant of integration.
Expressing this in exponential form:
\[
P = e^{kt + C} = e^C \cdot e^{kt}.
\]
Setting \( e^C = P_0 \), representing the initial population:
\[
P = P_0 e^{kt}.
\]
Final Answer (i):
The general solution is:
\[
P = P_0 e^{kt}.
\]
Was this answer helpful?
1
Top Questions on Absolute maxima and Absolute minima
