Step 1: Understanding the Concept:
The problem states that the rate of increase of a bacteria culture is proportional to its current population.
This is a classic application of first-order differential equations in biology, known as the law of exponential growth.
In such scenarios, if the population is small, it grows slowly, and as it becomes larger, the growth speed increases because more individuals are reproducing.
Mathematically, the "rate of increase" refers to the derivative of the population with respect to time, \( \frac{dN}{dt} \).
The term "proportional to the number present" implies a linear relationship, which we express using a constant of proportionality \( k \).
The solution to such a differential equation will always result in an exponential function.
We are required to find the population at the starting point, which corresponds to time \( t = 0 \).
Step 2: Key Formula or Approach:
Let \( N \) be the number of bacteria at any time \( t \).
The differential equation is:
\[ \frac{dN}{dt} = kN \]
By separating the variables, we get:
\[ \int \frac{dN}{N} = \int k \, dt \]
Integrating both sides yields:
\[ \ln |N| = kt + C \]
Converting from logarithmic form to exponential form:
\[ N(t) = e^{kt+C} = e^C \cdot e^{kt} \]
If we let \( N_0 = e^C \), then \( N_0 \) represents the initial population at \( t = 0 \).
The final general formula is:
\[ N(t) = N_0 e^{kt} \]
Step 3: Detailed Explanation:
We have two known conditions provided in the question.
Condition 1: At \( t = 3 \) hours, the population \( N = 10,000 \).
Substituting these values into our general formula:
\[ 10,000 = N_0 e^{3k} \quad \dots \text{(Equation 1)} \]
Condition 2: At \( t = 5 \) hours, the population \( N = 40,000 \).
Substituting these values into our general formula:
\[ 40,000 = N_0 e^{5k} \quad \dots \text{(Equation 2)} \]
To solve for \( N_0 \), we first need to eliminate one of the variables.
Dividing Equation 2 by Equation 1:
\[ \frac{40,000}{10,000} = \frac{N_0 e^{5k}}{N_0 e^{3k}} \]
\[ 4 = e^{5k - 3k} \]
\[ 4 = e^{2k} \]
Taking the square root of both sides gives us the value for \( e^k \):
\[ \sqrt{4} = \sqrt{e^{2k}} \implies 2 = e^k \]
This tells us that the bacteria population doubles every hour.
Now, substitute the value of \( e^k = 2 \) back into Equation 1 to find \( N_0 \):
\[ 10,000 = N_0 (e^k)^3 \]
\[ 10,000 = N_0 (2)^3 \]
\[ 10,000 = 8N_0 \]
Calculating \( N_0 \):
\[ N_0 = \frac{10,000}{8} = 1250 \]
Step 4: Final Answer:
The initial number of bacteria present at the beginning was 1250.
This corresponds to option (A).