Problem 8 - Thomas Calculus

Question

Growth of bacteria A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are \(10,000\) bacteria. At the end of 5 hours there are \(40,000\) . How many bacteria were present initially?

Step-by-Step Solution

Answer

Initially, there were 1,250 bacteria.

Step 1: Understanding the Exponential Growth Formula

The growth of the bacteria is described by the exponential formula \( P(t) = P_0 \cdot e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( k \) is the growth constant, and \( e \) is the base of natural logarithms. We need to find \( P_0 \).

Step 2: Write Equations with Given Data

From the problem, we are given two conditions: \( P(3) = 10,000 \) and \( P(5) = 40,000 \). Thus, we have:\[ 10,000 = P_0 \cdot e^{3k} \]\[ 40,000 = P_0 \cdot e^{5k} \]

Step 3: Solve for the Growth Constant \( k \)

Divide the second equation by the first equation to eliminate \( P_0 \):\[ \frac{40,000}{10,000} = \frac{P_0 \cdot e^{5k}}{P_0 \cdot e^{3k}} \]\[ 4 = e^{2k} \]Taking the natural logarithm of both sides gives:\[ \ln(4) = 2k \]Thus,\[ k = \frac{\ln(4)}{2} \]

Step 4: Determine the Initial Population \( P_0 \)

Substitute \( k = \frac{\ln(4)}{2} \) back into one of the original equations, say \( 10,000 = P_0 \cdot e^{3k} \):\[ 10,000 = P_0 \cdot e^{3 \cdot \frac{\ln(4)}{2}} \]\[ 10,000 = P_0 \cdot e^{\frac{3 \ln(4)}{2}} \]Solve for \( P_0 \):\[ P_0 = \frac{10,000}{e^{\frac{3 \ln(4)}{2}}} \]Simplifying, \( e^{\frac{3 \ln(4)}{2}} = 4^{1.5} = 4 \cdot \sqrt{4} = 8 \), therefore:\[ P_0 = \frac{10,000}{8} = 1,250 \]

Step 5: Final Answer and Verification

Verify the answer by substituting \( P_0 = 1,250 \) into the original equations and check that they satisfy both conditions for \( t = 3 \) and \( t = 5 \). For \( t = 3 \), \( P(3) = 1,250 \cdot e^{3 \cdot \frac{\ln(4)}{2}} = 10,000 \). For \( t = 5 \), \( P(5) = 1,250 \cdot e^{5 \cdot \frac{\ln(4)}{2}} = 40,000 \). Both conditions are satisfied, confirming \( P_0 = 1,250 \) is correct.

Key Concepts

Growth Constant

In exponential growth, the growth constant, often denoted by \( k \), plays a crucial role in defining how fast or slow a population increases over time. The formula we use for exponential growth is \( P(t) = P_0 \cdot e^{kt} \), where \( P(t) \) represents the population at a given time \( t \), \( P_0 \) is the initial population, and \( e \) is the base of the natural logarithm.

• The growth constant \( k \) determines the rate of growth; a larger \( k \) implies a faster-growing population.
• If \( k \) is positive, the population increases or grows over time. Conversely, if \( k \) were negative, it would indicate a decrease.
• To find the growth constant, we can solve for \( k \) when provided with population data over specific time intervals.

In our example with the bacteria, using the change in population from 10,000 to 40,000 over 2 hours, we use the relation \( 4 = e^{2k} \). Through applying the natural logarithm, we calculate \( k = \frac{\ln(4)}{2} \). This step highlights how \( k \) helps describe the rapid increase in population for the bacterial colony in optimal growth conditions.

Initial Population

Understanding the initial population, denoted as \( P_0 \), is key to solving exponential growth problems. The initial population represents the size of the population at the starting point, typically when \( t = 0 \). In the context of mathematical modeling of biological growth, it is essential to correctly determine this value to accurately predict future population sizes.

• \( P_0 \) is critical since it is the baseline multiplicative factor in the exponential growth formula.
• The entire growth trajectory of a population hinges on knowing where it starts from, thus requiring precise calculation or estimation of \( P_0 \).
• Knowing \( P_0 \) helps verify the accuracy of the growth constant and the correctness of the growth model itself.

In our bacteria problem, after determining the growth constant \( k \), we calculated \( P_0 \) by substituting back into one of our equations. Calculations showed that initially, there were 1,250 bacteria in the colony. This value aligns with the exponential growth model used given the observed populations at times 3 and 5 hours.

Natural Logarithm

The natural logarithm, indicated as \( \ln \), is fundamental in solving exponential growth equations. It helps translate exponential relationships into linear ones, simplifying the complexity of growth models. The base of the natural logarithm is \( e \), approximately equal to 2.71828, which is also the base used in continuous growth calculations.

• Natural logarithms allow for the conversion of exponential equations into linear form, making calculations more straightforward.
• In exponential growth problems, \( \ln \) is often used to solve for unknowns like the growth constant \( k \).
• Applying the natural logarithm can help in differentiating between rapid and slow growth scenarios.

In the example with the bacteria population, we used the natural logarithm to determine \( k \), starting from the equality \( 4 = e^{2k} \). Taking the natural logarithm on both sides transforms the problem, yielding \( \ln(4) = 2k \), hence allowing us to isolate and solve for \( k \). This powerful mathematical tool streamlines the process of solving exponential growth equations. By connecting such a transformation, it enhances our functional understanding of growth dynamics.