Problem 6 - Precalculus with Calculus Previews

Question

In genetic research a small colony of drosophila (small two-winged fruit flies) is grown in a laboratory environment. After 2 days it is observed that the population of flies in the colony has increased to \(200 .\) After 5 days the colony has 400 flies. (a) Find a model \(P(t)=P_{0} e^{k t}\) for the population of the fruit-fly colony after \(t\) days. (b) What will be the population of the colony in 10 days? (c) When will the population of the colony be 5000 fruit flies?

Step-by-Step Solution

Answer

(a) \( P(t) = 141.4 e^{\frac{\ln(2)}{3} t} \); (b) 1131 flies; (c) about 16 days.

Step 1: Define Initial Situation

The problem provides two data points: after 2 days, the population is 200 flies, and after 5 days, it is 400 flies. We can use these to find the unknowns in our model.

Step 2: Use the Equation to Express P(t)

The equation for the population is given by \( P(t) = P_{0} e^{k t} \), where \( P_{0} \) is the initial population and \( k \) is the growth rate. Use the state \( P(2) = 200 \) and \( P(5) = 400 \) to set up two equations.

Step 3: Solve for Initial Population \( P_0 \)

Assume \( t = 0 \) is the start of the observation. Use \( P(2) = 200 \): \[ 200 = P_{0} e^{k \cdot 2} \].

Step 4: Solve for Growth Rate \( k \)

Now, use \( P(5) = 400 \): \[ 400 = P_{0} e^{k \cdot 5} \]. Divide the equation from Step 3 with this equation to eliminate \( P_0 \):\[ \frac{400}{200} = \frac{P_0 e^{5k}}{P_0 e^{2k}} \Rightarrow 2 = e^{3k} \]. Solve for \( k \): \[ 3k = \ln(2) \Rightarrow k = \frac{\ln(2)}{3} \].

Step 5: Determine Initial Population \( P_0 \)

Substitute the value of \( k \) back into the equation from Step 3: \[ 200 = P_{0} e^{2 \cdot \frac{\ln(2)}{3}} \Rightarrow P_{0} = 200 \times e^{-\frac{2\ln(2)}{3}} \approx 141.4 \].

Step 6: Model the Final Population Function

Now, we have our model as:\[ P(t) = 141.4 e^{\frac{\ln(2)}{3} t} \].

Step 7: Predict Population at 10 Days

Plug \( t = 10 \) days into the model:\[ P(10) = 141.4 e^{\frac{\ln(2)}{3} \cdot 10} \approx 1131 \].

Step 8: Determine When Population Reaches 5000

Set \( P(t) = 5000 \) and solve for \( t \):\[ 5000 = 141.4 e^{\frac{\ln(2)}{3} t} \Rightarrow e^{\frac{\ln(2)}{3} t} = \frac{5000}{141.4} \approx 35.36 \], \[ \frac{\ln(2)}{3} t = \ln(35.36) \Rightarrow t = \frac{3 \ln(35.36)}{\ln(2)} \approx 15.69 \].

Key Concepts

Drosophila population dynamics

Understanding the dynamics of a Drosophila, or fruit fly, population is key in genetic research. Observing how this population grows over time gives researchers insight into genetic traits, experiments, or environmental impacts. Changes in population size are typically indicative of underlying biological processes. In a controlled laboratory setting, such growth can often be modeled mathematically.

This exercise revolves around the exponential growth of a Drosophila colony. The population is observed to increase rapidly, making it suitable for modeling with an exponential function. This model helps in predicting future populations and explaining how factors like initial population size and growth rates affect dynamism over time.

The exponential growth model used in this context is driven by two main variables - the initial population at the starting point and the growth rate over time. These variables are critical for projecting future population sizes and understanding the potential carrying capacity of the environment.

Initial population estimation

In mathematical modeling of population dynamics, estimating the initial population, denoted as \( P_0 \), is crucial. The initial population is the size of the population at the beginning of the observation period, which in this case is when \( t = 0 \).

To derive \( P_0 \), we start with the given conditions: after 2 days, the population increased to 200 flies. Using the formula \( P(t) = P_0 e^{kt} \), and substituting \( t = 2 \) with \( P(2) = 200 \), we establish a relationship that allows us to solve for \( P_0 \).

This step involves solving the following equation:

• \( 200 = P_0 e^{2k} \)
• Solve for \( P_0 \) by rearranging the equation as \( P_0 = \frac{200}{e^{2k}} \)

This gives a base size for the mathematical model prediction of further population dynamics after estimating the growth rate \( k \).

Population prediction

Predicting the future size of a population from an initial size is one of the powerful applications of the exponential growth model. Given an established model, \( P(t) = P_0 e^{kt} \), predictions for future time points become straightforward.

In this exercise, one prediction is for \( t = 10 \) days, using the values for \( P_0 \) and \( k \) obtained from previous steps. By substituting \( t = 10 \) into our equation, we can calculate how many fruit flies will be present:

• \( P(10) = 141.4 e^{\frac{\ln(2)}{3} \cdot 10} \approx 1131 \)

This result showcases the exponential nature of population growth, where the rate of increase becomes faster as the population grows, reflecting real-world biological patterns.

Growth rate calculation

Determining the growth rate \( k \) is central to understanding how quickly a population is growing. The growth rate influences how short-term changes impact long-term population predictions.

To calculate \( k \), we use the data points provided: 200 flies at day 2 and 400 flies at day 5. These provide clear time-dependent insights to calculate the rate of growth:

• Set up equations from the growth model for both points, as shown: \( 200 = P_0 e^{2k} \) and \( 400 = P_0 e^{5k} \)
• Divide these equations, simplifying to: \( 2 = e^{3k} \)
• Solving gives: \( 3k = \ln(2) \), thus \( k = \frac{\ln(2)}{3} \)

Substituting this \( k \) back into the population model refines our predictions, ensuring they accurately reflect the observed exponential growth trend in Drosophila populations.