Question

Assume the populations in Exercises 1-4 evolve according to the differential equation \(P^{\prime}=k P\). A colony of bacteria initially has \(10,000,000\) members. After 5 days, the population increases to \(11,000,000\). Estimate the population after 30 days.

Short answer

Question: Estimate the population of a bacteria colony after 30 days, given that it starts with 10,000,000 members and grows to 11,000,000 after 5 days. The population growth follows the differential equation \(P'(t) = kP(t)\). Answer: The estimated population of the bacteria colony after 30 days is approximately 16,105,212 members.

Step-by-step solution

Write down the given information

We know that the initial population is \(P(0) = 10,000,000\), and the population after 5 days is \(P(5) = 11,000,000\). We are asked to find the population \(P(30)\) after 30 days.

Set up the differential equation

The differential equation for the population is given as \(P'(t) = kP(t)\), which has the general solution in the form \(P(t) = P(0)e^{kt}\), where \(P(0)\) is the initial population.

Find the growth constant 'k'

To find the value of \(k\), use the known values of the population at \(t=0\) and \(t=5\). We have \(P(5) = 11,000,000 = 10,000,000e^{5k}\), which we can solve for \(k\). First, divide both sides by \(10,000,000\) to get: \[1.1 = e^{5k}\] Next, take the natural logarithm of both sides: \[\ln{1.1} = 5k\] Now, solve for \(k\): \[k = \frac{\ln{1.1}}{5}\]

Estimate the population after 30 days

Now that we have the value of \(k\), we can plug it into the general solution formula, \(P(t) = P(0)e^{kt}\), to estimate the population after 30 days (\(t=30\)). \[P(30) = 10,000,000e^{30\frac{\ln{1.1}}{5}}\] Calculate the value and round to the nearest whole number: \[P(30) \approx 16,105,212\] Therefore, the estimated population of the bacteria colony after 30 days is approximately 16,105,212 members.

Key concepts

Differential Equation

Differential equations are mathematical equations that relate a function with its derivatives. They are crucial in modeling any dynamic process, illustrating how changing conditions affect a quantity of interest over time. In the context of exponential growth, a differential equation helps us to understand how a population changes.
For a dynamic and growing population, the typical form of the differential equation is expressed as: \[ P'(t) = kP(t) \] This equation tells us that the rate of change of the population, \( P'(t) \), is proportional to its current size, denoted by \( P(t) \), with \( k \) acting as the constant of proportionality. This hints at a very natural, self-reinforcing growth pattern where the population grows faster as it becomes larger. Understanding differential equations allows one to connect the dots between initial conditions and future predictions.

Population Modeling

Population modeling is a mathematical way to describe how a population evolves over time. This can be extremely helpful in predicting the future size of populations, be it bacteria, animals, or people.
Using the context of our exercise, the initial population of bacteria was \( 10,000,000 \), and it increased to \( 11,000,000 \) after 5 days. The challenge lies in determining what the population will be at a later date, say after 30 days. By using the equation: \[ P(t) = P(0)e^{kt} \] We can predict the population at any future time \( t \). Here:

• \( P(0) \) represents the initial population count.
• \(e\) is the base of the natural logarithm, a constant approximately equal to 2.71828.
• \(k\) is the growth constant specific to the situation.
• \(t\) indicates the time elapsed.

Population models are incredibly insightful, helping researchers and scientists strategize for population sustainability and manage resource allocation efficiently.

Growth Constant

The growth constant denoted by \( k \) is a crucial element in exponential growth models. It signifies how fast a population is increasing at any point in time. In a situation where populations grow without bounds, \( k \) gives us a measure of the growth intensity.
To find \( k \) in the current problem, we analyzed data showing an increase from \( 10,000,000 \) to \( 11,000,000 \) in 5 days. We used the equation: \[ P(5) = 10,000,000e^{5k} \] By solving: \[ 11,000,000 = 10,000,000e^{5k} \] And simplifying to: \[ k = \frac{\ln{1.1}}{5} \] The constant \( k \) is not just another number in the equation, but a powerful parameter allowing us to calculate exponential growth. With \( k \), we can forecast the population many years into the future, making it an indispensable part of the growth equation.