Problem 35 - Calculus for Biology and Medicine

Question

In Problems 35-46, find the population sizes for \(t=0,1,2, \ldots, 5\) for each recursion. $$ N_{t+1}=2 N_{t} \text { with } N_{0}=3 $$

Step-by-Step Solution

Answer

For \(t=0\) to \(t=5\), populations are 3, 6, 12, 24, 48, 96 respectively.

Step 1: Understand the Recursion Formula

The given recursive formula is \(N_{t+1} = 2N_t\). This means that the population at time \(t+1\) is twice the population at time \(t\). We start with \(N_0 = 3\).

Step 2: Calculate Population at \(t=0\)

Since the initial population size at \(t=0\) is given, we have \(N_0 = 3\).

Step 3: Calculate Population at \(t=1\)

Using the recursion formula, \(N_1 = 2N_0 = 2 \times 3 = 6\).

Step 4: Calculate Population at \(t=2\)

Apply the formula again: \(N_2 = 2N_1 = 2 \times 6 = 12\).

Step 5: Calculate Population at \(t=3\)

Continuing the process: \(N_3 = 2N_2 = 2 \times 12 = 24\).

Step 6: Calculate Population at \(t=4\)

Again, using the recursion formula: \(N_4 = 2N_3 = 2 \times 24 = 48\).

Step 7: Calculate Population at \(t=5\)

Finally, \(N_5 = 2N_4 = 2 \times 48 = 96\).

Key Concepts

Population Growth

Population growth refers to the increase in the number of individuals in a population over time. In the exercise, we observe a doubling pattern, meaning each time period sees the population multiply by two. Population growth can be modeled using recursive formulas like the one in this exercise, which help in predicting future population sizes based on current data.

• **Constant Growth Rate**: The rate of growth remains consistent, as the population doubles consistently in each time period.
• **Exponential Growth**: This specific pattern where growth accelerates over time, as each additional individual contributes to more doubling.

A real-world example is bacteria populations growing in nutrient-rich environments. As resources are plentiful, besides doubling, the growth could become faster, but over longer periods, resources can become limiting, affecting continued exponential expansion.

Initial Value Problem

An initial value problem involves finding a solution to a differential equation given an initial condition. In this exercise, we started with an initial population, denoted as \(N_0 = 3\). This initial condition is crucial as it sets a baseline from which all future calculations arise. To solve an initial value problem:

• **Identify Initial Condition**: Start with known values, here \(N_0 = 3\).
• **Apply Recursion**: Use the given relationship (\(N_{t+1} = 2N_t\)) to iterate over each time step.
• **Forecast Future States**: Calculate subsequent populations at desired time intervals (\(t = 1, 2, 3, \text{etc.}\)).

The initial value effectively acts as the foundation for our entire predictive model. Without it, accurately projecting future values would be impossible.

Doubling Rule

The doubling rule is a guideline for understanding how a quantity grows when it consistently multiplies by two over fixed intervals. For our problem, the population of 3 at \(t=0\) doubles every unit of time. The doubling rule is observed through:

• **Time Doubling Steps**: Measure time intervals consistently to observe how rapidly values grow. Here, each unit \(t\) reflects a doubling.
• **Calculated Predictions**: Doubling the population sequentially as \(t\) increases. For example, moving from 3 to 6, and then from 6 to 12.
• **Recognizable Patterns**: Predicts the sequence as \(N_0 = 3, N_1 = 6, N_2 = 12\), etc., marking each subsequent number as a clear double of the previous.

Understanding this rule highlights how quickly exponential growth can occur, a phenomenon seen in technology advancements, financial investments, and even viral social trends.