Problem 41 - Calculus for Biology and Medicine

Question

In Problems , find the population sizes for \(t=0,1\), 2, \(\ldots, 5\) for each recursion. Then write the equation for \(N_{t}\) as a function of \(t\). $$ N_{t+1}=\frac{1}{5} N_{t} \text { with } N_{0}=31250 $$

Step-by-Step Solution

Answer

Population sizes: 31250, 6250, 1250, 250, 50, 10; Equation: \( N_t = \left( \frac{1}{5} \right)^t \times 31250 \).

Step 1: Understanding the Recursion Formula

The given recursion formula is \( N_{t+1} = \frac{1}{5} N_{t} \). This means that to find the population at time \( t+1 \), we need to multiply the population at time \( t \) by \( \frac{1}{5} \). The initial population at time \( t=0 \) is given as \( N_0 = 31250 \).

Step 2: Calculating Population Size at t=0

The initial population at \( t=0 \) is already given: \( N_0 = 31250 \).

Step 3: Calculating Population Size at t=1

To find \( N_1 \), use the recursion formula: \( N_1 = \frac{1}{5} N_0 = \frac{1}{5} \times 31250 = 6250 \).

Step 4: Calculating Population Sizes for t=2 to t=5

Continue applying the recursion formula:- \( N_2 = \frac{1}{5} N_1 = \frac{1}{5} \times 6250 = 1250 \).- \( N_3 = \frac{1}{5} N_2 = \frac{1}{5} \times 1250 = 250 \).- \( N_4 = \frac{1}{5} N_3 = \frac{1}{5} \times 250 = 50 \).- \( N_5 = \frac{1}{5} N_4 = \frac{1}{5} \times 50 = 10 \).

Step 5: Writing the Equation for N_t

The population at any integer time \( t \) can be expressed as \( N_t = \left( \frac{1}{5} \right)^t \times 31250 \). This formula comes from repeatedly applying the recursive relation, noting it follows a power function of \( \frac{1}{5} \).

Key Concepts

Recursion Formula

To solve problems involving changing populations, a recursion formula serves as a vital tool. A recursion formula provides a systematic way of determining the population at any given time, based on previous population sizes. Simply put, it sets a rule for how a sequence progresses. In our exercise, the recursion formula is given as:\[ N_{t+1} = \frac{1}{5} N_t \]This recursive relationship means that each succeeding population size is one-fifth of the previous one. Therefore, to understand and use the recursion formula effectively, remember:

• The base value: You need an initial value, \( N_0 \), which is given as 31,250 in this case. This acts as our starting point.
• The recursive rule: Here, you multiply the current population size by \( \frac{1}{5} \) to get the next one.

This formula can be applied iteratively to find subsequent population sizes, which builds an understanding of how populations can diminish over time in a structured way.

Population Size Calculation

When tasked with finding population sizes for specific times, such as \( t = 0 \) through 5, we can repeatedly use the recursion formula to do so. Let's walk through the process:Start with the initial population \( N_0 = 31,250 \), which establishes our ground zero for calculations. Then use the recursion formula step by step:

• At \( t = 1 \), calculate: \( N_1 = \frac{1}{5} \times 31,250 = 6,250 \)
• At \( t = 2 \), calculate: \( N_2 = \frac{1}{5} \times 6,250 = 1,250 \)
• At \( t = 3 \), calculate: \( N_3 = \frac{1}{5} \times 1,250 = 250 \)
• At \( t = 4 \), calculate: \( N_4 = \frac{1}{5} \times 250 = 50 \)
• Finally, at \( t = 5 \), calculate: \( N_5 = \frac{1}{5} \times 50 = 10 \)

These calculations reveal a consistent albeit rapid shrinkage of the population over time, showcasing the pattern enforced by the recursion formula. By doing this repetitively, you learn both the power of recursion and its application in real-world scenarios, like modeling populations.

Exponential Decay

The process observed in this exercise is a textbook example of exponential decay. Exponential decay describes a process where quantities reduce at a consistent percentage rate over equal time increments. In this case, each population size is one-fifth of the previous size:With the equation:\[ N_t = \left( \frac{1}{5} \right)^t \times 31250 \]This formula indicates how each increment of time \( t \) impacts the population size, making it shrink exponentially. Here's why it's crucial:

• It provides a clear function for predicting population sizes without iteration.
• It underscores the nature of change, where population halves as a consistent percentage (here, each step is \( \frac{1}{5} \) of the previous value).

Exponential decay applies in many fields, such as finance and radioactive decay, representing processes where outcomes reduce swiftly over time. Understanding this concept is vital, as it enables prediction and comprehension of natural declines.