Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Use your calculator and the 'Previous Answer' key or computer to compute \(Q_{1}, \cdots\) \(Q_{50}\) for $$ Q_{0}=10 \quad Q_{t+1}=Q_{t}+0.2 Q_{t}\left(1-\frac{Q_{t}}{50}\right) $$ (Type \(10,\) ENTER, \(\mathrm{ANS}+0.2 \times \mathrm{ANS} \times(1-\mathrm{ANS} / 50),\) ENTER (50 times)) You should find \(Q_{50}=49.99583\). Approximately what will be the values of \(Q_{51}, \cdots Q_{100} ?\)

Step-by-Step Solution

Answer

The values of \(Q_{51}, \ldots, Q_{100}\) will be close to 50, stabilizing around it.

Step 1: Initialize the Sequence

Start with the initial value given, which is \(Q_0 = 10\). You input this into the calculator by typing 10 and pressing ENTER. This establishes your first value in the sequence.

Step 2: Apply the Recurrence Relation

Use the formula \( Q_{t+1} = Q_{t} + 0.2 Q_{t} \left(1 - \frac{Q_{t}}{50}\right) \) for calculating subsequent terms. On the calculator, this is done by entering: \(\text{ANS} + 0.2 \times \text{ANS} \times (1 - \text{ANS} / 50)\), and then press ENTER.

Step 3: Iterate 50 Times

Continue to press ENTER to apply the formula a total of 50 times. This means you execute the formula repeatedly until you reach \(Q_{50}\). The instructions indicate this should give approximately \(Q_{50} = 49.99583\).

Step 4: Analyze Long-term Behavior

Given the pattern and the result at \(Q_{50}\), recognize that \(Q_t\) approaches the value 50 as \(t\) increases. This suggests that further terms \(Q_{51}, Q_{52}, \ldots, Q_{100}\) will be approaching closer to the value 50, following the observed stabilization near this target.

Key Concepts

Recurrence Relation

In math, a recurrence relation expresses how each term in a sequence is derived from previous terms. This is at the core of discrete dynamical systems. For the problem at hand, the recurrence relation given is \[Q_{t+1} = Q_{t} + 0.2 Q_{t} \left(1 - \frac{Q_{t}}{50}\right).\]This formula indicates how to calculate each subsequent term \( Q_{t+1} \) based on the preceding term \( Q_{t} \).Recurrence relations are useful as they provide a way to understand the evolution of a sequence over time. They model a step-by-step progression where each step depends on the outcome of the previous stage. This particular relation includes terms that control growth and limitations, maintaining the values within certain bounds.

Sequence Iteration

Sequence iteration involves repeating the process of applying the recurrence relation multiple times. In this exercise, you have started with an initial value of \( Q_0 = 10 \). Using a calculator, you input this and repeatedly use a set formula to arrive at further sequence terms.With each iteration, you substitute the most recent result to find the next value. This is made easier with the 'Previous Answer' function on calculators, allowing seamless calculation each time. By inputting the formula \( \text{ANS} + 0.2 \times \text{ANS} \times (1 - \text{ANS} / 50) \) and pressing ENTER repeatedly, each new term is computed swiftly.By iterating frequently, students can visualize how the sequence evolves. It gives practical insight into how discrete dynamical systems unfold over time.

Long-term Behavior Analysis

Long-term behavior analysis in sequences focuses on understanding where the terms will settle eventually. This involves looking closely at how the sequence behaves as it progresses towards infinity or a certain saturation point.For the given recurrence relation, calculations show that by the time you reach \( Q_{50} \), the sequence is close to \( 49.99583 \). This suggests that values after \( Q_{50} \) will approximate 50, indicating stabilization.When analyzing dynamical systems, noticing stable values or attracting points (like 50 in this case) is crucial. These signify a system's response to input over time, helping to predict future terms and understand system behavior thoroughly. Long-term behavior helps outline the sequence's destiny, offering a predictive glimpse into what further computations might reveal.