Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Use your calculator or computer to compute \(Q_{1}, \cdots Q_{20}\) for \(\begin{aligned} \text { a. } \quad Q_{0} &=5 \\ Q_{t+1} &=Q_{t}+0.1 Q_{t}\left(1-\frac{Q_{t}}{20}\right) \\ \text { c. } \quad Q_{0} &=5 \\\ Q_{t+1} &=Q_{t}+0.1 Q_{t}+0.02 * Q_{t}^{2} \\ \text { e. } \quad Q_{0} &=0.8 \\\ Q_{t+1} &=Q_{t}\left(2-Q_{t}\right) \\ \text { g. } \quad Q_{0} &=0.5 \\\ Q_{t+1} &=Q_{t}\left(3.5-Q_{t}\right) \\ \text { i. } \quad Q_{0} &=0.8 \\\ Q_{t+1} &=e^{-Q_{t}} \\ \text { k. } Q_{0} &=0.8 \\ Q_{t+1} &=\frac{Q_{n}+2 / Q_{n}}{2} \end{aligned}\) \(\begin{aligned} \text { b. } \quad Q_{0} &=5 \\ Q_{t+1} &=1.1 Q_{t}\left(1-\frac{Q_{t}}{20}\right) \\ \text { d. } \quad Q_{0} &=0.5 \\\ Q_{t+1} &=\cos \left(Q_{t}\right) \\ \text { f. } \quad Q_{0} &=0.8 \\\ Q_{t+1} &=Q_{t}\left(3-Q_{t}\right) \\ \text { h. } \quad Q_{0} &=0.6 \\\ Q_{t+1} &=Q_{t}\left(3.5-Q_{t}\right) \\ \text { j. } \quad Q_{0} &=0.8 \\\ Q_{t+1} &=\left(\sqrt{Q_{t}}\right) \times e^{-Q_{t}} \\ \text { l. } \quad Q_{0} &=0.8 \\ Q_{t+1} &=\frac{Q_{n}+3 / Q_{n}}{2} \end{aligned}\)

Step-by-Step Solution

Answer

Compute each sequence using their respective formulas iteratively from \(Q_0\) to \(Q_{20}\).

Step 1: Understanding the Calculation

To solve this exercise, we need to calculate the values of \(Q_t\) over 20 iterations with different initial conditions \(Q_0\) and corresponding recursive formulas. Each formula will be used to compute \(Q_t\) from \(Q_0\) to \(Q_{20}\), using either a calculator or computational software.

Step 2: Calculation for Part (a)

Use \(Q_0 = 5\) and the recurrence relation \(Q_{t+1} = Q_t + 0.1 Q_t \left(1 - \frac{Q_t}{20}\right)\). Calculate \(Q_1, Q_2, \ldots, Q_{20}\). Starting at \(Q_0 = 5\), iterate by using the given formula to find each subsequent \(Q_t\).

Step 3: Calculation for Part (b)

Use \(Q_0 = 5\) and the recurrence relation \( Q_{t+1} = 1.1Q_t\left(1-\frac{Q_t}{20}\right) \). Calculate \( Q_1, Q_2, \ldots, Q_{20} \). Like Step 2, start with \(Q_0 = 5\) and iteratively apply the formula to compute each \( Q_t \).

Step 4: Calculation for Part (c)

Use \(Q_0 = 5\) and the recurrence relation \( Q_{t+1} = Q_t + 0.1 Q_t + 0.02 Q_t^2 \). Iterate from \( Q_1 \) to \( Q_{20} \) using this formula to see the changes over 20 iterations.

Step 5: Calculation for Part (d)

Use \(Q_0 = 0.5\) and the recurrence relation \( Q_{t+1} = \cos(Q_t) \). Compute \( Q_1 \) through \( Q_{20} \) by evaluating the cosine of the previous value.

Step 6: Calculation for Part (e)

Use \(Q_0 = 0.8\) and the recurrence relation \(Q_{t+1} = Q_t(2 - Q_t)\). Compute \(Q_1\) to \(Q_{20}\) through iteration.

Step 7: Calculation for Part (f)

Use \(Q_0 = 0.8\) and the recurrence relation \(Q_{t+1} = Q_t(3 - Q_t)\). Iterate to find \(Q_1\) through \(Q_{20}\).

Step 8: Calculation for Part (g)

Use \(Q_0 = 0.5\) and the recurrence relation \(Q_{t+1} = Q_t(3.5 - Q_t)\). Calculate \(Q_1\) to \(Q_{20}\) iteratively.

Step 9: Calculation for Part (h)

With \(Q_0 = 0.6\) and the recurrence \(Q_{t+1} = Q_t(3.5 - Q_t)\), iterate for results \(Q_1\) to \(Q_{20}\).

Step 10: Calculation for Part (i)

Start with \(Q_0 = 0.8\) and the formula \( Q_{t+1} = e^{-Q_t} \). Compute values from \(Q_1\) to \(Q_{20}\) by substituting the exponential of the negative of the current term.

Step 11: Calculation for Part (j)

With \(Q_0 = 0.8\) and the formula \( Q_{t+1} = \sqrt{Q_t} \cdot e^{-Q_t} \), compute each term up to \(Q_{20}\), taking the square root multiplied by the exponential factor.

Step 12: Calculation for Part (k)

Using \(Q_0 = 0.8\) and formula \( Q_{t+1} = \frac{Q_t + 2/Q_t}{2} \), iterate this process through \(Q_{20}\), applying the formula at each step.

Step 13: Calculation for Part (l)

Start with \(Q_0 = 0.8\) and iterate using the formula \( Q_{t+1} = \frac{Q_t + 3/Q_t}{2} \) to calculate up to \(Q_{20}\). Adjust the formula to the current iteration value.

Key Concepts

Recurrence Relations

Recurrence relations form the backbone of iterative calculations in calculus, especially when solving problems connected to growth or decay processes. Think of them as a way to calculate future values based on current ones. Recurrence relations are expressed in the form \(Q_{t+1} = f(Q_t)\), where the next term in the sequence \(Q_{t+1}\) is defined by a function of the current term \(Q_t\).

• They are heavily used in modelling biological systems, like population growth and the spread of diseases.
• For example, in the exercise given, recurrence relations like \(Q_{t+1} = Q_t + 0.1Q_t(1 - Q_t/20)\) model a growth process where growth rate changes relative to the current value.
• This allows calculations without a direct equation for \(Q_t\) based on time \(t\), instead building on previously computed terms.

By understanding how these relations evolve over iterations, you can predict long-term behavior of the system being modeled.

Iterative Solutions

Iterative solutions help solve recurrence relations by repeatedly applying the same process. Here's how it works: you start with an initial value, say \(Q_0 = 5\), and apply the recurrence relation to compute \(Q_1\), then \(Q_2\), and so on until you've solved for all desired terms.

• This involves simple operations done multiple times, hence the term 'iteration', making it more computationally feasible than solving complex equations directly.
• Each iteration depends on the outcome of the previous one, so accuracy in each calculation is essential.
• Iterative methods are popular when analytical solutions are complex or impossible to derive.

Through iteration, you can visualize how initial conditions influence the system's evolution over time, which is particularly meaningful in biological contexts.

Computational Software

To handle calculations for recurrence relations, especially when they extend to multiple iterations, computational software becomes invaluable. Programs like MATLAB, Python with libraries like NumPy, or specialized calculation tools such as Mathematica efficiently solve these problems.

• They automate the iteration process, significantly reducing human error and computation time.
• By entering initial conditions and recurrence relations, these tools calculate all terms in just seconds.
• They also allow for visualization, such as plotting the sequence \(Q_t\) as a curve, aiding in dynamic analysis and understanding.

Incorporating software into solving life sciences calculus problems allows for more sophisticated analysis and exploration of complex models that would otherwise be cumbersome to solve manually.

Exponential Functions

Exponential functions play a key role in life sciences calculus due to their relevance in natural processes like growth rates and decay phenomena. These functions are depicted by \(e^x\), where \(e\) represents the base of natural logarithms, approximately equal to 2.718.

• The exercise includes equations such as \(Q_{t+1} = e^{-Q_t}\), which illustrates concepts of decay, modeling phenomena like radioactive breakdown or the progression of certain population dynamics.
• Exponential relations are vital for understanding compounding growth and decay, providing insights into their dynamics over consecutive iterations.
• When recurrence relations involve exponential terms, they reflect variations swiftly, underscoring why exponential functions are crucial for dynamic life science models.

Understanding these functions assists in capturing the essence of rapid changes within biological systems, effectively translating these into mathematical models.