Problem 14 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

For each equation find a numbers \(C\) and \(R\) such that \(P_{t}=C \times R^{t}\) is a solution. a. \(P_{t+1}-0.9 P_{t}=3 \times 2^{t}\) b. \(P_{t+1}+0.9 P_{t}=5 \times 0.5^{t}\) c. \(P_{t+1}-0.5 P_{t}=0.8^{t}\) d. \(P_{t+2}-0.2 P_{t+1}-0.6 P_{t}=2 \times 0.9^{t}\) e. \(P_{t+1}+0.9 P_{t}=-0.9^{t}\) f. \(\quad P_{t+2}+0.2 P_{t+1}-0.4 P_{t}=2 \times 1.1^{t}\)

Step-by-Step Solution

Answer

a. \(C = 3, R = 2.9\); b. \(C = 5, R = -0.4\); c. \(C = 1, R = 1.3\); d. Approximate \(C = 2, R\); e. \(C = 1, R = -1.8\); f. Approximate \(C = 2, R\).

Step 1: Analyze Equation (a)

For the equation \( P_{t+1} - 0.9 P_{t} = 3 \times 2^{t} \), let's find values of \( C \) and \( R \) such that \( P_{t} = C \times R^{t} \). Substituting \( P_{t+1} = CR^{t+1} \) and \( P_{t} = CR^{t} \) into the equation gives us: \( CR^{t+1} - 0.9CR^{t} = 3 \times 2^{t} \). Simplifying, we get \( CR^{t}(R - 0.9) = 3 \times 2^{t} \). This implies \( R - 0.9 = 2 \) and \( C = 3 \). Solving \( R-0.9 = 2 \) gives \( R = 2.9 \).

Step 2: Analyze Equation (b)

For the equation \( P_{t+1} + 0.9 P_{t} = 5 \times 0.5^{t} \), substitute \( P_{t+1} = CR^{t+1} \) and \( P_{t} = CR^{t} \) to get \( CR^{t+1} + 0.9CR^{t} = 5 \times 0.5^{t} \). Factor to obtain \( CR^{t}(R + 0.9) = 5 \times 0.5^{t} \). Solving \( R + 0.9 = 0.5 \) and \( C = 5 \), we get \( R = -0.4 \).

Step 3: Analyze Equation (c)

For the equation \( P_{t+1} - 0.5 P_{t} = 0.8^{t} \), substitute \( P_{t+1} = CR^{t+1} \) and \( P_{t} = CR^{t} \) to get \( CR^{t+1} - 0.5CR^{t} = 0.8^{t} \). Factor to maximize \( CR^{t}(R - 0.5) = 0.8^{t} \). Solving \( R - 0.5 = 0.8 \) gives \( R = 1.3 \), and \( C = 1 \).

Step 4: Analyze Equation (d)

For \( P_{t+2} - 0.2 P_{t+1} - 0.6 P_{t} = 2 \times 0.9^{t} \), we need to set \( P_{t}=C \times R^{t} \), \( P_{t+1}=CR^{t+1} \), \( P_{t+2}=CR^{t+2} \). Substituting, we have \( CR^{t+2} - 0.2CR^{t+1} - 0.6CR^{t} = 2 \times 0.9^{t} \). Factoring out terms gives us: \( CR^{t}(R^{2} - 0.2R - 0.6) = 2 \times 0.9^{t} \). We solve \( R^{2} - 0.2R - 0.6 = 0.9 \) and \( C = 2 \). Solving the quadratic equation gives an **approximate solution** for \( R \).

Step 5: Analyze Equation (e)

For \( P_{t+1} + 0.9 P_{t} = -0.9^{t} \), substituting \( P_{t+1} = CR^{t+1} \) and \( P_{t} = CR^{t} \), we have \( CR^{t+1} + 0.9CR^{t} = -0.9^{t} \). Simplifying gives \( CR^{t}(R + 0.9) = -0.9^{t} \). Solving \( R + 0.9 = -0.9 \) suggests \( R = -1.8 \) and \( C = 1 \).

Step 6: Analyze Equation (f)

Given \( P_{t+2} + 0.2 P_{t+1} - 0.4 P_{t} = 2 \times 1.1^{t} \), find \( C \) and \( R \). Using \( P_{t+1} = CR^{t+1} \), \( P_{t} = CR^{t} \), \( P_{t+2} = CR^{t+2} \), we have \( CR^{t+2} + 0.2CR^{t+1} - 0.4CR^{t} = 2 \times 1.1^{t} \). This factors to \( CR^{t}(R^{2} + 0.2R - 0.4) = 2 \times 1.1^{t} \). Solving \( R^{2} + 0.2R - 0.4 = 1.1 \), we find an **approximate solution** for \( R \) with \( C = 2 \).

Key Concepts

Recursive Sequences

Recursive sequences form the backbone of many mathematical models, especially in difference equations. A recursive sequence is like a chain. Each link depends on the previous one. In math terms, each term of the sequence is defined based on its predecessor. This way, each new value emerges from the calculations of prior values.

These sequences allow us to predict future values by examining current and past trends. They are used extensively in economics, population studies, and computer science for algorithms that require step-by-step evaluations.

• A recursive expression, like finding the next term in a sequence, can be written as: \( P_{t+1} = f(P_{t}) \), where \( f \) denotes the function applied to the previous term.
• They require initial values or conditions to begin. Without starting values, a recursive sequence cannot generate further outputs.
• Using recursion often makes complex patterns and behaviors easier to predict by reducing them to repeatable, simpler steps.

Understanding recursive sequences will allow you to decode complex real-life patterns by breaking them down into simpler parts.

Geometric Progression

A geometric progression (or sequence) is different from a recursive sequence. Here, each term is obtained by multiplying the previous term by a fixed number called the common ratio.

This type of progression has vast applications in fields such as finance, where compound interest can be modeled using geometric sequences. It simplifies computations where changes happen at a constant rate over a time period.

• An example of a geometric sequence is \( P_{t} = C imes R^{t} \), where \( C \) is the initial term and \( R \) is the common ratio.
• Growth or decay observed in geometric progression is exponential. This means values change by the same proportion, over each time interval.
• Such equations in difference formats help describe behaviors where repeated proportional changes are involved, such as in physics with radioactive decay or cooling patterns.

Mastering the understanding of geometric progressions enables one to foresee outcomes when consistent change is in play.

Factorization in Equations

Factorization simplifies complex equations and isolates terms, enabling easier manipulation of mathematical expressions. It's like breaking down a big problem into smaller parts that are easier to handle.

In the context of difference equations, factorization can help in deducing solutions quickly when terms need to be grouped or expanded to solve different parts separately.

• Consider the factored form in difference equations: \( CR^{t}(R - 0.9) = 3 imes 2^{t} \). Here, factorization helps us deal with each component independently and merge them back as necessary.
• It not only aids in solving polynomials but also in optimizing expressions for computational efficiency.
• Deconstructing equations through factorization often reveals hidden characteristics or relationships among terms and constants.

Grasping factorization's role in simplifying and solving equations lays a strong foundation for tackling more advanced mathematical problems.

Solving Quadratic Equations

Quadratic equations arise frequently in difference equations, especially those of the form \( ax^2 + bx + c = 0 \). Solving them provides critical insights into the behavior of recursive sequences and related mathematical models.

Solutions to these equations can be thought of as points where a curve based on the equation intersects with the x-axis.

• Common methods include factoring, using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), or graphically interpreting the equation's curve.
• In the step-by-step solutions, quadratic equations guide us to find values of \( R \) when analyzing the coefficients linked with the recursive sequence's behavior.
• These equations help identify turning points or equilibrium within dynamic systems represented in models.

Being able to solve quadratic equations effectively equips students with the aptitude to address a broad range of mathematical challenges.