Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

The equation, \(B_{t}-B_{t-1}=r B_{t-1},\) carries the same information as \(B_{t+1}-B_{t}=r B_{t}\) a. Write the first four instances of \(B_{t}-B_{t-1}=r B_{t-1}\) using \(t=1, t=2, t=3,\) and \(t=4\). b. Cascade these four equations to get an expression for \(B_{4}\) in terms of \(r\) and \(B_{0}\). c. Write solutions to and compute \(B_{40}\) for (a.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.2 B_{t-1}\) (b.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.1 B_{t-1}\) (c.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.05 B_{t-1}\) (d.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=-0.1 B_{t-1}\)

Step-by-Step Solution

Answer

(a) 5321.43; (b) 2262.95; (c) 351.95; (d) 0.70.

Step 1: Write Equations for Each Time Step

For each provided time step, substitute the corresponding value of \(t\) into the equation \(B_t - B_{t-1} = r B_{t-1}\). 1. For \(t = 1\): \[ B_1 - B_0 = r B_0 \] 2. For \(t = 2\): \[ B_2 - B_1 = r B_1 \] 3. For \(t = 3\): \[ B_3 - B_2 = r B_2 \] 4. For \(t = 4\): \[ B_4 - B_3 = r B_3 \]

Step 2: Cascade the Equations

Now substitute each equation from step 1 into the next to express \(B_4\) in terms of \(B_0\). \(B_1 = B_0 + rB_0 = B_0(1+r)\) \(B_2 = B_1 + rB_1 = B_1(1+r) = B_0(1+r)^2\) \(B_3 = B_2 + rB_2 = B_2(1+r) = B_0(1+r)^3\) \(B_4 = B_3 + rB_3 = B_3(1+r) = B_0(1+r)^4\)

Step 3: General Formula and Computation for B_40

The expression for \(B_t\) is \(B_t = B_0(1+r)^t\). Use this to compute \(B_{40}\) for various values of \(r\). 1. \(B_0 = 50, r = 0.2\): \[ B_{40} = 50(1+0.2)^{40} \approx 50(106.429) \approx 5321.43 \]2. \(B_0 = 50, r = 0.1\): \[ B_{40} = 50(1+0.1)^{40} \approx 50(45.259) \approx 2262.95 \]3. \(B_0 = 50, r = 0.05\): \[ B_{40} = 50(1+0.05)^{40} \approx 50(7.039) \approx 351.95 \]4. \(B_0 = 50, r = -0.1\): \[ B_{40} = 50(1-0.1)^{40} \approx 50(0.014) \approx 0.70 \]

Step 4: Consolidating Results

Based on the calculations, present the results clearly:(a) For \(r = 0.2\), \(B_{40} \approx 5321.43\).(b) For \(r = 0.1\), \(B_{40} \approx 2262.95\).(c) For \(r = 0.05\), \(B_{40} \approx 351.95\).(d) For \(r = -0.1\), \(B_{40} \approx 0.70\).

Key Concepts

Geometric Progression

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This concept is central to the problem we're examining. The basic formula to compute the nth term in a geometric progression is:

• General Term: \( a_n = a_1 imes r^{n-1} \)

where:

• \( a_n \) is the nth term,
• \( a_1 \) is the first term, and
• \( r \) is the common ratio.

In our problem, the common ratio \( r \) determines how each subsequent value of \( B \) is calculated from the previous one. The progression describes how an initial quantity grows or shrinks over time, which is essential in understanding exponential growth and its behavior under different conditions. With this understanding, students can predict future terms in the sequence easily by applying the formula. The exponential growth behavior seen in the equation \( B_t = B_0(1+r)^t \) is a direct application of this progression.

Recursive Equation

Recursive equations are equations where subsequent terms are defined in relation to the term preceding them. The problem uses a simple, first-order recursive equation, \( B_t - B_{t-1} = r B_{t-1} \), to model exponential growth. Let's break down how this works:In the notation of recursive formulas, each term is constructed based on its predecessor. Here, \( B_t \) is computed as:

• \( B_t = B_{t-1} + rB_{t-1} \)
• Or simply, \( B_t = B_{t-1}(1 + r) \)

This means that each term is the previous term incremented by a fixed ratio \( r \). It's helpful when modeling situations where the quantity increases by a certain fraction or percent over each time interval. Recursive equations are powerful because they not only describe a mathematical process but also help concretize the idea of growth stages in sequenced steps. By bridging consecutive terms, recursive relations illustrate a close-up view of the transition between quantities, giving insight into how exactly an entity changes.

Initial Value Problem

An initial value problem is a type of problem where you need to find a future value based on an initial or starting value and other conditions. The equation \( B_0 \) in our exercise stands as the initial value that we use to compute future values of \( B_t \). This kind of problem frequently appears in fields that rely on differential equations and modeling real-world scenarios that evolve over time.The initial value problem described here is solved by understanding how this initial setting interacts with the recursive formula. The given initial amount, \( B_0 = 50 \), and a specified rate \( r \) determine the trajectory of the progression.To solve the problem, the following steps are generally taken:

• Identify the starting point or initial value (here \( B_0 = 50 \)).
• Apply the given recursive formula for each required step or term.
• Use this initial value to generate further values within the equation stating \( B_t = B_0(1+r)^t \).

The method allows students to explore how different rates \( r \) would impact future outcomes. Understanding initial value problems is vital for modeling exponential growth and predicting future states based on current data.